Chemistry A Molecular Approach, 6th Edition Nivaldo J. Tro Solution Manual.pdf (58)

Extended Figure Description for Chapter 15 Problem 37 Figure 1b Rate= k [A] plugging in any of the values from the graph 0.002 M·s−1 0.2 M 0.002 M·s−1 k = 0.01 s−1 = k [0.2 M]k = Check: The plot has a shape that matches the one in the text for first-order plots. c. Given: part (a) Find: write a rate law and estimatek Conceptual Plan: Using the result from part (a), the slope of the plot is the rate constant.

Solution:Slope =ΔΔy x = 0.010Ms−0.00 Ms 1.0 M−0.0 M =0.010 s −1 so Rate= k[A] 1or Rate=k[A] or Rate=0.010 s−1 [A] Check: The units (s−1) are correct. The magnitude of the answer (10−2 s−1) makes physical sense because of the rate and concentration data. Remember that concentration is in units of M; so plugging the rate constant into the equation has the units of the rate as M·s−1, which is correct.

15.38 a. Given: Rate versus [A] plot Find: reaction order Conceptual Plan: Look at shape of plot and match to possibilities.

Solution: The plot is a linear plot that is horizontal, so rate is independent of [A] or the reaction is zero-order with respect to A. Check: The order of the reaction is a common reaction order. b. Given: part (a) Find: sketch plot of [A] versus time Conceptual Plan: Using the result from part (a), shape plot of [A] versus time should be a straight line with [A] decreasing. Use 1.0 M as initial concentration.

Solution:

Extended Figure Description for Chapter 15 Problem 38 Figure 1b Check: The plot has a shape that matches the one in the text for zero-order plots. c. Given: part (a) Find: write a rate law and estimatek Conceptual Plan: Using the result from part (a), the rate is equal to the rate constant. −1 k or0 Rate= or Rate=0.011 k Solution:Rate=[A] M·s −2 1 Check: The units (M·s −1 ) are correct. The magnitude of the answer (10 M·s −1 ) makes physical sense because of the rate and concentration data. Plugging the rate constant into the equation, we find that the rate has the units−of , M·s which is correct.

15.39 Given: reaction order: (a) first-order, (b) second-order, and (c) zero-order Find: units of k Conceptual Plan: Using rate law, rearrange to solve fork .

Rate=k[A]n, where n=reaction order Solution: For all cases, rate has units of M·s−1 and [A] has units of M. a. Rate= k[A] 1 = k[A] so k = Rate[A] = MsM =s −1 2 k so = k Rate[A]2 = MsM·M =M −1 ·s −1 b. Rate=[A] −1 k 0 k c.

Rate = [A] = = M · s Check: The units (s−1 , M −1 ·s −1 , and M·s −1 ) are correct. The units for k change with the reaction order so that the units on the rate remain as M·s−1.

15.40 Given:k=0.053/s and [NO]=0.055 M; reaction order: (a) first-order, (b) second-order, and zero-order (change units on k as necessary) Find: rate 2 5 Conceptual Plan: Using rate law, substitute values to solve for Rate.

Rate = k[N 2O5],n where n=reaction order Solution: For all cases, Rate has units of M·s−1 and [A] has units of M. Use the results from Problem 15.37 to choose the appropriate units for k. a. Rate=k[NO]1= k[N2O5]= 0.053 s ×0.055 M=2.9×10 −3Ms 25

2

b. Rate=k[NO]2= 25

0.053M s×(0.055 M)=1.6×10

−4 Ms

Rate=k[NO]0=

k =5.3×10−2 Ms Check: The units (M·s−1) are correct. The magnitude of the rate changes as the order of the reaction changes because we are multiplying by the concentration a different number of times in each case. The higher the order, the lower the rate because the concentration is less than 1 M. 15.41 i; Rate=k[A][B]. For i: Rate=k(6)(6)=36 k; for ii: Rate=k(4)(8)=32 k; for iii: Rate=k(8)(4)=32 k.

15.42 ii; Rate=k[Y]. 2For i: Rate=k(6) 2 =36 ;kfor ii: Rate=(8)k 2 =64 k; for iii: Rate=k(4)2=16 k. Note that the concentration of X does not impact the rate.

15.43 Given: A, B, and C react to form products. Reaction is first-order in A, second-order in B, and zero-order in C. Find: (a) rate law; (b) overall order of reaction; (c) factor change in rate if [A] doubled; (d) factor change in rate if [B] doubled; (e) factor change in rate if [C] doubled; and (f) factor change in rate if [A], [B], and [C] doubled Conceptual Plan: a. Using general rate law form, substitute values for orders.

2 5

Rate = k[A] m[B]n[C]p, where m, n , and p =reaction orders b. Using rate law in part (a), add up all reaction orders. overall reaction order=m+n+p c. Through (f), using rate law from part (a), substitute concentration changes.

Rate 2Rate 1 = k [ A ] 2 [ B ] 1

Solution: a. m=1, n =2, and p =0 so Rate=k[A]1[B]2[C]0 or Rate=k[A][B]2 b. overall reactionorder =m+n+p=1+2+0=3, so it is a third-order reaction overall c. Rate 2Rate 1 = k [ A ]2 [ B ] 2k[A]1 [B] 1and [A]2=2[A]1, [B]2=[B]1, [C]2=[C]1, so 1

2

d. Rate 2Rate 1 = k [ A ]2 [B] 1

Rate 2Rate 1 = k( 2 [ A ) 1 ] 1 [ B ]

1 [B] and 1 [A] 2=[A]1, [B]2=2[B]1, [C]2=[C]1, so 2

Rate 2Rate 1 = k[ A ] 1 ( 2 [ B ) 1 ]

1 [B] and 1 [A] 2=[A]1, [B]2=[B]1, [C]2=2 [C]1, so 2

Rate 2Rate 1 = k[ A ] 1 [ B ]

1

2k[A] 2

2

1

2k[A] 2

e. Rate 2Rate 1 = k [ A ]2 [B] 1

1

1

2

1

2

k[A] 1 [B]

2 2

2k [ A ] 1 [ B ] 1 1 2

k[A] 1 [B] 1

1 2

1

1 2

=2 so the reaction rate doubles (factor of 2) =2 2 =4 so the reaction rate quadruples (factor of 4)

1k [ A ] 1 [ B ] 1=1 so the reaction rate is unchanged (factor of 1) 1 2

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f. Rate 2Rate 1=k[A]2 [B]2k[A]1 [B] 1and [A]2=2[A]1, [B]2=2[B]1, [C]2=2 [C]1, so 1

2

)1 (2[B]

Rate 2Rate 1=k(2[A]1

1

2

)2

k [A] 1 [B ] 1

1

1

2

=2×2 2 =8 so the reaction rate goes up by a factor of 8

Check: The units (none) are correct. The rate law is consistent with the orders given, and the overall order is larger than any of the individual orders. The factors are consistent with the reaction orders. The larger the order, the larger the factor. When all concentrations are changed, the rate changes the most. If a reactant is not in the rate law, then changing its concentration has no effect on the reaction rate.

15.44 Given: A, B, and C react to form products. Reaction is zero-order in A, one-half-order in B, and second-order in C. Find: (a) rate law; (b) overall order of reaction; (c) factor change in rate if [A] doubled; (d) factor change in rate if [B] doubled; (e) factor change in rate if [C] doubled; and (f) factor change in rate if [A], [B], and [C] doubled Conceptual Plan: a. Using general rate law form, substitute values for orders.

Rate = k[A] m[B]n[C]p, where m, n , and p =reaction orders b. Using rate law in part (a), add up all reaction orders. overall reaction order= m + n + p c. Through (f), using rate law from part (a), substitute concentration changes.

Rate 2Rate 1 = k[B]

[C] 2k[B]

2 1/2

2

Solution: a. m=0, n=1/2, and p=2 so Rate=k[A]0[B]1/2[C]2 or Rate=k [B]1/2[C]2 b. overall reaction order=m+n+p=0+1/2+2=5/2=2.5, so it is a two-and-a-half-order reaction overall 2=2[A]1, [B]2=[B]1, [C]2=[C]1, so Rate 2Rate 1=k[B] 1 [C]1k[B] c. Rate 2Rate 1=k[B] 2 [C]2k[B] 1 [C] 1and [A] 1/2

2

1/2

d. Rate 2Rate 1=k[B]1/22 [C] 2k[B] 2

1/2

2

1

f. Rate 2Rate 1=k[B] 2 [C]2k[B] 1/2

2

2

2

1 [C]

=21/2 so the reaction rate increases by a factor of 21/2or 2 or 1.414

21

√

1/2

2

2

1and [A]2=[A]1, [B]2=[B]1, [C]2=2[C]1, so

1 [C]

1 2

=2 2 =4, so the reaction rate quadruples (factor of 4)

1 [C] 1and [A]2=2[A]1, [B]2=2[B]1, [C]2=2[C]1, so 1/2

Rate 2Rate 1=k(2[B]1 )1/2 (2[C]

1/2

[C]1k[B] 2

1/2

1/2

[C] 1

1 [C]1=1 so the reaction rate is unchanged (factor of 1).

2 1/2

2

1/2

)2 k[B]

1/2

1and [A]2=[A]1, [B]2=2[B]1, [C]2=[C]1, so Rate 2Rate 1=k(2[B]1)

1 [C]

e. Rate 2Rate 1=k[B] 2 [C]2k[B] Rate 2Rate 1=k[B] 1 (2[C]

2

[C] 1/21

1 1/2

1

)2

2

k[B] [C] 21 1/21

=21/2×22=25/2, so the reaction rate goes up by a factor of 25/2 or 4 2 or 5.66

√

Check: The units (none) are correct. The rate law is consistent with the orders given, and the overall order is larger than any of the individual orders. The factors are consistent with the reaction orders. The larger the order, the larger the factor. When all concentrations are changed, the rate changes the most. If a reactant is not in the rate law, then changing its concentration has no effect on the reaction rate.

15.45 Given: table of [A] versus initial rate Find: rate law andk Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order.

Rate 2Rate 1= k[A] 2k[A] 1 n

n

Then use one of the concentration/initial rate pairs to determine k .

Rate = k[A]n k (0.200 M) 1 Solution: Rate 2Rate 1=k[A]2k[A] Comparing the first two sets of data,0.112 M/s0.112 M/s= n

n

n

nk

n

0.112 M/s0.112 M/s= (0.100 M) and 1.00=2 so n =0. If we compare the first and the last data sets,

k(0.300 M)nk( 0.100 M)n

n

and 1.00=3 so n

=0

.

This second comparison is not necessary, but it increases our confidence in the reaction order. So Rate=k[A]0=k. Selecting the second data set and rearranging the rate equation, k =Rate=0.112Ms=0.112 M·s−1, so Rate=0.112 M·s−1 Check: The units (none and M·s−1) are correct. The rate law is a common form. The rate is not changing with the concentration, so zero order is consistent. The rate constant is consistent with the units necessary to get the rate as M/s, and the magnitude is reasonable because we have a zero-order reaction.

15.46 Given: table of [A] versus initial rate Find: rate law andk Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order.

Rate 2Rate 1= k[A] 2k[A] 1 n

n

Then use one of the concentration/initial rate pairs to determine k .

Rate = k[A]n k Solution: Rate 2Rate 1=k[A]2[A]1 Comparing the first two sets of data,0.220 M/s0.055 M/s = n

k(0.400 M)nk(0.200 M)n

and 4=2; n

n

so n=2. If we compare the first and last data sets, 0.880 M/s0.055 M/s=k(0.800 M)nk(0.055 M)n and 16=4n, n=2.

This second comparison is not necessary, but it increases our confidence in the reaction order. So Rate=k[A]2.

Selecting the second data set and rearranging the rate equation, k =Rate[A]2=0.220Ms(0.400 M)2=1.38 M−1·s−1; so Rate=1.38 M−1·s−1[A].

Check: The units (none and M−1 ·s −1 ) are correct. The rate law is a common form. The rate is changing more rapidly than the concentration which is consistent with second order. The rate constant is consistent with the units necessary to get the rate as M/s, and the magnitude is reasonable because we have a second-order reaction.

15.47 Given: table of [A] versus initial rate Find: rate law andk Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order.

Rate 2Rate 1= k [A] 2k [A] 1 n

n

Then use one of the concentration/initial rate pairs to determine k. n nk (0.150 M)n Solution: Rate 2Rate 1=k [A]2k [A] Comparing the first two sets of data:0.0546 M/s0.0410 M/s = k (0.200 M)nk (0.150 M) and 1.33_33=1.33_33 so nn=1. If we compare the first and the last data sets: 0.0683 M/s0.0410 M/s=k (0.250 M) 1 n

n

and 1.66_67=1.6_667n so n=1.

This second comparison is not necessary, but it increases our confidence in the reaction order. So Rate=k [A].

Selecting the second data set and rearranging the rate equation k =Rate[A]=0.0546 Ms0.200 M=0.273 s−1 so Rate=0.273 s−1 [A] Check: The units (none and s−1) are correct. The rate law is a common form. The rate is changing proportionately to the concentration, so first-order is consistent. The rate constant is consistent with the units necessary to get rate as M/s and the magnitude is reasonable since we have a first-order reaction.

15.48 Given: table of [A] versus initial rate Find: rate law andk Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order.

Rate 2Rate 1 = k [A] 2k [A] 1 n

n

Then use one of the concentration/initial rate pairs to determine k .

Rate=k [A]n Solution: Rate 2Rate 1=k [A]2k [A] n

n n k (0.075 M) nk (0.050 M) 0.225 M/s0.100 M/s= and 2.25=1.5 so 1 Comparing the first two sets of data: n

n =2. If we compare the first and the last data sets:0.324 M/s0.100 M/s=

k (0.090 M)n(.050 k 0 M)n

and 3.24=1.8

n

so n=2. This second comparison is not necessary, but it increases our confidence in the reaction order. So Rate=k [A]2. Selecting the second data set and rearranging the rate equation k =Rate[A]2=0.225 Ms(0.075 M)2=40. M−1· s−1 so Rate=40. M−1·s−1 [A]2 Check: The units (none and M−1·s −1) are correct. The rate law is a common form. The rate is changing more rapidly than the concentration, so second order is consistent. The rate constant is consistent with the units necessary to get the rate as M/s, and the magnitude is reasonable since we have a second-order reaction.

2 versus initial rate Find: rate law,k, and overall order and [F] 15.49 Given: table of [NO] 2 Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order of each reactant. Be sure to choose data that changes only one concentration at a time.

Rate 2Rate 1 = k[NO2] 2 [F2] 2k[NO2] 1 [F2] 1 m

n

m

n

Then use one of the concentration/initial rate pairs to determine k . m

Rate = k[NO2] [F2]n

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Solution: Rate 2Rate 1=k[NO2]2 [F2]2k[NO2]1 [F2] 1Comparing the first two sets of data, m

n

m

n

0.051 M/s0.026 M/s=k(0.200 M)m (0.100 M)nk(0.100 M)m (0.100 M)n and 1.9_615=2m; so m=1. If we compare the second and third data sets, 0.103 M/s0.051 M/s=k(0.200 M)m (0.200 M)nk(0.200 M)m (0.100 M)n and 2=2n, n=1. Other comparisons can 0.411Ms (0.400 M)(0.400 M)=2.57 M−1·s−1; so 2][F2]. Selecting the last data set and rearranging the rate equation, k=Rate[NO][F]= 2 2 made, but they are not necessary. They should reinforce these values of the reaction orders. So Rate=k[NO Rate=2.57 M−1·s−1 [NO 2][F2], and the reaction is second-order overall.

Check: The units (none and M−1·s −1) are correct. The rate law is a common form. The rate is changing as rapidly as each concentration is changing, which is consistent with first-order in each reactant. The rate constant is consistent with the units necessary to get the rate as M/s, and the magnitude is reasonable because we have a second-order reaction.

15.50 Given: table of [CHCl] and [Cl]2versus initial rate Find: rate law,k, and overall order 3 Conceptual Plan: Using general rate law form, compare rate ratios to determine reaction order of each reactant. Be sure to choose data that changes only one concentration at a time.

Rate 2Rate 1 = k[CH3Cl] 2 [Cl2] 2k[CH3Cl] 1 [Cl2] 1 m

n

m

n

Then use one of the concentration/initial rate pairs to determine k .

Rate = k[CH3Cl]m[Cl2] n Solution: Rate 2Rate 1=k[CH3Cl]2 [Cl2]2k[CH3Cl]1 [Cl2].1 Comparing the first two sets of data, m

n

m

n

we compare the second and third data sets, 0.041 M/s0.029 M/s=k(0.100 M)m (0.100 M)nk(0.100 M)m (0.050 M)n and 1.4_14=2n; so n=1/2. Other 0.029 M/s0.014 M/s=k(0.100 M)m (0.050 M)nk(0.050 M)m (0.050 M)n and 2.0_714=2m; soIfm=1.

1/2 3Cl][Cl2] . Selecting the last data set and rearranging the rate equation, comparisons can be made, but they are not necessary. reinforce these values of the reaction orders. So Rate=k[CH 1/2 They should )(0.200 M) k=Rate[CH3Cl] [Cl2] 1/2=0.115Ms(0.200M =1.29 M −1/2 ·s −1 ; so Rate=1.29 M −1/2 ·s −1 [CH3Cl] [Cl2]1/2 , and the reaction is one-and-a-half order overall.

Check: The units (none and M−1/2·s−1) are correct. The rate law is not as common as others, but it is reasonable. The rate is changing as rapidly as the 3ClCH concentration is changing, which is consistent with first order in this reactant. The rate is 2 concentration, which is consistent with half order in this reactant. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a one-and-a-half order changing a bit more slowly than the Cl reaction.

The Integrated Rate Law and Half-Life 15.51 a. The reaction is zero order. Because the slope of the plot is independent of the concentration, there is no dependence of the concentration of the reactant in the rate law. b. The reaction is first order. The expression for the half-life of a first-order reaction is t1/2=0.693k, which is independent of the reactant concentration. c. The reaction is second order. The integrated rate expression for a second-order reaction is 1[A]t=kt+1[A]0, which is linear when the inverse of the concentration is plotted versus time.

15.52 a. The reaction is second order. The expression for the half-life of a second-order reaction, t 1k[A] 0, shows that the half-life decreases as concentration increases.

1/2 = t=−kt+ln[A]0, which is linear when the natural log of the concentration is plotted versus time. b. The reaction is first order. The integrated rate expression for a first-order reaction is ln[A] c. The reaction is zero order. The expression for the half-life of a zero-order reaction, t1/2=[A]02 k, shows that the half-life increases as concentration increases.

15.53 Given: table of [AB] versus time Find: reaction order, k, and [AB] at 25 s Conceptual Plan: Look at the data and see if any common reaction orders can be eliminated. If the data does not show an equal concentration drop with time, zero order can be eliminated. Look for changes in the half-life (compare time for concentration to drop to one-half of any value). If the half-life is not constant, the first order can be eliminated. If the half-life is getting longer as the concentration drops, this might suggest second order. Plot the data as indicated by the [ ]25 s by using the appropriate integrated rate expression. appropriate rate law. Determinekfrom the slope of the plot. Finally, calculate theABat Solution: By the preceding logic, we can eliminate both the zero-order and first-order reactions. (Alternatively, you could make all three plots and only one should be linear.) This suggests that we should have a second-order reaction. Plot 1/[AB] versus time. −1 ·s1,−and the rate law is Because 1[AB]t=kt+1[AB] 0, the slope will be the rate constant. The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel. Thus, the rate constant is 0.0225 M Rate=0.0225 M−1·s−1[AB]2.

Extended Figure Description for Chapter 15 Problem 53 Figure 1 Finally, use 1[AB]t =kt+1[AB]0; substitute the values of [AB] 0, 25 s, and k; and rearrange to solve for [AB] at 25 s.

[AB] = 1kt+ 1[AB] t

0=1(0.0225 M−1·s−1)(25 s)+(10.950 M)=0.619 M

Check: The units (none, M−1·s−1, and M) are correct. The rate law is a common form. The plot was extremely linear, confirming second-order kinetics. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a second-order reaction. The [AB] at 25 s is between the values at 0 s and 50 s.

2 5 versus time Find: reaction order,k, and [NO] 15.54 Given: table of [NO] 2 5 at 250 s Conceptual Plan: Look at the data and see if any common reaction orders can be eliminated. If the data does not show an equal concentration drop with time, zero order can be eliminated. Look for changes in the half-life (compare time for concentration to drop to one-half of any value). If the half-life is not constant, the first order can be eliminated. If the half-life is getting longer as the concentration drops, this might suggest second order. Plot the data as indicated by the ] 250 s by using the appropriate integrated rate expression. appropriate rate law. Determinekfrom the slope of the plot. Finally, calculate theN[ 2O5at t versus time. Because ln[A]=− kt +ln[A]0, the negative of the Solution: By the preceding logic, we can see that the reaction is most likely first order. It takes just under 75 s for the concentration to be cut in half for any concentration. Plot 2 5]ln[NO slope will be the rate constant. The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel. Thus, the rate constant is 0.00780 s−1, and the rate law is Rate=0.007802O5]. s−1 [N Finally, use substitute the values of [N2O], 2505 s, and k; and rearrange to solve for [N2O5] at 250 s. ln[N2O] ln[N2O 5]=−kt+ln[N2O]; 5 0 t 0

5 250 s =−(0.00780 s

−1

)(250 s)+ln[1.000]0 then [N 2O]5

250 s

= e−1.95 =0.142 M

Extended Figure Description for Chapter 15 Problem 54 Figure 1 Check: The units (none, s−1, and M) are correct. The rate law is a common form. The plot was extremely linear, confirming first-order kinetics. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a first-order reaction. The [NO] 2 5 at 250 s is less than the value at 200 s.

4 8 versus time 15.55 Given: table of [CH] M Find: reaction order,k, and reaction rate when [CH]=0.25 48 Conceptual Plan: Look at the data and see if any common reaction orders can be eliminated. If the data does not show an equal concentration drop with time, zero order can be eliminated. Look for changes in half-life (compare time for concentration to drop to one-half of any value). If the half-life is not constant, the first order can be eliminated. If the half-life is getting longer as the concentration drops, this might suggest second order. Plot the data as indicated by the . appropriate rate law. Determinekfrom the slope of the plot. Finally, calculate the reaction rate whenC[ 4H8] = 025Mby using the rate law. t 4H8]ln[C versus time. Because ln[A]=− kt +ln[A]0, the negative of the slope Solution: By the preceding logic, we can see that the reaction is most likely first-order. It takes about 60 s for the concentration to be cut in half for any concentration. Plot will be the rate constant. The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel. Thus, the rate constant is 0.0112 s−1, and the rate law is Rate=0.01124H8]. s−1 [C Finally, use Rate=0.0112 s−1[CH48] and substitute the values of [C4H8]: Rate=0.0112 s

−1

[0.25 M]=2.8×10

−3

M·s −1

Extended Figure Description for Chapter 15 Problem 55 Figure 1 Check: The units (none, s−1, and M·s−1) are correct. The rate law is a common form. The plot was extremely linear, confirming first-order kinetics. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a first-order reaction. The rate when4H8]=0.25 [C M is consistent with the average rate using 90 s and 100 s.

15.56 Given: table of [A] versus time Find: reaction order, k, and reaction rate when [A]=0.10 M Conceptual Plan: Look at the data and see if any common reaction orders can be eliminated. If the data does not show an equal concentration drop with time, zero order can be eliminated. Look for changes in half-life (compare time for concentration to drop to one-half of any value). If the half-life is not constant, the first order can be eliminated. If the half-life is getting longer as the concentration drops, this might suggest second order. Plot the data as indicated by the . appropriate rate law. Determinekfrom the slope of the plot. Finally, calculate the reaction rate whenA[ ] = 010Mby using the rate law.

[A]=− kt +[A] 0, the negative Solution: By the preceding logic, we can see that the reaction is most likely zero-order. There is a difference of about 0.085 M between each data point, so the rate is independent of the [A]. Plot [A] versus time. Because t of the slope will be the rate constant. The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel. Thus, the rate constant is 3.41×10−3 M·s−1, and the rate law is Rate=3.41×10−3 M·s−1.

Finally, because the rate is independent of concentration, Rate=3.41×10−3 M·s−1 at 0.10 M and all other concentrations. Note: A plot is not necessary because the kinetics are so simple.

Extended Figure Description for Chapter 15 Problem 56 Figure 1 Check: The units (none, M·s −1 , and M·s −1 ) are correct. The rate law is a common form. The plot was extremely linear, confirming zero-order kinetics. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a zero-order reaction. The rate is the same as any average rate that can be calculated using the data.

15.57 a. 1st order; The plot of ln[A] versus time is linear, so the reaction is first order. b. k= Slope of the plot of ln[A] versus time, so k=−−4.3−(−2.3) 200. s−0. s or k=0.010 s−1. c. Rate=0.010 s−1 [A] 15.58 a. 2nd order; The plot of 1/[X] versus time is linear, so the reaction is second order. b. k=−Slope of the plot of 1/[X] versus time, so k=−50. M−1−10. M−1 c. Rate=0.20 M−1·s−1 [X]2

200. s−0. s

or k=0.20 M−1·s−1

15.59 Given: plot of ln[A] versus time has slope=−0.0105/s; [A] 0=0.150 M Find: (a) k, (b) rate law, (c) t1/2, and (d) [A] after 125 s Conceptual Plan: a. A plot of ln[A] versus time is linear for a first-order reaction. Using ln [A]t=−kt+ln [A]0, the rate constant is the negative of the slope. b. Rate law is first-order. Add rate constant from part (a). c. For a first-order reaction, t 1/2=0.693k. Substitute k from part (a). d. Use the integrated rate law, ln [A] t=−kt+ln [A]0, and substitute k, t, and the initial concentration.

Solution: a. Because the rate constant is the negative of the slope, k=0.0105 s−1. b. Because the reaction is first-order, Rate=k[A]=0.0105 s−1 [A]. c. t1/2=0.693 k =0.6930.0105/s=66.0 d. ln[A]t=−kt+ln[A]0 and substitute k and the initial concentration. So ln[A]=−(0.0105/s)(125 s) +ln 0.150 M=−3.209¯62 and [A]1 25 s = e−3.209¯62=0.0404 M t Check: The units (s−1, none, s, and M) are correct. The rate law is a common form. The rate constant is consistent with value of the slope. The half-life is consistent with a small value of k. The concentration at 125 s is consistent with being about two half-lives.

15.60 Given: plot of 1/[AB] versus time has slope=0.25/M·s Find: (a) k, (b) rate law, (c) t1/2 when [AB]0=0.42 M, and (d) [AB]0=0.150 M [A] and [B] after 55 s Conceptual Plan: a. A plot of 1/[AB] versus time is linear for a second-order reaction. Using 1[AB]t=kt+1[AB]0, the rate constant is the slope. b. Rate law is second order. Add rate constant from part (a). c. For a second-order reaction, t1/2=1k[AB]0. Substitute k from part (a). d. Use the integrated rate law, 1[AB]t=kt+1[AB]0, and substitute k, t, and the initial concentration to get the [AB] t. Then [AB]0, [AB]→[A], [B]. Δ[AB]=[AB] 0 s −[AB] 75 s with 1 mol A1 mol AB and 1 mol B1 mol AB Solution: a. Because the rate constant is the slope, k=0.25 M−1·s−1. b. Because the reaction is second order, Rate=k[A]2=0.25 M−1·s−1 [AB]2. =9.5 s c. t1/2 =1k[AB]0 so t1/2 =1(0.25 M−1·s−1)(0.42 M) d. 1[AB] t =kt+1[AB] so0 [AB]=1kt+1[AB]= t

0

− (55 1(0.25 M−1·s1) s)+

(0.150 )=0.048_9796 M M 1

then Δ[AB]=[AB] 0 s−[AB]55 s=0.150 M−0.048_9796 M=0.101_020 M AB L L ×1 mol A1 mol AB=0.101 M A and 0.101_020 mol AB ×1 mol B1 mol AB=0.101 M B so 0.101_020 mol AB of kconcentration at 55 s is consistent with being over . The Check: The units (M−1·s−1, none, s, and M) are correct. The rate law is a common form. The rate constant is consistent with value of the slope. The half-life is consistent with a small value five half-lives. decomposition of SOCl, first order; k=1.42×10−4 s−1 15.61 Given: 2 2 t decrease to 25% of [SOCl 2], (c) t to 0.78 M when [SOCl]=1.00 M, and (d) [SOCl] after 2.00×102 s and 5.00×102 s when [SOCl]=0.150 M Find: (a) t1 / 2 , (b) to 2

2

0

20

2

2

2

20

Conceptual Plan: a. k → t 1/2t1/2= 0.693 k b.

[SO 2Cl2], 25% of [SO2Cl2], k→t 0

ln[A]t = − kt + ln[A]0

0

c.

[SO 2Cl2], [SO2Cl2], k→t 0

ln[A]t = − kt + ln[A]0

t

d.

[SO2Cl2], t, k→[SOCl] 2 0

ln[A]t = − kt + ln[A]0

2 t

Solution: −4 −1 a. t1 / 2 = 0.693 k =0.6931.42×10 s =4.88×103 s b. [SOCl]=0.25 [SO 2Cl2]0 Because ln[SO2Cl2]=−kt+ln[SO2Cl2], rearrange to solve for t.

2 2 t t

0

[SOCl] 2 2] 0 = − 11.42×10 −4 s−1ln0.25 [SOCl] 2 20 2 20 t= − 1kln [ S O2 C l2]t [SOCl

=9.8×10 s 3

rearrange to solve for t. t=−1k c. [SO 2Cl 2] t=0.78 M; [SO2Cl2]0=1.00 M Because ln[SO2Cl2]=−kt+ln[SO2Cl2], t 0 M and 2.00×102 s in d. [SOCl]=0.150 2 2 0

2 l2] 0 = − 11.42×10 ln [SO2Cl2] t [SOC

−4 −1

s

ln 0.78 M1.00 M =1.7×103 s

ln[SO Cl ] = − 22

t

2 (1.42×10 −4 −1 s )(2.00×10 s)+ln 0.150 M=−1.92_552→

[SO2Cl 2] = e −1.92_552=0.146 M t

[ S O C l ] = 0 . 1 5 0M and 5.00×10 2s in 2

2 0

ln[SO Cl ] = − 22

t

(1.42×10

−4 −1

2

s

)(5.00×10 s)+ln 0.150 M=−1.96_812→

[SO 2Cl2]=e−1.96_812=0.140 M t

Check: The units (s, s, s, and M) are correct. The rate law is a common form. The half-life is consistent with a small value of k. The time to 25% is consistent with two half-lives. The time to 0.78 M is consistent with being less than one half-life. The final concentrations are consistent with the time being less than one half-life.

15.62 Given: decomposition of XY, second order in XY; k=7.02×10−3 M−1·s−1 M and 0.200 M, (c) t to 0.062 M when [XY]=0.150 M, and (d) [XY] after 5.0×101 s and 5.50×102 s when [XY]0=0.050 M Find: (a) t1/2 when [XY]0=0.100 M, (b) t to decrease to 12.5% of [XY]=0.100 0 0 Conceptual Plan: a. t1/2 = 1 k [ A ] 0

[XY]0, k→t1/2 b.

[XY]0, 12.5% of [XY]0, k→

1 [ A ] t = kt + 1 [ A ] 0

t

c.

[XY]0, [XY]t, k→

1 [ A ] t = kt + 1 [ A ] 0

t

d.

1 [A] t = kt + 1 [ A ] 0 t k [XY],0 , →[XY] t Solution: −3 a. t1/2=1k[XY]0=1(7.02×10 M−1 s−1)(0.100 M)=1.42×103 s b. [XY]t=0.125 [XY]0 =0.125×0.100 M=0.0125 M. Because 1 [ X Y ] t =kt + 1 [ X Y ] 0, rearrange to solve for t.

(

1[XY] 0 t = 1k 1[XY]t −

)

)

(

) 10.0125 M−10.100 M =9.97×103 s and =1(7.02×10−3 M−1·s−1

[XY] t=0.125 [XY]0 =0.125×0.200 M=0.0250 M. Because t=1k

(

1[XY]t− 1 [ X Y ] 0

)

= 1(7.02×10

−3

M− 1· s − 1)

c. [XY] t=0.062; [XY]0 =0.150 M. Because t=1k

(

t 1[XY]−1[XY]

0

)

= 1(7.02×10

−3

1 [ X Y ] t =kt +1[XY]0, rearrange to solve for t.

(

)

10.0250 M −10.200 M =4.99×103 s

1 [ X Y ] t = kt + 1 [ X Y ] 0, rearrange to solve for t.

M −1 ·s −1 )

)

(

10.062 M−10.150 M=1.3×103 s

d. [XY] 0=0.050 M and 5.0×101 s in 1[XY]t=kt+1[XY]0→ 1[XY]

−3

−1

−1

1

)(5.0×10 t=(7.02×10 M ·s s in 0=0.050 M and 5.50×102 1[XY] s)+10.050 M=20_.351M so [XY]=0.049 M and 2 t=(7.02×10 −3 M −1 ·s −1 )(5.50×10 s)+10.050 M=23_.861M [XY] so [XY]=0.042 M Check: The units (s, s, s, s, M, and M) are correct. The rate law is a common form. The half-life is consistent with a small value of k. The time to 12.5% is consistent with three half-lives, where the half-life time is increasing. The next time (5000 s) is shorter because the initial concentration is higher. The last time is the shortest because it is less than a half-life with an intermediate concentration. The final concentrations are consistent with the time being much less than one halflife.

15.63 Given: t1/2 for radioactive decay of U-238=4.5 billion years and independent of [U-238]0 Find: t to decrease by 20%; number U-238 atoms today, when 3.2×1018 atoms formed 13.8 billion years ago Conceptual Plan: t1/2 independent of concentration implies first-order kinetics, t1/2→kt 1/2= 0.693k 90% of [U-238] 0, k→t and [U-238]0, t, k→[U-238]t

then

ln[A]t=−kt+ln[A]0 ln[A]t=−kt+ln[A]0 = 0.6934.5×109 yr=1.5_4×10−10 yr−1 then

Solution: t1/2=0.693 k Rearrange to solve for k. k=0.693t1/2 [U-238]t=0.80 [U-238]0 Because ln[U-238]t=−kt+ln[U-238]0, rearrange to solve for t. −1

t= −1k ln [U-238][U-238] 0= − 11.5_4×10−10 yr ln 0.80 [U-238]0 [U-238] 0 t and [U-238]0=3.2×1018 atoms; t=13.8×109 yr ln[U-238] t=−kt+ln[U-238]0=−(1.5_4×10 −10 yr −1 )(13.8×10 yr) 9

=1.4×10 yr9

+ln(3.2×1018 atoms)=40.48_448→ [U-238] t=e40.48_448=3.8×10 17 atoms Check: The units (yr and atoms) are correct. The time to 20% decay is consistent with less than one half-life. The final concentration is consistent with the time being about three half-lives.

15.64 Given: t1/2 for radioactive decay of C-14=5715 years Find: t to decrease by 25.00%; mmol C-14 atoms left, after 2725 yr in sample initially contains 1.20 mmol C-14 Conceptual Plan: Radioactive decay implies first-order kinetics, t1/2→kt 1/2= 0.693k

then

ln[A]t=−kt+ln[A]0

75% of [C-14]0, k→t and [C-14]0, t, k→[C-14]t

ln[A]t=−kt+ln[A]0

Solution: t1/2= 0.693 k Rearrange to solve for k. k=0.693t1/2=0.6935715 yr=1.21¯260×10−4 yr−1 then [C-14] t=0.7500 [C-14]0 Because ln[C-14]t=−kt+ln[C-14]0, rearrange to solve for t.

11.21¯260×10−4 yr−1ln0.7500[C-14]0 [C-14] 0

t 0 =− t=−1kln[C-14][C-14] t=2725 yr ln[C-14]t=−kt+ln[C-14]0=−(1.21¯260×10 [C-14] t=e−0.148¯112=0.862 mmol

−4

yr

=2.37×10 yr3 and [C-14]

0=1.20 mmol

−1

)(2725 yr)+ln(1.20 mmol)=−0.148¯112→

Check: The units (yr and mmol) are correct. The time to 25% decay is consistent with less than one half-life. The final concentration is consistent with the time being less than one half-life.

The Effect of Temperature and the Collision Model 15.65

Extended Figure Description for Chapter 15 Problem 65 Figure 1 15.66

Extended Figure Description for Chapter 15 Problem 66 Figure 1 15.67 Given: activation energy =44.2 kJ/mol, frequency factor=1.9×1011/s, 25 °C Find: rate constant Conceptual Plan: °C→K and kJ/mol→J/mol then E Solution: T =25 °C+273.15=298 K and

44.2 kJmol

a, T, A→k K=°C+273.15 =4.42×104Jmol then

k=Ae−Ea/RT

1000 J1 kJ

× 1000 J1 kJ

4Jmol JK·mol ( )

8.314 / RT 298 K=3.4×103s−1 =(1.9×1011 s−1) e −4.42×10 k = Ae− Ea Check: The units (s−1) are correct. The rate constant is consistent with a large activation energy and a large frequency factor.

15.68 Given: 32°C, rate constant =0.041/s, and frequency factor=1.4×1013/s Find: activation energy

Conceptual Plan: °C → K then E a, T, A → k then J/mol → kJ/molK=°C+273.15 Solution: T =32 °C+273.15=305 K then k=A e−Ea/RT Rearrange to solve for E a.

Ea=−RT ln

k=A e−Ea/RT

1 kJ1000 J

(kA)=−8.314JK mol×305 K×ln(0.041 s−11.4×1013 s−1)=8.49

Check: The units (kJ/mol) are correct. The activation energy is consistent with a modest rate constant and a large frequency factor.

15.69 Given: plot of ln k versus 1/T (in K) is linear with a slope of −7012 K Find:a E Conceptual Plan: Becauseln k =−EaR Solution: Because the slope

(1T)+ ln A plot of ln k versus 1/ Twill have a slope = − Ea/R.

=−7012 K=−E a/R,

a=−(slope)R=−(−7012 K)(8.314 JK·mol)(1 kJ1000 J)=58.30 kJmol

E

Check: The units (kJ/mol) are correct. The activation energy is typical for many reactions.

15.70 Given: Plot of ln k versus 1/T (in K) is linear with a slope of −1.85×104 K. Find: Ea Conceptual Plan: Becauseln k =−EaR

(1T)

+ln A, a plot of ln k versus 1/ T will have a slope = − Ea/R.

Solution: Because the slope =−1.85×104 K=− 4

Ea=−( slope) R=−(−1.85×10 K)

Ea / R,

(8.314 kJ1000 J)=154 kJmol. )(1JK·mol

Check: The units (kJ/mol) are correct. The activation energy is typical for many reactions.

15.71 Given: table of rate constant versus T Find: aEand A Conceptual Plan: Because ln k=−EaR

(1T) +ln A, a plot of ln k versus 1/T will have a slope=− intercept = lnA.

Solution: The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel.

Because the slope=−301_89 K=−E a/R, Ea = − (slope)R = −(−301_89 K)

(8.314JK·mol )(1 kJ1000 J)

=251kJmol and intercept=27.39_9=ln A then A = eintercept = e27.39_9 = 7.93 × 1011s −1.

Check: The units (kJ/mol and s−1) are correct. The plot was extremely linear, confirming Arrhenius behavior. The activation energy and frequency factor are typical for many reactions.

Extended Figure Description for Chapter 15 Problem 71 Figure 1 15.72 Given: table of rate constant versus T Find: Ea and A Conceptual Plan: Because ln k=−EaR

(1T)+ln A, a plot of ln k versus 1/T will have a slope=− intercept =ln A.

Solution: The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel.

Because the slope=−102_83 K=−E

a/R, Ea=−(slope)R=−(−102_83 K)

(8.314 )( ) and = 85.5 JK·mol

1 kJ1000 J

Check: The units (kJ/mol and s−1) are correct. The plot was extremely linear, confirming Arrhenius behavior.

The activation energy and frequency factor are typical for many reactions.

kJmol and intercept =29.96_7=ln

A then A = e intercept = e29.96_7= 1.03 × 1013 s − 1.

Extended Figure Description for Chapter 15 Problem 72 Figure 1 15.73 Given: table of rate constant versus T Find: Ea and A

(1T) +ln A, a plot of ln k versus 1/T will have a slope=−

Conceptual Plan: Because ln k=−EaR

intercept =ln A.

Solution: The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel.

Because the slope=−276_7.2 K=−E a/R, Ea = − (slope)R =−(−276_7.2 K)

(8.314JK·mol )(1 kJ1000 J)

=23.0 kJmol and intercept=25.11_2=ln A then A=eintercept=e25.11_2=8.05×1010 s−1.

Check: The units (kJ/mol and s−1) are correct. The plot was extremely linear, confirming Arrhenius behavior.

The activation energy and frequency factor are typical for many reactions.

Extended Figure Description for Chapter 15 Problem 73 Figure 1 15.74 Given: table of rate constant versus T Find: Ea and A

(1T)+ ln A, a plot of ln k versus 1/T will have a slope=−

Conceptual Plan: Because ln k=−EaR

intercept =ln A.

Solution: The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel.

Because the slope=−116_24 K=−E a/R, Ea = − (slope)R =−(−116_24 K)

(8.314JK·mol )(1 kJ1000 J)

=96.6 kJmol and intercept=32.05_5=ln A then A=eintercept=e32.05_5=8.34×1013 s−1.

Check: The units (kJ/mol and s−1) are correct. The plot was extremely linear, confirming Arrhenius behavior.

The activation energy and frequency factor are typical for many reactions.

Extended Figure Description for Chapter 15 Problem 74 Figure 1 15.75 Given: rate constant =0.0117/s at 400.0 K and 0.689/s at 450.0 K Find: (a) Ea and (b) rate constant at 425 K Conceptual Plan: a. k1, T1, k2, T2, → Ea then J/mol→kJ/mol

ln

(k2k1 ( ) = E a R 1T1 − 1T2

1 kJ1000 J

b.

() ( )

ln k2k1

E a, k1, T1, T2→ k2

= Ea R

1T1 − 1T2

Solution:

( ) (1 )

a. ln k2k1

= EaR

T1 − 1T2

Rearrange to solve forEa.

( ) ( )=

Ea = R ln k2k1

1T1−1T 2

8.314 JK·mol ln

1T2

( ) (1 ) ) (T−T

b. ln k2k1

= EaR

R ln k2 = Ea 11

T1 −

12

(

0.689 s − 10.0117 s − 1

with k1,=0.0117/s,

+ln k1 = 1.22×10

5Jmol

)(

1400.0·K − 1450.0·K

) =1.22×10

5 Jmol

× 1 kJ1000 J =122 kJmol

T1=400.0 K, T 2=425 K Rearrange to solve for k2.

8.314 JK·mol

(1400.0·K−1425·K)+ln 0.0117 s−1=−2.290_2→k=−2.290_2 2

e

=0.101 s −1

Check: The units (kJ/mol and s−1) are correct. The activation energy is typical for a reaction. The rate constant at 425 K is between the values given at 400 K and 450 K. 15.76 Given: rate constant=0.000122/s at 27 °C and 0.228/s at 77 °C Find: (a) Ea and (b) rate constant at 17 °C Conceptual Plan: a.

( )= ( )

°C → K then k1 , T1, k2, T2 → E a then J/mol → kJ/mol K = °C + 273.15

ln k2k1

Ea R

b. °C → K then E a, k1, T1, T2 → k2K = °C + 273.15

Solution: T1=27 °C+273.15=300. K and T a. a. Rearrange to solve for E

( )= (1T1 ) ( ) (1 )

ln k2k1

2=77 °C+273.15=350. K then ln

k2k1

EaR

− 1T2

= Ea R

T1 − 1T2

1T1 − 1T2

1 kJ1000 J

Ea = R ln

b. ln

( )( ) k2k1

1T1 − 1T2

= 8.314 JK·mol ln

(

0.228 s − 10.000122 s − 1

) (

1300. K − 1350. K

)

= 1.32 × 105Jmol × 1 kJ1000 J =132 kJmol

(k2k1)=EaR ) with k(1T1−1 T2

1 =0.000122/s,

T1=300. K, T 2=17 °C+273.15=290 K

Rearrange to solve for 2k. ln k2 =EaR

(1T−1T ) 1

2

5Jmol

+ln k1 =1.32×10

(1300. K−1290. K)+ln 0.000122 s−1

8.314 JK·mol

=−10.83_64→

k2 =e−10.83_64=0.0000197 s−1=1.97×10−5 s−1 Check: The units (kJ/mol and s−1) are correct. The activation energy is typical for a reaction. The rate constant at 17 °C is smaller than the values given at 27 °C. 15.77 Given: rate constant doubles from 10.0 °C to 20.0 °C Find:aE Conceptual Plan:°C → K then

k1, T1, k2,

T2 →Ea then J/mol → k J/molK=°C+273.15

Solution: T1=10.0 °C+273.15=283.2 K and T

2=20.0 °C+273.15=293.2 K and k2=2 k1 then

() ( )

ln k2k1

= EaR

1T1 − 1T2

1 kJ1000 J

( ) ( ). Rearrange to solve for

ln k2k1

= Ea R 1T1 − 1T2

Ea = R ln

(k2k1 ) (1T1−1T2)=8.314 JK·mol ln(2 k1k1)(1283.2 K−1293.2 K)

Ea.

=4.78_51×10

4Jmol

× 1 kJ1000 J =47.85 kJmol

Check: The units (kJ/mol) are correct. The activation energy is typical for a reaction.

15.78 Given: rate constant triples from 20.0 °C to 35.0 °C Find:aE Conceptual Plan:°C → K then

() ( )

k1, T1, k2 , T2 →Ea then J/mol → kJ/molK=°C+273.15

Solution: T1=20.0 °C+273.15=293.2 K and T

ln k2k1

= EaR

1T1 − 1T2

1 kJ1000 J

2=35.0 °C+273.15=308.2 K and k2=3 k1 then

( ) ( ) Rearrange to solve for

ln k2k1

= E a R 1T1 − 1T2

Ea =R ln

(k2k1 ) (1T1−1T2)=8.314 JK·mol ln(3 k1k1)(1293.2 K−1308.2 K)

Ea.

=5.502×104 Jmol×1 kJ1000 J=55.02 kJmol

Check: The units (kJ/mol) are correct. The activation energy is typical for a reaction.

15.79 Reaction (a) would have the faster rate because the orientation factor, p, would be larger for this reaction because the reactants are symmetrical.

15.80 Reaction (b) would have the smaller orientation factor because we are reacting an asymmetric molecule with a homonuclear diatomic molecule (symmetrical); so the orientation is important. In reaction (a), both reacting species are symmetrical; so orientation is unimportant.

Reaction Mechanisms 15.81 The overall reaction is the sum of the elementary steps in the proposed mechanism. Because the first reaction is the slow step, it is the rate-determining step. Use this first step to determine the rate law, Rate=k [AB]2 this is the observed . Because 1

rate law, this mechanism is consistent with the experimental data.

15.82 a. The reaction cannot occur in a single step in which X and Y collide because the rate law would be Rate=k[X][Y].

This is not consistent with the stated rate law of Rate=k[X]2[Y]. b. The overall reaction is the sum of the elementary steps in the proposed mechanism. Because the second step is the rate-determining step, Rate=k3[X2] [Y]. X2 is an intermediate; so its concentration cannot appear in the rate law. Using the fast 2 equilibrium in the first step, we see that1k[X]2= k2[X] or [X]=k 1k2[X] .

2 2

Substituting this into the first rate expression, we get Rate=k3 k1k2 [X2] [Y]. Simplifying this expression, we see that Rate=k [X]2[Y], which is consistent with the experimentally derived rate law.

15.83

a. The overall reaction is the sum of the steps in the mechanism: →k3HCl(g)+CCl3(g)Cl(g)+CCl3(g) →k4CCl4(g)Cl2(g)+CHCl3(g) → HCl(g)+CCl4(g) Cl2(g) k1 ⇄k22 Cl(g)Cl(g)+CHCl3(g) b. The intermediates are the species that are generated by one step and consumed by other steps. These are a Cl(g) and CCl3(g). c. Because the second step is the rate-determining step, Rate=k 3 [Cl][CHCl3]. Because Cl is an intermediate, its concentration cannot appear in the rate law. Using the fast equilibrium in the first step, we see that kk 12[Cl2]. Substituting this into the first rate expression, we get Rate=k3 kk12[Cl2] 1/2 [CHCl3]. Simplifying this expression, we see that Rate=k[Cl2]1/2 [CHCl3].

k[Cl]=k[Cl]2 or [Cl]= 1

2

2

√

√

15.84 a. The overall reaction is the sum of the steps in the mechanism: NO2(g)+Cl2(g) →k1 ClNO2(g)+Cl(g)NO2(g)+Cl(g) →k2 ClNO2(g)¯2 NO2(g)+Cl2(g) → 2 ClNO2(g) b. The intermediates are the species that are generated by one step and consumed by other steps. This is Cl(g). c. Because the first step is the rate-determining step, Rate=k 1[NO2] [Cl2]. Because both of these species are reactants, this is the predicted rate law.

Catalysis 15.85 Heterogeneous catalysts require a large surface area because catalysis can only happen at the active sites on the surface. A greater surface area means greater opportunity for the substrate to react, which results in a speedier reaction.

15.86 The initial and final energies (reactants and products) remain the same. The activation energy drops from 75 kJ/mol to a smaller value (e.g., 30 kJ/mol). There are usually more steps in the reaction progress diagram.

Extended Figure Description for Chapter 15 Problem 86 Figure 1 15.87 Assume rate ratio ∝k ratio (because concentration terms will cancel each other) and k=Ae−Ea/RT. T=25 °C+273.15=298 K, Ea1=1.25×105 J/mol, and Ea =5.5×104 J/mol. The ratio of rates will be 2 4Jmol 5Jmol =Ae−Ea/RTAe−Ea/RT 298 K 22298_ .K199e−50.4_53=1012 e−1.25 × 10 (8.314JK·mol=e−) =e−5.5 × 10(8.314JK·mol) kk 21

2

1

Note that no additional significant figures are quoted since the exponent has its last significant digit in the 1’s place.

15.88 Assume rate ratio∝k ratio (because concentration terms will cancel each other) and k=Ae−Ea/RT. =1.08×105 J/mol. The ratio of rates will be T=25 °C+273.15=298 K and E a 1

kk 1 =106=

− a / − aRT / AeERTAeE 2 1

=

2(

e−Ea

8.314 JK·mol

)

( ol=− JK·me)E 1.17×10−19→−Ee2.47_756 × 103Jmol 2982 .K47756 × 10 =1.17×10−13 →−E a 2.47_756×103Jmol= ln(1.17 × 10−13)=−29.7_766→E a

298 K−1.08 × 105Jmol8.314

e

a2

3Jmol

2

2

4Jmol =73.8 kJmol a2 = 7.38 × 10

Cumulative Problems 15.89 Given: table of [CH3CN] versus time Find: (a) reaction order, k; (b) t1/2; and (c) t for 90% conversion Conceptual Plan: (a) and (b) Look at the data and see if any common reaction orders can be eliminated. If the data does not show an equal concentration drop with time, zero order can be eliminated. Look for changes in the half-life (compare time for concentration to drop to one-half of any value). If the half-life is not constant, the first order can be eliminated. If the half-life is getting longer as the concentration drops, this might suggest second order. Plot the data as

indicated by the appropriate rate law, or if it is first order and there is an obvious half-life in the data, a plot is not necessary. Determine k from the slope of the plot (or using the half-life equation for first order). (c) Finally, calculate the time to 90% conversion using the appropriate integrated rate equation.

Solution: (a) and (b) By the preceding logic, we can see that the reaction is first-order. It takes 15.0 h for the concentration to be cut in half for any concentration (1.000 M to 0.501 M, 0.794 M to 0.398 M, and 0.631 M to 0.316 M), so t1/2=15.0 h.

Then use t1/2=0.693 k and rearrange to solve for k. k=0.693t1/2 =0.69315.0 h=0.0462 h−1 (c) [CH3CN]=0.10 [CH 3CN] Because ln[CH 3CN]=− kt +ln[CHCN], rearrange to solve for . t 3 t 0 0 [CHCN] t= −1k ln [CHCN][CHCN] =49.8 h 3 t 3 0 = −10.0462 h−1 ln 0.10[CHCN] 3 0 3 0 Check: The units (none, h−1, h, and h) are correct. The rate law is a common form. The data showed a constant half-life very clearly. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a first-order reaction. The time to 90% conversion is consistent with a time between three and four half-lives.

15.90 Given: table of [XY] 2 versus time Find: (a) reaction order,k; (b) t1 / 2 at initial concentration; and (c) [X] at 10.0 h Conceptual Plan: (a) Look at the data and see if any common reaction orders can be eliminated. If the data does not show an equal concentration drop with time, zero order can be eliminated. Look for changes in the half-life (compare time for concentration to drop to one-half of any value). If the half-life is not constant, the first order can be eliminated. If the half-life is getting longer as the concentration drops, this might suggest second order. Plot the data as indicated by the appropriate rate law. Determinekfrom the slope of the plot. (b) Calculate the half-life with the appropriate equation. (c) Finally, calculate theX [ 2Yat ] 10.0 h using the appropriate integrated rate expression and convert this to a [ ] then to X using [ ] the reaction stoichiometry. change inX2Yand Solution: a. By the preceding logic, we can eliminate both the zero-order and first-order reactions. (Alternatively, you could make all three plots, and only one should be linear.) This suggests that we should have a second-order reaction.

2 Plot 1/[XY] versus time. Because 1[X2Y]t= kt + 1[X 2Y]0, the slope will be the rate constant. The slope can be determined by measuring Δy/Δx on the plot or by using functions such as “add trendline” in Excel. Thus, the rate constant is 1.68_27 M−1·h−1, and the rate law is Rate=1.68 M−1·h−1[X 2Y].2 t so b. t1/2=1k[X2Y]0 1(1.68_27 M−1·h−1 )(0.100 M)=5.94 h 1/2= Finally, use t=kt+1[X2Y]0; substitute the values of [X2Y], 10.0 h, and k; and rearrange to solve for [XY] c.

2 at 10.0 h.

0 1[X2Y] −1 1.0 M 2 2Y]=1kt+1[XY]=1(1.68_27 M−1 0 ·h )(1 0.0 h)+ [X

(010 )=0.0372_759 M then

t

Δ[X

2Y] = [X2Y] − [X 2Y] 0

10.0 h

0.062_724 mol X2Y

L

=0.100 M−0.0372_759 M=0.062_724 M then ×2 mol X1 mol X2Y=0.13 M X

Extended Figure Description for Chapter 15 Problem 90 Figure 1c Check: The units (none, M−1 ·h −1 , h, and M) are correct. The rate law is a common form. The plot was extremely linear, confirming second-order kinetics. The rate constant is consistent with the units necessary to get rate as M/s, and the magnitude is reasonable because we have a second-order reaction. The half-life is consistent with the data table, which indicates that the half-life is a little over 5 h. The [X] at 10 h s is consistent with the changes that we see in the data table through 5 h.

15.91 G i v e n :Rate= k [ A ] [ C ] 2 [ B ] 1 / 2=0.0115 M/s at certain initial concentrations of A, B, and C; double A and C concentration and triple B concentration Find: reaction rate Conceptual Plan:[A]1, [B]1, [C]1, Rate 1, [A]2, [B]2, [C]2→Rate 2 Solution:Rate 2Rate = 1 k[A]2[C]2[B] 2

1/2

Rate 2Rate 1 =k [A]2[C]2[B] 2k [A]1[C] 1[B] 2

1/2

2

1 1/2

1 2k[A]1[C]1[B] Rearrange to solve for Rate 2. Rate 2=k[A]2[C]2[B] 2k[A]1[C]1[B] 2

1/2

2

1/2

1

2

1/2

Rate 1 [A]2=2[A]1, [B]2=3[B]1, [C]2=2[C]1,and Rate 1=0.0115 M/s so (3[B)]1/2 1 1 1 1 [B] k[A] [C] Rate 2=k2[A] 1(2[C)]2 0.10115Ms = 2331/2 0.0115 Ms=0.0531Ms 2

Check: The units (M·s−1) are correct. They should increase because we have a factor of eight (23) divided by the square root of three (1.73). k 23] [O2]; initially, 1.0 mol O 3 and 1.0 mol O2 in 1.0 L 15.92 Given: Rate= [O Find: fraction O3 reacted when reaction rate is cut in half Conceptual Plan: 2

reacted then O 2]1 [O] , [O, Rate 1, Rate 2→[O] [ 3 ,]O1 [ →O3 fraction mol, L→M3 1then 3 3] 2 Solution: M=

mol L

so [O]= 3

1

Let x =Δ[O 3] so [O]3

2

=[O]3

1.0 mol1.0 L

1 −x

2

2

[ 23 ]fraction 1 O reacted= [ O3 ] 1 − [O3 ]2

[ O3 ] 1

=1.0 M and [O2]=1.0 mol1.0 L=1.0 M. Rate 1=2 Rate 2

and [O]2 2 =[O]2

1

2 2 . Substitute values into Rate 2Rate 1=k[O3] 2 [O]k[

1 + 3 / 2x

√(−2.7_5)2−(4)(1.0)(0.50) 2(1.0)

(1.0 M−x) [1Oan2d]1 rearrange to solve for x. Rate 22 Rate 2=k (1.0 M+3/2x)k (1.0 M)2 (1.0 M)→0.50 M (1.0 M+3/2x)=(1.0 M−x)2→ 2

O3 ]

2

0.50 + 0.75x = 1.0 − 2.0x + x2 →0=x2 −2.7_5x+0.5_0 Solve with quadratic equation So x=2.7_5 ±

Rate 2Rate 1 = k [ O3 ] 2 [ O2 ] 2 k [ O3 ] O1

M=molL

2

(

2

x = − b ± √ b −4 ac

)

2a

=2.7_5 ± √5.5_625 2.0

=2.7_5 ± 2.3_585

2.0

=0.1_9575 M or 2.5_5 M

The answer must be 0.1_9575 M because the other answer is larger than our initial concentration (and therefore is impossible). _ [O]31 M 1.0 M O3 fraction reacted=[O3]−[O3] = x[O]3 1= 0.19575 =0.2 1 2 Check: The units (unitless) are correct. The concentration is reasonable because two forces are slowing down the reaction: (1) the decrease in the reactant and (2) the increase of the product (which appears in the rate law). The calculation can be doublechecked by substituting the value of X; the resulting rate =0.5 k.

15.93 Given: table of PTotal versus time Find: rate law, k, andTotal P at 2.00×104 s Conceptual Plan: Because 2 moles of gas are generated for each mole of CH3CHO decomposed, this PCHCHO= P Total −(PTota−P l 3 °

Total ). Look at the data and see if any common reaction orders can be eliminated. If the data does not °

show an equal PTotal rise (or PCHCH Odrop) with time, zero order can be eliminated. There is not enough data to look for changes in the half-life (compare time for P CH3CHO to drop to one-half of any value). It does appear that the half-life 3 is getting longer, so the first order can be eliminated. Plot the data as indicated by the appropriate rate law. Determine k from the slope of the plot. Finally, calculate the P CH3CHO at 2.00×104 s using the appropriate integrated rate expression and convert this to PTotal using the reaction stoichiometry.

Solution: Calculate PCH 3CHO = P P ).

Total − (PTotal − Total °

°

Time(s)

P Total(atm)

PCH3CHO(atm)

0

0.22

0.22

1000

0.24

0.20

3000

0.27

0.17

7000

0.31

0.13

By the preceding logic, we can eliminate both the zero-order and the first-order reactions. (Alternatively, you could make all three plots, and only one should be linear.) This suggests that we should have a second-order reaction.

Plot versus 1/P CH CHO 3

3H time. Because 1PCHCO

−4 = kt +1P CH3CHO, the slope will be the rate constant. The slope can be determined by measuring y Δ/Δ x on the plot or by using functions such as “add trendline” in Excel. Thus, the rate constant is 4.5×10 atm− 1 · s − 1, and the °

rate law is Rate=4.5×10−4 atm − 1 · s−1PCH

3CHO.

, 2.00×104 s, and k and rearrange to solve forP

Finally, use 1PCHCO= 3H kt +1P CH3CHO; substitute the values ofP °

P CH3CHO =1kt+1P°

CH3CHO °

CH 3CHO

= 1( 4.5×10

−4

am t −1 ·s − 1 )(2.00×104 s)+

(

10.22 atm

CH3CHO

at 2.00×104 s.

°

)=0.073_8255 atm=0.074 atm

P Total =2P Total − P CH3CHO =2(0.22 atm)−0.073_8255 atm= 0.36_6175 atm=0.37 atm.

Finally, from the first equation in the solution, °

Extended Figure Description for Chapter 15 Problem 93 Figure 1 Check: The units (none, atm−1·s−1, and atm) are correct. The rate law is a common form. The plot was extremely linear, confirming second-order kinetics. The rate constant is consistent with the units necessary to get rate as atm/s, and the magnitude is reasonable because we have a second-order reaction. TheTotal P at 2.00×104 s is consistent with the changes we see in the data table through 7000 s.

15.94 Given: table of P H C O versus PTotal at 20,000 s Find: rate law and k 2 2 4

°

Conceptual Plan: Because 2 moles of gas are generated for each mole of H2C2O4 decomposed, this P H2C2O4 = P H2C2O − (P Total − P H2C2O ) and the Rate =(P 4

H2C2O4 −

4

°

°

P Total)/20,000 s. Using the general rate law form, compare rate

°

ratios to determine reaction order.

Rate 2Rate 1= k[A] 2k[A] 1 n

n

Then use one of the concentration/initial rate pairs to determine k .

Rate = k[A]n Solution: Calculate Rate=( P H2C2O4− P Total)/20,000 s. °

Experiment 1

Experiment 2

Experiment 3

65.8

92.1

111

94.6

132

160

0.00144_

0.0019_ 95

0.0024_ 5

P H2C2O (mmHg) 4

°

P Total at 20,000 s (mmHg) Rate (mmHg/s)

Rate 2Rate 1=k[A]2k[A]1 Comparing the first two sets of data, n

n

0.0019_95 mmHg/s0.00144 mmHg/s=k(92.1 mmHg)nk(65.8 mmHg)n and 1.3_854=1.39_97n; so n=1. If we compare the first and third data sets, 0.0024_5 mmHg/s0.00144 mmHg/s=k(111 mmHg)nk(65.8 mmHg)n and 1.7_0139=1.6_8693n; so n=1. This second comparison is not necessary, but it increases our confidence in the reaction order. So Rate=k P . Selecting the first data set and rearranging the rate equation, H2C2O4 . s −5 −1 −5 −1 k=RatePH2C2O4 =0.00144 mmHg 65.8 mmHg =2.19×10 s ; so Rate=2.19×10 s P H2C2O4 Check: The units (none and s−1) are correct. The rate law is a common form. The rate is changing proportionately with the initial pressure, so first order is consistent. The rate constant is consistent with the units necessary to get rate as mmHg/s, and the magnitude is reasonable because we have a first-order reaction.

15.95 Given: N2O5 decomposes to NO2 and O2, first order in [N2O5]; t1/2=2.81 h at 25 °C; V=1.5 L, P N O =745 torr Find: PO2after 215 minutes 2 5

°

t1 / 2 →k then °C→K and torr→atm then

Conceptual Plan: Write a balanced reaction. Then

P N

, V, T→ / n V then min→h then [N 2O], 5 t, k→[N2O5] thent [NO], [NO]→[O 2 5 25t

2O5

0

0

N2O5→2 NO2+ 12 O2

2]

PV = nRT

t

t 1 / 2 = 0.693 k 1 h60 min

K=°C+273.15

ln [A]t = − kt +ln [A] 0

°

then [O2], V, T→ P

(

)

= [N2O 5]0 − [N 2O ]t × 1/25 mol O21 mol N2O5 [O2] t

760 torr1 atm

PV = nRT

O 2 and finally atm→torr

t

1 atm760 torr

°

Solution: t1/2 =0.693 k and rearrange to solve for k. k=0.693t1/2=0.6932.81 h=0.246_619 h−1 then T=25 °C+273.15=298 K. 745 torr× nV=PRT=0.980_263 atm0.08206 L·atmK·mol×29=8 K0.0400_862 M then 215 min×1 h60 min=3.58_333 h Because ln[NO]=−kt+ln[NO] 2 5

25

t044

[N2O5]=e −4.10_ t

0

=−(0.246_619 h

1 atm760 torr =0.980_263 atm then PV = nRTRearrange to solve forn/V.

−1 )(3.58_333 h)+ln(0.0400_862 M)=−4.10_044→

=0.0165_654 M then

[O2]=([N2O5)−[NO]]×2 5 t 0 t

1/2 mol O12 mol NO25=

(

0.0400_862molN25O

L

L

2 −0.0165_654 mol NO5

)

× 1/2 mol O21 mol N2O5=0.0117_604 M O 2

then finally PV=nRT and rearrange to solve for P =219 torr P=nV RT=0.0117_604 molL×0.08206 L atmK·mol×298 K=0.287587 atm×760 torr1 atm Check: The units (torr) are correct. The pressure is reasonable because it must be less than one-half of the original pressure.

3H6) reacts,(Cfirst order in [C3H6]; Ck=5.87×10−4/s at 485 °C; V=2.5 L, P 15.96 Given: cyclopropane H =722 torr 3 6

°

Find: t to P C3H6 =100.0 torr P Conceptual Plan: Because P

M, we do not need to convert P to M.

C3H6

, P C3H6, k → tln[A] t = − kt + ln[A]0

°

∝

3H6] = −[C3H6] kt 0+ lnRearrange to solve for t.

Solution: ln[C t 3 t=−1k ln [C3H6][C3H] 60 =−1k ln P C3H6P C3H6=−15.87×10−4 s−1 ln 100. torr722 torr =3.37×10s=56.1 min t °

Check: The units (s or min) are correct. The time is reasonable because it is about three half-lives (pressure dropped to 14% of original pressure). second order in I;k =1.5×1010 M−1·s−1 , [I]0=0.0100 M 15.97 Given: I2 formation from atoms, I Find: t to decrease by 95% Conceptual Plan: [I]0, [I]t, k→t

1[A]t=kt+1[A]0

1 [ I ] t = kt + 1 [ I ] 0 Rearrange to solve for t.

Solution: [I] t=0.05 [I]0 =0.05×0.0100 M=0.0005 M t = 1k

(1[I]−1[I])=1(1.5×1010 )(10.0005 M−10.0100 M−1·s−1M)=1_.267×10−7s=1×10−7s t

0

Check: The units (s) are correct. We expect the time to be extremely small because the rate constant is so large.

Find: m(C 12H ) hydrolyzed 22O [CHO]; k=1.8×10−4 at 25 s−1 °C; V=2.55 L, [C12H22O11]=0.150 M, and 195 min

15.98 Given: sucrose hydrolysis, first order in 122211

11

0

0 then 60 s1 min then0 [C12H22O11], t, k→[C12H22O11] t

Conceptual Plan:min →s

ln[A]=−kt+ln[A] t

molC12H22O11 hydrolyzed → gC12H22O11 hydrolyzed [C12H22O11]0, [C12H22O11]→ t

V,

mol C12H22O11 =

11]−[C 11]) ([C12H22O

12H22O t × V 342.30 g C12H22O11 1 mol C12H22O11

0

Solution: 195 min× 60 sec1 min =117_00 s Because 12H22O11]= − 11] =kt + − ln[C t 0

[

lnC 12H22O

[C12H22O11]t =e

−4.00_312

(1.8 ×)( 10)+ln (0.150 M)=−4.00_312→ s

117_00 s

=0.0182_586 M then

])

11 = ([11]−[C

mol C12H22O

−4 −1

C12H22O

0

12H22O11 t × V =

(0.150mol C12H22O11 )×2.55 L=0.335_941 mol C12H22O11 L

−0.0182_586mol C12H22O11 L

=115 g C12H22O11 0.335_941 mol C 12H22O11×342.30 g C12H22O111 mol C12H22O11 Check: The units (g) are correct. The mass is reasonable because it must be less than the original amount in solution (131 g). The amount is close to the original amount in solution because the final sucrose concentration is so low because we have gone over three half-lives.

Total=755.1 mmHg, and V=200.0 mL; P H2 O =23.8 mmHg Find: t 15.99 Given: AB(aq)→A(g)+B(g); k=0.0118 M−1·s−1; 250.0 mL of 0.100 M AB; collect gas over water T=25.0 °C, P °

→ PA + P B then mmHg→atm and mL→L Conceptual Plan: P Total, PHO 2 and °C→K P, V, T→n

A+B→∆nAB K=°C+273.15 PV=nRT

1 atm760 mmHg

PTotal= PHO + PA + P B 2

△nAB =12n A+B

1 L1000 mL

= kt+ then k,[AB] → t1[AB]t then [AB]0, VAB, ∆nAB→[AB][AB]=[AB]-△nV×1 L1000 mL 0

Solution: PTotal=PHO+PA+PB Rearrange to solve for PA+PB. PA+PB=PTotal−P

1[AB] 0

ABAB

H2O =755.1 mmHg −23.8 mmHg=731.3 mmHg

2

PA+PB=731.3 mmHg×1 atm760 mmHg=0.9622_3684 atm V=200.0 mL×1 L1000 mL=0.2000 L T=25.0 °C+273.15=298.2 K PV=nRT Rearrange to solve for n. n=PVRT K mol A+B Because 1 mole each of A and B is generated for each mole of AB reacting,ΔnAB = 12nA + B = 12 (0.007864_5309 mol A+B)=0.003932_2655 mol AB then =0.007864_5309 nA+B =0.9622_3684 atm×0.2000 L0.08206L·atmmol·K×298.2 1 L1000 mL =0.100 M−0.003932_2655 mol AB250.0 mL×1 L1000 mLM−0.01572_9062 M [AB]=[AB]− 0 Δn V × =0.100 =0.084_2709 M Because 1[AB]t=kt+1[AB]0, rearrange to solve for t.

(

t = 1k 1 [ X Y ] t − 1 [ X Y ] 0

)

=1

(10.084_2709 M− )

(0.0118· s M ) −1

−1

10.100 M

=158_.1773 s=160 s

Check: The units (s) are correct. The magnitude of the answer (160 s) makes sense because the rate constant is 0.0118 M−1 s−1 and a small volume of gas is generated.

15.100 Given: 2 HO(aq)→2 HO(l)+O(g); k=0.00752 s−1 ; 150.0 mL of 30.0 % H2O2 by mass, d=1.11 g/mL; collect gas over water T=20.0 °C, PTotal=742.5 mmHg, and t=85.0 s; P 22

2

2

H2O

=17.5 mmHg

Find: VO

°

→ gH2O→mol H2O and mL solution → L solution Conceptual Plan: mL solution →g solution 2 2 then mol H2O, 2L solution → M H2O2 thenk,

→ [H2O2]t [H2O2], t0 M= amount solute (moles)volume solution (L)

P [H2O2], [H2O2], 0 solution→ Lt ∆ n H O → nOthen P Total,PH 2O→OthenmmHg 22

and mL→L and °C→K then

30.0 g H2O2100 g solution 1 mol H2O234.02 g H2O2

1.11 g1 mL

2

P, n, T → VO 2

2

1 L1000 mL K=°C+273.15

→ atm

ΔnH2O2 =

1 L1000 mL

ln[A]t = − kt + ln[A]0 then

[22]) ([22]0 HO

−HO

× V H2 O2

ΔnH2O2 = nO2

P Total = P H2O + P O 2

PV=nRT

g solution ×1 mol H2O234.02 g H2O2=1.46_8254 mol 22 H and Solution: 150.0 mL solution× 1.11 g solution1 mL solution × 30.0 g HO100 2O2 150.0 mL solution× 1 L solution1000 mL solution=0.1500 L solution then _ M = amount solute (moles)volume solution (L) = 1.468254 mol HO 22 0.1500 L solution =9.78_8360 M H 2O2

(

)( )+ln(9.78_8360 M)=1.64_1994→

−1 85.0 s 2O2ln[H 2O2] = − 0.00752 s Because ]t = − kt + ln[H 0

1 L1000 mL VH2O2×2 [H 2O ]t = e1.64_1994 2 =5.16_5459 M, ΔnH2O = ([H 2O 2]0− [H])× 2O 2

(

ΔnH2O2 =9.78_8360 M−5.16_5459 M then Δn

)× (0.1500 L)=0.693_4352 mol H2O2

P P H2O + P O2 Rearrange to solve forP O2.

H2O2 =0.5n O2 =0.34672 mol O 2then Total=

P O=P Total−P H2O =742.5 mmHg−17 .5 mmHg=725.0 mmHg 2 P O=725.0 mmHg×1 atm760 mmHg=0.9539_4737 atm, T=20.0 °C+273.15=293.2 K, PV=nRT 2

Rearrange to solve for V. V=nRTP and ×293.2 K VO=0.34672 mol O×0.08206L·atmmol·K 2

0.9539_4737 atm

=8.74480 L O 2 =8.75 L O 2

Check: The units (L) are correct. The magnitude of the answer (9 L) makes sense because about 0.4 mole of gas is generated (22.4 L=1 mole gas at STP).

15.101 a. There are two elementary steps in the reaction mechanism because there are two peaks in the reaction progress diagram. b.

Extended Figure Description for Chapter 15 Problem 101 Figure 1b c. The first step is the rate-limiting step because it has the higher activation energy. d. The overall reaction is exothermic because the products are at a lower energy than the reactants.

15.102 a.

The first step is the rate-limiting step because it has the higher activation energy.

k1[ ][

b.Because the first step is the rate-determining step, Rate= HCl HC=CH 2

2]. The reaction will be second-order overall.

c. The overall reaction is exothermic because the products are at a lower energy than the reactants.

15.103 Given: n-butane desorption from single crystal aluminum oxide, first order; k =0.128 s−1 at 150 K; initially completely covered Find: (a) t1/2; (b) tfor 25% and for 50% to desorb; (c) fraction remaining after 10 s and 20 s

1 atm760 mmHg

2

()

] ()[ 10], [C4H

a k→t

Conceptual Plan:

Solution: a. t1/2 = 0.693 k

1 / 2 b C4H

10

0

t

10] (c)[C4H

, k→t

0

[

, t, k→C4H

10

]t t1 / 2 = 0.693 k

ln[A]t = − kt + ln[A]0

ln[A]t = − kt + ln[A]0

=0.6930.128 s−1=5.41 s

] t [C4H ] = − 10.128 s ln 0.75 [C4H ] [C4H ] =2.2 s. For 50% desorbed, =0.75 CH and t= − 1k ln [C4H Rearrange For to 25% solvedesorbed, for [ 4[CH 10]0 [ 4H10] 4 10 ]tt.

0 10.128 s1 0.50 [C4H CH and t=−1kln[C4H1] t[ ] ] [C4H10 ] 0=− ln =5.4 s [C4H10]t = 0.50 [C4H10] 0

[

]

b. lnCH4 10 t = − k t + l n C

10

0

[

c. For 10 s, lnC 4H10

−

410

(

) )

(

( )

10

−1

10

0

10

0

10

0

0

0

−1 10 s + ln(1.00) = − 1.2_ 8 → 4H10]= − 0.128 s ]t = − kt + ln[C 0

−1.2_8 =0.28=fraction covered [C4H 10 ]=t e −1 20 s + ln(1.00) = − 2.5_ 6 → 4H10]= − 0.128 s = − kt + ln[C For 20 4Hs,10 ]ln[C 0 t

[CH]= 4 10 t

e −2.5_6 =0.077=fraction covered

Check: The units (s, s, s, none, and none) are correct. The half-life is reasonable considering the size of the rate constant. The time to 25% desorbed is less than one half-life. The time to 50% desorbed is the half-life. The fraction at 10 s is consistent with about two half-lives. The fraction covered at 20 s is consistent with about four half-lives.

15.104 Given: 120 nm film n-pentane evaporation from single crystal aluminum oxide, zero order; k=1.92×1013 2 at 120 K; initially, coverage=8.9×1016 molecules/cm2 molecules/cm·s Find: (a) t1/2; (b) fraction remaining after 10 s Conceptual Plan: a.

) (

1/2 t1 / 2 = [ A ] 2k [C5H 12 ]0, k → t [C5H , then [C5 H12] b.

12 ]t → fraction remaining 12 ]0, t, k → [C5H12] t 0 [C5H 0

Solution: C5H[12 ] a. t1/2 = 0 b.

16 molecules cm2

2k

= 8.9 × 10

C5H12 ] 12 ] t [ fraction remaining= [C5H 0

[A] t = − kt +[A] 0

13 molecules c 2 × 1.92 × 10 =m22·.s3×103 s for half of the film to evaporate

=−(1.92×10 )(10·s)+8.9×1016molecules

= −kt+[C5H12] [C5H12] t 0

13 moleculescm2 s

cm2

= 8. 8_ 808 × 1016molecules cm2 ecules cm2 8.9×1016moleculescm2 _ fraction remaining=[C 5H12t[C5H12] ] 0= 8.8808×1016mol =0.99_784=1.0; so within experimental error, all are remaining on the surface.

Check: The units (s and none) are correct. The half-life is reasonable considering the size of the rate constant. The fraction at 10 s is reasonable given the fact that the time is very, very small compared to the half-life.

15.105 a. Given: table of rate constant versus T Find: aEand A =−EaR

Conceptual Plan: First, convert temperature data into kelvin(°C+273.15=K).Becauseln k

(1T)+ln A , a plot of ln k versusl/T will have a slope

Solution:The slope can be determined by measuring ∆/∆ y x on the plot or by using functions such as “add trendline” in Excel. Because theslope =−107_59 K=− Ea = −

= − Ea / R and an intercept =ln A.

Ea/R,

(slope)R

(8.314JK·mol )(1 kJ1000 J)

=−(−107_59 K)

=89.5 kJmol and intercept=26.76_9=ln A then A=eintercept=e26.76_9=4.22×1011 s−1 Check: The units (kJ/mol and s−1) are correct. The plot was extremely linear, confirming Arrhenius behavior. The activation energy and frequency factor are typical for many reactions.

Extended Figure Description for Chapter 15 Problem 105 Figure 1a b. Given: part (a) results Find: k at 15 °C Conceptual Plan: °C →K then T, E , A→k°C+273.15=K a

ln k = − Ea R

() 1T

+ln A

Solution: 15 °C+273.15=288 K then ln k = − Ea R

(1T)

+ln A = −89.5 kJmol× 1000 J 1 kJ

Check: The units

8.314 JK·mol

10 ( ) (4.22 × )=−10_.610→k=e 1288 K

+ln

−10_.610

11s − 1

= 2.5 × 10

−5

M

−1

·s

−1

(M−1·s−1) are correct. The value of the rate constant is less than the value at 25 °C.

c. Given: part (a) results, 0.155 M CHBr and 0.250 M OH−at 75 °C Find: initial reaction rate 25 Conceptual Plan: ° C →K then T, E a, A→k then k,

reaction rate °C+273.15=K ln k = ] → initial [C2H5Br], [OH−

− Ea R

( ) +A 1T

] ] [C 2H[OH− 5Br

Rate = k

Solution: 75 °C+273.15=348 K then ln k= − Ea R

(1T)

+ln A = −89.5 kJmol× 1000 J 1 kJ

k = e−4.1 _656 =1.5_521×10−2 M−1·s

8.314 JK·mol

(1348 K)

+ ln(4.22 × 1011s − 1)=−4.1_656→

−1

[C2H5Br ][ ] = (1.5_ 521)(× )( 10 )=6.0×10

Rate = k

−2

OH

Check: The units

M

−1

·s

−1

0.155 M 0.250 M

−4

M·s

−1

(M·s) are correct. The value of the rate is reasonable considering the value of the rate constant (larger than

15.106 Given: k =2.35×10−4 s

−1

− 1 at 293 K and k=9.15×10−4 s−1 at 303 K

Conceptual Plan: °C → K then k1, T1, k , T2 →Ea then k2, T2 , Ea→A 2

Solution: T1=293 K and k1 = 2.35 × 10

Find: A

K=°C+273.15 ln k

= − Ea R

−4 −1

s

; T 2=303 K and k2 =9.15×10−4 s−1 then

( ) ( ) Rearrange to solve for

ln k2k1

= Ea R

1T1 − 1T2

Ea.

( )+ln A 1T

E /RT

k= Ae− a

Ea =R ln

(1293 K−1303 K) (kk1 ln(9.15×10−4 s−12.35×10−4 s−1 ) (1T−1T)=8.314 )JK·mol 2

1

=1.0_0334×105Jmol

2

· mol) 303s−1 K a a Because k=Ae−E/RT, rearrange to solve for A. A=keE/RT=9.15 =1.8×1014 ×10 −4 s −1e1.0_0334 × 105Jmol(8.314 JK Check: The units

(s−1) are correct. The frequency factor is typical for a reaction.

15.107

(k=Ae−Ea/RT) will be independent of temperature.

a. No, because the activation energy is zero. This means that the rate constant

b. No bond is broken, and the two radicals (CH3) attract each other. of adiatomic gasessofrom atomic gasesenergy to break the N≡N bond than the H—H bond.

15.108 c. a. Formation Nitrogen has triple bond, it will take more b. Given: Ea=315 kJ/mol for reaction 1 and Ea=23 kJ/mol for reaction 2, frequency factor similar, and 25 °C Find: ratio of rate constants Conceptual Plan: °C→K then T, E, a1 Ea2, → A

kk=Ae−Ea/RT1Ae−Ea/RT 2 12

k1/ k2°C+273.15=K

Ratio of rate constants will be1k2=Ae k

kJ/mol. a1 =315 kJ/mol and Ea=23 2

Solution: T=25 °C+273.15=298 K. E

−Ea/RTAe−Ea/RT2 1

kJmol × 1000 J1 kJ

=e −315

(8.314) 298 Ke −23 (8 =6.5×)10 298 K−52 .

JK·mol

kJmol × 1000 J1 kJ

.314 JK·mol

Check: The units (none) are correct. Because there is a large difference between the activation energies, we expect a large difference in the rate constants.

15.109 Given: t1/2 for radioactive decay of C-14=5715 years; bone has 19.5% C-14 in living bone.

Find: age of bone 0 0, k→tt 1/2 =.693k Conceptual Plan: Radioactive decay implies first-order kinetics, t1/2→k then 19.5% of [C−14] Solution: t1/2=0.693 k Rearrange to solve for k. k=0.693t1/2 =0.6935715 yr=1.21¯26×10−4 yr−1 then [C-14] t=0.195 [C-14]0 Because ln[C-14]t=−kt+ln[C-14]0, rearrange to solve for t. −11.21¯26×10−4 yr−1 ln 0.195[C-14]0 [C-14] 0 =1.35×104 yr t= −1k ln [C-14][C-14]0= t

ln[A]t=−kt+ln[A]0

Check: The units (yr) are correct. The time to 19.5% decay is consistent with the time being between two and three half-lives.

15.110 Given: t1/2 for radioactive decay of U-238=4.5 billion years; rock has 83.2% of original U-238 Find: age of rock 1/2= 0.693k ln[A]t=−kt+ln[A]0 Conceptual Plan: Radioactive decay implies first-order kinetics, t1/2→k then 82.3 of [U−238]0, k→t t 9 Solution: t1/2=0.693k Rearrange to solve for k. k=0.693t1/2 =0.6934.5×10 yr=1.54×10−10 yr−1 then [U-238]t=0.823[U-238]0 Because ln[U-238]t=−kt+ln[U-238]0, rearrange to solve for t. ln 0.832[U-238]0 [U-238] 0 =1. 19×109 yr t= −1k ln [U-238][U-238] t 0 = −11.54×10−10 yr−1 Check: The units (yr) are correct. The time to 82.3% decay is consistent with the time being less than one half-life.

15.111

(Rate = k[H2][I2]).

2][I2], is valid.

For the first mechanism, the single step is the overall reaction.

The and rate the lawmechanism is determined by the stoichiometry; so Rate=k[H

a. For each, check that all steps sum to overall reaction and that the predicted rate law is consistent with experimental data

For second mechanism, therate-determining¯ overallHI(g)H2(g)+I2(g) reaction isstep, the Rate=k[H sum the steps the in the I2(g)the k2⇄k12 I(g)H2(g)+2 I(g)→k32 →of 2 HI(g)So summechanism: matches the overall reaction.

Because the second step is the 2 2 ]2. Because I is an intermediate, its concentration cannot appear in the rate law. Using the fast equilibrium in the first step, we see1[I2]=k2[I] that k or [I]=k1k2[I2].

3 2][I 3k1k2[H2][I2], and the mechanism is valid.

Substituting this into the first rate expression, we get Rate=k 2I(g), b. To distinguish between mechanisms, you could look for the buildup of I(g), the intermediate in the second mechanism, or for the buildup of I(g) and the H intermediates in the third mechanism.

15.112 For this mechanism, the overall reaction is the sum of the steps in the mechanism: Cl2(g) k1⇄k2 2 Cl(g)Cl(g)+CO(g) k3⇄k4 ClCO(g)ClCO(g)+Cl2(g) →k5 Cl2CO(g)+Cl(g)¯CO(g)+2 Cl2(g) → Cl2CO(g)+2 Cl(g)

5 [ClCO] [Cl2]. Because ClCO is an intermediate, its concentration cannot appear in th

No overall reaction is given. Because the third step is the rate-determining step, Rate=k

k3[Cl] [CO]=k4[ClCO] or [ClCO]=k3k4[Cl] [CO]. Substituting this expression into the first rate expression, we get Rate=k5 k3k4[Cl] [CO] [Cl2]. Because Cl is an intermediate, its concentration cannot appear in the rate law. Using the fast 5 3k4 k equilibrium in the first step, we see that k[Cl]=k[Cl]2 or [Cl]= k1k2[Cl k1k2[Cl2] [CO] [Cl2]=k5 k3k4 k1k2[CO] [Cl2]3/2. Simplifying this expression, we 2]. Substituting this expression into the last first rate expression, we get Rate=k 1

2

2

√

√

√

. [CO] [Cl]3/2 see that Rate=k 2 15.113 The steps in the mechanism are as follows: Br2(g) k1⇄k−1 2 Br(g)Br(g)+H2(g) →k2 HBr(g)+H(g)H(g)+Br2(g) →k3 HBr(g)+Br(g)

[Br2 ]

2 [H2] [Br]. Because Br is an intermediate, its concentration cannot appear in the rate law. Using the fast equilibrium in the first step, we see that k1= k − 1 [Br]2 or Because the second step is the rate-determining step, Rate=k [Br]=

1/2 k1k [H]Br B 2] .rSubstituting this into the first rate expression, we get Rate=k 2[ 2]1/2. The rate law is 3/2 order overall.

1k−1[ 2

k

√

√

−1

15.114 The sum of the steps in the mechanism is as follows: I2(g) k1⇌k−1 2I(g)I(g)+H2(g) k2⇌k−2 H2I(g)H2I(g)+I(g) →k3 2 HI(g)H2(g)+I2(g) → 2 HI (g)

3[H2I] [I]. Because H2I and I are intermediates, their concentrations cannot appear in the rate law.

Because the third step is the rate-determining step, Rate=k

I .

[I]=k−1 [I]2 or [I]= 1k−1[2] 1/2

k

Using the fast equilibrium in the first step, we see that1 k 2

√

Using the fast equilibrium in the second step, we see that2[H2] k [I]=k−2 [H2I] or [H2I]= k 2k − 2 [H2] [I]. k2k − 2 [H2] Substituting these into the first rate expression, we get Rate=k k2k−2 [H] [I] k1k−1 [I] 1/2. This expression contains I, so another substitution of the first equilibrium expression needs to be done to give Rate=k 3

√

3

2

[ 2]1/2k k [I2]1/2. Simplifying, we get Rate=k 3 k k k k [H2] [I2] or Rate= kobs[H2] 2[I].

k1 k − 1 I

√

2

√

1

−1

1 −1

2 −2

15.115 a. For a zero-order reaction, the rate is independent of the concentration. If the first half goes in the first 100 minutes, the second half will go in the second 100 minutes. This means that none, or 0%, will be left at 200 minutes. b. For a first-order reaction, the half-life is independent of concentration. This means that if half of the reactant decomposes in the first 100 minutes, half of this (or another 25% of the original amount) will decompose in the second 100 minutes.

This means that at 200 minutes, 50%+25%=75% has decomposed or 25% remains. c. For a second-order reaction, t1/2=1k[A]0=100 min and the integrated rate expression is 1[A]t=kt+1[A]0. We can rearrange the first expression to solve for k as k=1100 min [A]0. Substituting this and 200 minutes into the integrated rate expression, we get 1[A]t=200 min100 min [A]0+1[A]0→1[A]t=3[A]0→[A]t[A]0=13 or 33% remains.

15.116 Given: t1/2 for radioactive decay of Pu-239=24,000 years; 1 mole initially to 1 atom Find: t Conceptual Plan: Radioactive decay implies first-order kinetics, t 1>2 → k then [Pu−239]0, [Pu−239]t, k→tt1/2=

0.693k

Solution: t1/2=0.693 k Rearrange to solve for k. k=0.693t1/2 = 0.69324000 yr=2.8_875×10 −5 yr −1 then [Pu-239]t=1 and [Pu-239]0=6.022×10 −5

−1

23

ln[A]t=−kt+ln[A]0 23

Because ln[Pu-239]t=−kt+ln[Pu-239]0, rearrange to solve for t.

6 atom Pu-239=1.9×10yr

_ yr ln1 atom Pu2396.022×10 t=−1k ln[Pu-239][Pu-239] t 0 =− 12.8875×10 Check: The units (yr) are correct. The time to decay to 1 atom is consistent with the time being 79 half-lives, which makes sense because 279=6.09×1023.

15.117 Using the energy diagram shown and using Hess’s law, we can see that the activation energy for the decomposition is equal to the activation energy for the formation reaction plus the heat of formation of 2 moles of HI, or

(HI). So Ea formation=185 kJ+2 mol(−5.65 kJ/mol)=174 kJ.

Ea formation=Ea decomposition+2ΔH f °

Check: Because the reaction is endothermic, we expect the activation energy in the reverse direction to be less than that in the forward direction.

Extended Figure Description for Chapter 15 Problem 117 Figure 1 14 15.118 Given:first-order reaction,Ea =249 kJ/mol, A =1.6×10 s −1 , and 710 K Find: ,kfraction decomposed in 15 min and T for double the reaction rate Conceptual Plan: T, A, E a→k then min→ sthen k, t → fraction decomposed then k2/k1, T , E a→T2 1

then 15 min× 60 s1 min =90_0 s in ln

C2H5Cl] t [C2H5Cl] = e − kt = e − fraction remaining→[ 0

() k2k1

= 249

= Ae−Ea/RT

60 s1 min

( )=( )

ln k2k1

ln[A]t = − kt + ln[A]0

−5 −1 −5 (8.314) 710 K=7.6_657×10 s = 7.7 × 10 s − 1, − kt +ln][C2H5Cl] . Rearrange to solve for [C 2Ht = 5Cl 0

Solution: Because k =Ae−Ea/RT=1.6×1014 s − 1e

1−ln T2 = Ea R EaRT

k

kJmol × 1000 J 1 kJ

8.314 JK·mol

Ea R

1T1 − 1T2

JK · mol

−249 kJmol × 1000 J 1 kJ

(7.6_657×10 ) ( )=0.93_333; thus, fraction decomposed=1−0.93_333=0.06_66 ( )=(1 ). Rearrang −5 −1

s

(

90_0 s

EaR

1=710 K, k 2/k1 =2, and ln k2k1

T1 − 1T2

T2.

)1−.l8n6( 2K)=720 K

249 kJmol × 1000 J 1 kJ 8.314 JK·mol

=×77102 K_

Check: The units (s−1, none, and K) are correct. The reaction rate is reasonable considering the frequency factor, the activation energy,. The and T fraction decomposed is reasonable because 900 s is a small fraction of the half-life. The temperature is reasonable because many reactions double their rate with an increase in temperature of 10 K.

Challenge Problems 15.119 a. Because the rate-determining step involves the collision of two molecules, the expected reaction order would be second-order. b. The proposed mechanism is CH3NC*+CH3NC(fast)CH3NC*k3→ CH3CN(slow)CH3NC → CH3CNSo the sum matches the overall CH3NC+CH3NC k1⇌k2molecule.

CH3NC* is the activated Because the second step is the rate-determining step, R¯ate=k3[CH3NC*].

Because CH3NC* is anreaction. intermediate, its concentration cannot appear in the rate law. Using the fast equilibrium in the first step,

we see1 that k[CHNC]2 =k2[CH3NC*][CH3NC] or [CH3NC*]=k1k2[CH3NC]. Substituting this into the first rate expression, we get 3 observation of first order, and the mechanism is 15.120 =−k[A]1/2. Moving the A terms to the left and the t and constants to the right, we have d[A][A]1/2=−kdt. Integrating, we get ∫

a. valid. Rate=k[A]1/2 and Rate=−d[A]dt, so d[A]dt

[A] 0

d[A][A]1/2=−∫kdt.

When we evaluate 0 t

[A]

t − 2[A] 0 = − kt → 2[A] t = − kt + 2[A] 0, which is the desired integrated rate law. [ A ] 0 = − kt 0 → 2[A] 1/2 1/2 1/2 1/2 t

this integral, 2[A] 1/2[A]

b. To derive the half-life, set [A]t=1/2[A]0. Substituting this into 2[A] t = − kt + 2[A] 1/2

15.121 Rate=k [A]2 and Rate=−d [ A ] dt , so d [ A ] dt

0 1/2

, we get 2

(1/2[A] 0)1/2 = − kt1/2 + 2[A] 0 → t1/2 = 2 [ A ] 1/2

0−2 1/2

(1/2[A]0 )1/2

[ Ad] [ A ] 2 =−k[A]2. Moving the A terms to the left and the t and constants to the right, we have = − kdt. Integrating, we get∫

[A] 0 [A]

integral, −[A]

[A] 0

= − kt 0 → − [ A ]

−1[A]

t − (−[A] 0)

= − kt →[A]

−1

t

−1

t −1

= kt +[A]

0 or −1

→ t1/2 =

k

[A] (2−2√2)

d[ A ] [ A ] 2 = −

0 1/2

k

.

∫ 0kdt. When we evaluate this t

1 [ A ] t = kt + 1 [ A ] 0, which is the desired integrated rate law.

15.122

2O5 decomposes to ];NO2 and O2, first order in [N2O5

a. Given: N

k = 7.48 × 10

−3 −1

s

; P N2O =0.100 atm 5 °

Find: t to PTotal=0.145 Conceptual Plan: Write aatm balanced reaction. Assume that x is the amount of N2O5 that has reacted.

N2O5→2 NO2+ 12 O2 Then write the expression for P Totallet =x P N2O5 reactedP Total = P N2O5+ P NO2 + P

(

O 2

in terms of amount reacted then x,

) () ( )

Solution: PTotal = PN P O2 = 0.100 atm− x + 2x + 1/2x =0.100 atm+1.5 Set x 2O5 + PNO+ 2 generated; so the pressure increases by a factor of 1.5.

1.5 x=0.145 atm−0.100 atm =0.030 atm then P 0.100 atm−0.030 atm=0.070 atm. Because P =0.100 atm− x = N2O 5 t= −

ln [N2O5] t[NO25]0

k

= −ln

(

0.070 atm0.100 atm

)

7.48×10−3 s − 1

P N2O , k → tln[A] t = − kt + ln[A]0 5

°

P Total =0.145 atm=0.100 atm+1.5 x

and solve for x. Note that for every 1 mole of reactant decomposing, 2.5 moles of product are

2O5]. Rearrange to solve for t.

[25]t = −kt + ln[N 0

n/V or M, lnNO

∝ =47_.684 s=48 s

Check: The units (s) are correct. The time is reasonable because it is less than one half-life and the amount decomposing is less that 50%.

2O5 decomposes to NO2 and O2, first order [N2O5]; k=7.48×10−3 s−1; P

b. Given: N

N2 O5 =0.100 atm

°

Find: t to PTotal=0.200 atm Conceptual Plan: Write a balanced reaction.

12 O2 N25→2 O NO2+

Then write the expression for PTotal in terms of amount reacted thenx, P N2O5, k → t °

let x=P N2O5 reactedP

(

P O2 Total =P N2O5 +P NO+ 2

Solution: PTotal=PN2O5 +P NO2 +P O2 = 0.100 atm−x

ln[A]=− kt +ln[A] 0

) () ( )

for + 2x + 1/2x =0.100 atm+1.5 xP Total =0.200 atm=0.100 atm+1.5. Solve x

[N2O5]=−kt+lnNO [ 2 5]0 t

−x=0.100 atm−0.066_667 atm=0.033_333 atm. Because P 2O5 ] Nt [ 2O5] t = − ln[N

k

= − ln

(

0.3 0 3_33at 3 m 0.100 atm

n/V or M, ln

. Rearrange to solve for t.

∝

)7.48×10 −3

s −1

=14_6.875 s=150 s=1.5×102 s

Check: The units (s) are correct. The time is reasonable because it is between one and two half-lives and the amount decomposing is 67%.

x. x = 0.200 atm−0.100 atm

1.5

=0.066_667 atm then P N2O =0.100 atm 5

2O5 decomposes to ];NO2 and O2, first order in [N2O5 k = 7.48 × 10 − 3 s − 1; P N2O =0.100 atm 5

c. Given: N

°

Find: P Totalafter 100 s Conceptual Plan:

P N O , k, t → P N2O then write a balanced reaction. Then P N2O , P N2O5 →x ln[A]=t 5 5 2 5 °

°

− kt + ln[A]0

N2O5→2 NO2+ 12 O2

x = P N2O5 − P N2O5 °

°

then

write the expression for P Total in terms of amount reacted P Total = P P + P O2 let x = NO N2O5+ NO2 5 P 2 reacted Solution: Because P n/V or M,

(7.48×10−3 ) )( ) ( s−1

∝ = − kt + ln[N O ] = − ln[N 2O5] t 0 P

= e −3.05_059

25

100 s +ln0.100 atm

=−3.05_059

=0.0473_310 atm so x N2 O5 = P N2O −PN2O5 =0.100 atm−0.0473_310 atm=0.052_669 atm 5

°

finally, P Total = P N2O5+ P NO2 + P O2 = 0.100 atm+1.5

(0.100)+atm−x )(2x + )=0.100 (1/2x atm+1.5x=

(0.052_669 atm)=0.179 atm

Check: The units (atm) are correct. The pressure is reasonable because the time is between those for parts (a) and (b).

2O3 decomposes to NO2 and NO, first order in [N2O3], table of [NO2] versus time, at 50,000 s=[N2O3]=0 Conceptual Plan: Write a balanced reaction. the expression for [NO 2] inThen termswrite [N2O3], then 15.123 Given: N

N2O3 → NO2 + NO

Find: k

[N2O3] = 0.784M − [NO2]

2O3] versus time. Because ln[A]t=−kt+ln[A]0, the negative of the slope will be the rate constant. plot ln[N Solution: [N2O3]=0.784 M−[NO2]

ΔyΔx / will on the be plot orthe by using functions such as “add trendline” Excel. Thecan last point s) Plot 2O3] ln[N versus time. Because ln[A]t=−kt+ln[A]0, the negative of the slope rate constant.

The inslope be(50,000 determin

[N 2O 3 ].

− 1s cannot be plotted because the concentration is 0 and the ln0 is undefined. Thus, the rate constant is 3.20×10−4 s−1, and the rate law is Rate=3.20×10−4

Check: The units

(s−1) are correct. The rate constant is typical for a reaction.

Extended Figure Description for Chapter 15 Problem 123 Figure 1 15.124 For the elementary reaction 2 NOCl() g k1 k −1 2 NO()+Cl(), g g see that 2we 2 [NO]=2[Cl]. 2Substituting this into the first expression, ⇌ we getk1[NOCl]=k 3 3

[Cl2]=

√

k14 k− 1 [NOCl]2/3 =

7.8 × 10−2L2mol2· s 4

√

=0.0084 M Cl 2 and [Cl2]=2[NO]=2

k1[NOCl]2= k−1[NO]2[Cl2]. For each mole of NOCl that reacts, 1 mole of NO and 0.5 mole of Cl2 are generated. Because before any reaction only NOCl is present,

( [ ])2 [Cl]→k[NOCl]2 =4k−1[Cl 32]. Rearranging and substituting the specific values into this expres 2 1

−1 2Cl 2

) (4.7×10 )(0.12 2 L2mol·s

molL

2/3

=0.0084_2235 molL Cl2

(0.0084_ ) 2235 molL

= 0.016_ 8445 molL NO=0.017 M NO

Conceptual Problems b has the b=k(4)(2)1/2=5.7 fastest rate.

15.125 Because 3Rate=k [CHCl][Cl]1/2,Rate a=k(3)(3)1/2=5.2=4k. k,SoRate k, and Rate c=k(2) (4)1/2 2 15.126 Since reaction A is first order (half-life is independent of initial concentration); reaction B is zero order (half-life increases as concentration increases); and reaction C is second order (half-life increases as concentration decreases), the most consistent statement is statement b. Two of the three reaction orders are correct. No statement is totally consistent with the observations.

(1[A]t). Reaction B is

15.127 Reactant concentrations drop more quickly for first-order reactions than for second-order reactions, so reaction A must be second-order. A plot of 1/[A] versus time will be linear = kt + 1 [ A ] 0

versus time willtbe = linear − kt +(ln[A] ln[A]0).

15.128 A reaction that slows down as the reaction proceeds and has a half-life that is dependent on the concentration is a second-order reaction. Statement (a) is false because it describes a first-order reaction. Statement (b) is true because it describes a second-order reaction. Statements (c) and (d) are false because they describe a zero-order reaction.

Questions for Group Work [A]1, so 15.129 It is very tempting to think that the same descriptor that relates the concentration change to the rate change is the order of the reaction. This statement is incorrect because the reaction order is first order. For a first-order reaction, Rate=k

(22 = 4).

doubling the concentration of A doubles the reaction rate. For a second-order reaction, Rate=k[A]2, so doubling the concentration of A quadruples the reaction rate 15.130 For [A] versus t:

Extended Figure Description for Chapter 15 Problem 130 Figure 1 For ln[A] versust:

Extended Figure Description for Chapter 15 Problem 130 Figure 2 For 1/[A] versus t:

Extended Figure Description for Chapter 15 Problem 130 Figure 3 Since the plot of ln[A] versus t is the only linear plot, the reaction is first order. −1 −1 −1 0.5728 s ,sfor 25.0 °C, 35.0 °C, and 45.0 °C, respectively. kt + ln[A]0of , 0.3959 s , and ). The rate constant constants are 0.2711 a. Using Excel, the slope is the negative the rate (ln[A]t=−

b. Since ln k = − Ea R

( )+ln A a plot of ln k 1T

versus 1/T will have aslope = − Ea/Rand an intercept=ln A.

Extended Figure Description for Chapter 15 Problem 130 Figure 1b Since the slope = −3547.1 K = − Ea/ Rthen

a=−(−3547.1 K)(8.314 J)=29.5 JK·mol kJmol )(1 kJ1000

E

c. A catalyst will speed up a reaction by reducing the activation energy. The reaction is still first order, as seen by the linear plots of ln[A] versus t.

0. The rate 3.246 constant s−1 , 4.215 s −1 , and 5.269 s − 1, for 25.0 °C, 35.0 °C, and 45.0 °C, respectively.

Using Excel, we see that the slope is the t=−kt+ln[A]) negative ofconstants the are rate (ln[A]

Extended Figure Description for Chapter 15 Problem 130c Figure 1 Since ln k = − Ea R

( )+ln A a plot of ln k 1T

versus 1/T will have aslope = − Ea/Rand anintercept =ln A.

Extended Figure Description for Chapter 15 Problem 130c Figure 2 Since the slope =−2298.6 K=− Ea/ Rthen

a=−(−2298.6 K)(8.314 J) JK·mol )(1 kJ1000

E

=19.1 kJmol The activation energy has been cut by about a third by adding a catalyst.

Data Interpretation and Analysis 15.131 a. Since the reaction rate doubles when the concentration of each of the reactants is doubled, the reaction is first order in both methane and ozone. b. Since ln k = − Ea R

(1T)+ln A a plot of ln k versus 1/ T will have aslope = − Ea/Rand anintercept =ln A.

a/R then

Since the slope =−7695.8 K=− E Ea

=

−(−769_5.8 K)

(8.314 )( JK ) · mol 1 kJ1000 J

Since the intercept=18.9_02=ln A, then A=eintercept=e18.9_02=1.61_821×108 s

−1

=

63.98_ 288 kJmol = 63.98 kJmol.

=1.62×108 s − 1.

Extended Figure Description for Chapter 15 Problem 131 Figure 1b c.

Since ln k=−EaR

(1T)+ln (8.314JK·mol A ) (1 kJ1000 (J)1 273 K)+18.9_02 −63.98_288 kJmol

=

−5

then k=e−9.28774=9.25_517×10

−1

=

−28.18_ 974 + 18.9_ 02 = − 9.2_ 8774

−1

M ·s . Convert reactant concentrations from ppm to molar: 6 L air × 1 mol CH22.4 L CH4=8.0_357×10−8 M CH 4 4 [CH4]=1.8 ppm CH4= 1.8 L CH10 4 and 6 [O]=5.0 ppm O L O 3 =2.2¯321×10−7 310 L air × 1 mol O22.4 =5.0 LO 3 M O3.

3 3 Rate =

k[CH4] [O3]=

(9.25_ 517 )( × 10 )( −5

M

−1

−1

·s

8. 0_ 357 × 10

−8

M CH4

2. 2_ 321 × 10

−7

M O3

)

=

1.6_ 6008 × 10

−18

M/s=1.7×10

−18

M/s.

d.

Since ln k = − Ea R −4.9_2401

k=e

(1T)+ln (8.314JK·mol A ) (1 kJ1000 (J)1323 K)+18.9_02 =

= 7.26_ 995 × 10

t1/2= 1 k [ A ] o =

1(

−3

M

_ −63.98288

−1

7.26995×10−3 M−1·s _

·s

kJmol

=

−23.82_601+18.9_02=

−4.9_2401 then

−1

−1

. The reaction is second order overall,

)(

−7

5.0 × 10

M

)

=

−2.7_5105×108 s×1 min60 s × 1 h60 min × 1 day24 h × 1 year365 days =8.7 years.

16 Chemical Equilibrium Review Questions 16.1 Like adult hemoglobin, fetal hemoglobin is in equilibrium with oxygen. However, the equilibrium constant for fetal hemoglobin is larger than the equilibrium constant for adult hemoglobin, meaning that the reaction tends to go farther in the direction of the product. Consequently, fetal hemoglobin loads oxygen at a lower oxygen concentration than does adult hemoglobin. In the placenta, fetal blood flows in close proximity to maternal blood, without the two ever mixing. Because of the different equilibrium constants, the maternal hemoglobin unloads oxygen that the fetal hemoglobin then binds and carries into its own circulatory system. Nature has thus evolved a chemical system through which the mother’s hemoglobin can in effect hand off oxygen to the hemoglobin of the fetus.

16.2 Dynamic equilibrium in a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction. Dynamic equilibrium is called dynamic because the forward and reverse reactions are still occurring; however, they are occurring at the same rate.

16.3 The general expression of the equilibrium constant is K=[C]c[D]d[A]a[B]b.

16.4 The equilibrium constant for a reaction is defined as the ratio—at equilibrium—of the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients. If the equilibrium constant for a reaction is large, the equilibrium point of the reaction lies far to the right—the concentration of products is large, and the concentration of reactants is small. If the equilibrium constant for a reaction is small, the equilibrium point of the reaction lies far to the left—the concentration of products is small, and the concentration of reactants is large. reverse = 1Kforward 16.5 If you reverse the equation, invert the equilibrium constant.

K If you multiply the coefficients in the equation by a factor, raise the equilibrium constant to the same factor. That is, if you multiply the reaction by n, raise the equilibrium constant K to the n. K′=Kn 16.6 If you add two or more individual chemical equations to obtain an overall equation, multiply the corresponding equilibrium constants by each other to obtain the overall equilibrium constant. For two reactions with equilibrium 1constants of K and K2, the equilibrium constant for the combined reaction K3=K1K2.

16.7 Kcis the equilibrium constant with respect to concentration in molarity.

Kpis the equilibrium constant with respect to partial pressures in atmospheres. The two can be related as follows: Kp=KcΔn (RT) .

16.8 For the equilibrium constant, concentrations are expressed in terms of molarity, and partial pressures are expressed in terms of atmospheres. The equilibrium constant usually is written with no units. Formally, the values of concentration or partial pressures that we substitute into the equilibrium constant expressions are ratios of the concentration or pressure to a reference concentration or reference pressure. So as long as concentration units are expressed in molarity or atmospheres, we simply enter the quantities directly into the equilibrium expression, dropping their corresponding units.

16.9 The concentration of a solid does not change because a solid does not expand to fill its container. Its concentration, therefore, depends only on its density, which is constant as long as some solid is present. Consequently, solids are not included in the equilibrium expression. Similarly, the concentration of a pure liquid does not change, so liquids are also excluded from the equilibrium expression.

16.10 For any reaction, the equilibrium concentrations of the reactants and products will depend on the initial concentration. However, the equilibrium constant will always be the same at a given temperature, regardless of the initial concentrations.

16.11 When we know the initial concentrations of the reactants and products and the equilibrium concentration of one reactant or product, the other equilibrium concentrations can be deduced from the stoichiometry of the reaction. From the initial and equilibrium concentration of the one reactant, we can find the change in concentration for that reactant. Using the stoichiometry of the reaction, we can determine the equilibrium concentrations of the other reactants and products. We generally use an ICE table (I=Initial C=Change E=Equilibrium) to keep track of the changes.

16.12 The definition of the reaction quotient takes the same form as the definition of the equilibrium constant, except thereac rea c) is the ratio—at any pointthat in the concentrations of the products raised to their stoichiometric coefficients divided by the concentration of the reactants raised to their stoichiometric coefficients. The value of Q relative to K is a measure of the progress of the reaction to-ward equilibrium.

At equilibrium, the reaction quotient is equal to the equilibrium constant.

16.13 Standard states are defined as 1 M concentration, or 1 atm pressure. So the value cofand K Kp when the reactants and products are in their standard states is 1.

16.14

a. When Q<K, the reaction goes to the right (toward products). b. When Q<K, the reaction goes to the left (toward reactants). c. When Q=K, the reaction is at equilibrium. c, predict 16.15 To solve a problem given initial concentrations and the equilibrium constant, you would prepare an ICE table, calculate Q, compare Q and K the direction of the reaction, represent the change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine the reactant and product concentration.

16.16 When adding or subtracting a small value of x, we can often ignore x because of significant figures. In addition and subtraction, the number of significant figures depends on the number with the fewest decimal places. Therefore, a small value of x would not be significant. In multiplication and division, the number of significant figures depends on the number with the fewest significant figures; therefore, we cannot ignore x in this case.

16.17 According to Le Châtelier’s principle, when a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance.

16.18 If a system is at equilibrium: increasing the concentration of one or more of the reactants causes the reaction to shift to the right. increasing the concentration of one or more of the products causes the reaction to shift to the left. decreasing the concentration of one or more of the reactants causes the reaction to shift to the left. decreasing the concentration of one or more of the products causes the reaction to shift to the right.

16.19 If a chemical system is at equilibrium: decreasing the volume causes the reaction to shift in the direction that has the fewer moles of gas particles. increasing the volume causes the reaction to shift in the direction that has the greater number of moles of gas particles. if a reaction has an equal number of moles of gas on both sides of the chemical equation, then a change in volume produces no effect on the equilibrium. adding an inert gas to the mixture at a fixed volume has no effect on the equilibrium.

16.20 If the temperature of a system at equilibrium is changed, the system shifts in a direction to counter that change. So if the temperature is increased, the reaction shifts in the direction that tends to decrease the temperature and vice versa. In an exothermic chemical reaction: increasing the temperature causes the reaction to shift left and the value of the equilibrium constant decreases. decreasing the temperature causes the reaction to shift right and the value of the equilibrium constant increases.

In an endothermic chemical reaction: increasing the temperature causes the reaction to shift right and the equilibrium constant increases. decreasing the temperature causes the reaction to shift left and the equilibrium constant decreases.

Problems by Topic Equilibrium and the Equilibrium Constant 16.21 The equilibrium constant is defined as the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

a. K=[SbCl3][Cl2] [SbCl5] [BrNO]2 b. K=[NO]2[Br] c. K=[CS] [H]4[CH] [HS]2 d. K=[CO]2[CO]2 [O2] 2

2

2

4

2

2

16.22 a. The equilibrium constant is defined as the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

K=[H]2[S] [HS]2 2

2

2

b. The equilibrium constant is defined as the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

K = [COCl2 ] [CO ] [Cl2 ] 16.23 With an equilibrium constant of 1.4×10−5, the value of the equilibrium constant is small; therefore, the concentration of reactants will be greater than the concentration of products. This is independent of the initial concentration of the reactants and products.

16.24 Figure a at equilibrium has 8 2H4Cl2, C 2 Cl2, and 2 C2H4.

Figure b at equilibrium has 6 2H4I2, 2CH4Br2,7 4I2,Br2, 4 C2H4. andand 7 C2H4.

Figure c at equilibrium has 3 C >K Br2 >KKI2.

Because the equilibrium constant is concentration of products/concentration of reactants, the equilibrium situation that has the largest concentration of products will have the largest equilibrium constant. clTherefore, 2 16.25 i. has 10 H2 and 10 I2 ii. has 7 H2 and 7 I2 and 6 HI iii. has 5 H2 and 5 I2 and 10 HI iv. has 4 H2 and 4 I2 and 12 HI v. has 3 H2 and 3 I2 and 14 HI vi. has 3 H2 and 3 I2 and 14 HI a. Concentrations of (v) and (vi) are the same, so the system reached equilibrium at (v). b. If a catalyst was added to the system, the system would reach the conditions at (v) sooner because a catalyst speeds up the reaction but does not change the equilibrium conditions. c. The final figure (vi) would have the same amount of reactants and products because a catalyst speeds up the reaction but does not change the equilibrium concentrations.

16.26 The equilibrium constant gives us the ratio of products to reactants at equilibrium; it does not say how long it takes to reach equilibrium. So after 15 minutes, if the smaller equilibrium constant has more products, the kinetics of that reaction are faster.

16.27 For the reaction A (g)⇌2 B (g) K=[B]2[A]. At equilibrium [A]=0.10 M and [B]=0.20 M, so K=(0.20 M)2(0.10 M)=0.40.

16.28 For the reaction 2 A (g)⇌B (g) K=[B][A]2. At equilibrium [A]=0.25 M and [B]=0.20 M, so K=(0.20 M)(0.25 M)2=3.2.

16.29 a. If you reverse the reaction, invert the equilibrium constant. So K′=1K p= 12.26×104 =4.42×10 −5 .

The reactants will be favored.

41/2 2 =(2.26×10) =1.50×10. The products will be favored.

(K)1/2 p

K′ So = b. If you multiply the coefficients in the equation by a factor, raise the equilibrium constant to the same factor. c. Begin with the reverse of the reaction and invert the equilibrium constant.

Kr e v e r s e1K =p = 12.26×104= 4 . 4 2 × 1−50 to the second power.

Then multiply the reaction by 2 and raise the value ofreverse K

K′ =reverse(K )2=(4.42×10 −5)2 =1.95×10 −9. The reactants will be favored.

16.30 a. The reaction is multiplied by 1/2, so raise the value of the equilibrium constant to 1/2.

(

)

1 0 3products will be favored.

K ′ = K 1 / 2 = 2 . 2 × 1 0 6 1 / 2 = 1 . 5 ×. The

( p)

b. The reaction is multiplied by 3, so raise the value of the equilibrium constant to 3.

( ) (

)

0 1 9 products will be favored.

K ′ = Kp 3 = 2 . 2 × 1 0 6 3 = 1 . 1 × 1. The c. Begin with the reverse of the reaction and invert the equilibrium constant. p = 12.2×106 Kreverse = 1 K = 4.5 × 10− 7 to the second power.

Then multiply the reaction by 2 and raise the value ofreverse K K′=

. The reactants will be favored.

(K reverse )2=(4.5×10−7)2=2.0×10−13

16.31 To find the equilibrium constant for reaction 3, you need to combine reactions 1 and 2 to get reaction 3. Begin by reversing reaction 2; then multiply reaction 1 by 2 and add the two new reactions. When you add reactions, you multiply the values of K. 2 −31 −29 2 NO (g) constant K1=1K=12.1×1030=4.76×10−31 p reaction 3, you need to¯ combine reactions 16.32 find the⇌equilibrium for g 1 gand 2 to get reaction g 3.KBeginK2 by multiplying reaction 1gb¯y 2; then g reverse g reaction 2 andgadd the K twoKK new 12=(4.76×10 reactions. When you¯ )(28.09)=1.3×10 add reactions, y¯ou multiply the values + O2(g) N2(g)To of K. 2 A(s)

.1×10 −3 B()+2 g C() g

K=(K)2=(0.0334)2=1 1 p

B(g)+2 C(g)

K2= 1Kp= 12.35 =0.4262 A()s

3 D()g

3 D()g

⇌

⇌

Kp , Kc , and Heterogeneous Equilibria

⇌

16.33 5 6.2×10 T=298 K Find: Kc KRTΔn

a. Given: Kp =

Conceptual Plan: Kp→KKcp= c( ) Solution: Δn=mol product gas−mol reactant gas=2−4=−2 Kc = Kp (RT)

∆n

]

[

−2

= 6.2×105 0.08206 L·atmmol·K×298 K = 3.7 × 108

Kpof .

Check: Substitute into the equation and confirm that you get the original value

(0.08206 )

Kp = Kc (RT)Δ n = (3.7 × 108)

−2

L · atmmol · K ×298

= 6.2 × 105

b. Given: Kp = 4.10×10 −31 T=298 K Find: Kc Conceptual Plan: Kp→K c Kp= KRTΔn c( ) Solution: Δn=mol product gas−mol reactant gas=2−2=0 Kc = Kp (RT)

Δn

−31

= 4.10 × 10

( 0.08206 = )4.10 × 10 0

L · atmmol · K ×298 K

−31

of Kp. Check: Substitute into the equation and confirm that you get the original value Kp = KcRT ( )Δn=(4.10×10−31)

(0.08206L · atmmol · K × 298)

c. Given: K =0.14 T=298 K Find: K p

0

= 4.10 × 10

−31

c

Δn Conceptual Plan: Kp → KcKp = Kc(RT) Solution: Δn=mol product gas−mol reactant gas=2−1=1

Kc = Kp (RT)

Δn

= 0.14

)

(

1

0.08206 L · atmmol · K ×298 = K 5.7 × 10 − 3

Kpof .

Check: Substitute into the equation and confirm that you get the original value Kp = Kc (RT)Δ n = (5.7 × 10 − 3)

(0.08206 ) L · atmmol · K ×298

1

= 0.14

16.34 a. Given: Kc=81.9 T=298 K Find: Kp Conceptual Plan: Kp→KKcp= KRTΔn c( ) Solution: Δn=mol product gas−mol reactant gas=2−2=0

= 81.9

(0.08206 )

= 81.9

Kp = Kc (RT)Δ n = 1.3 × 1022

(0.08206 )

= 7.8 × 1024

Δn

Kp = Kc (RT)

L · atmmol · K × 298K

0

Kpof .

Check: Substitute into the equation and confirm that you get the original value Δn

(

)0

Kc = Kp( RT ) =81.9 0.08206 L · atmmol · K ×298 =K81.9 K b. Given: c=1.3×1022 T=298 K Find: Kp Conceptual Plan: Kp→K cKp=Kc(RT)Δ n Solution: Δn=mol product gas−mol reactant gas=4−2=2 L · atmmol · K × 298K

2

Kpof .

Check: Substitute into the equation and confirm that you get the original value Kc = Kp (RT)

Δn

( 0.08206 = )1.3 × 10 2

= 7.8 × 24

22

L · atmmol · K ×298 K

c. Given: Kc = 2.56×10 −23 T=298 K Find: Kp Conceptual Plan: Kp→K cKp= KRTΔn c( ) Solution: Δn=mol product gas−mol reactant gas=2−1=1 Δn

Kp = Kc (RT)

= 2.56 × 10

−23

(0.08206 ) L · atmmol · K × 298K

1

= 6.26 × 10

Kpof .

Check: Substitute into the equation and confirm that you get the original value Kc = Kp (RT)

Δn

−22

= 6.26 × 10

( 0.08206 = )2.56 × 10 L · atmmol · K ×298 K

16.35 a. Because HO K= 2 is a liquid, it is omitted from the equilibrium expression.eq

[HCO3 ] [HO−]

2− [CO 3 ]

−

[ 2]3

Keq = O b. Because KClO3 and KCL are both solids, they are omitted from the equilibrium expression.

[ ] F−[ ] [ ] d. Because HO K= [NH ] OH− 2 is a liquid, it is omitted from the equilibrium expression. eq HO c. Because HO 2 is a liquid, it is omitted from the equilibrium expression. Keq=

+

[HF]

[NH3 ] 16.36 Because PCl3 is a liquid, it is omitted from the equilibrium expression. Keq=[Cl2 ] [PCl5] 4

+

Relating the Equilibrium Constant to Equilibrium Concentrations and Equilibrium Partial Pressures Kc Find:[H M, [CH3OH]=0.185 M 16.37 Given: at equilibrium: [CO]=0.125 M, 2 ]=0.124 Conceptual Plan: Balanced reaction → equilibrium expression→

Kc

1

−23

−22

K' = KK 12=(1.1×10

−3

)(0.426)=4.69×10

−4

Solution:K c = [CHOH][CO][H2]2 3 = (0.18 5)(0.125)(0.124)2 =96.3 Check: The answer is reasonable because the concentration of products is greater than the concentration of reactants and the equilibrium constant should be greater than 1.

16.38 Given: at equilibrium: NH =0.228 M, HS =0.335 M Find:Kc

[ 3]

[2]

Conceptual Plan: Balanced reaction → equilibrium expression→K c Solution: Because NHHS is a solid, it is omitted from the equilibrium expression.

4

= 0.0764 ][ )(0.335) c = [NH3 H2S] = (0.228

K

Check: The answer is reasonable because the concentration of products is less than 1 M and the equilibrium constant is less than 1.

16.39 At 500 K : Given: 2]=0.115 at equilibrium: M, [H2]=0.105 [N M, and [NH3]=0.439 M Find: Kc Conceptual Plan: Balanced reaction → equilibrium expression →K

c

3]2[N]2 [H2] = (0.439)2(0.115)(0.105)3 =1.45×10 3 Solution: Kc= [NH Check: The value is reasonable because the concentration of products is greater than the concentration of reactants.

3

2]=0.110 M, [NH3]=0.128 M, Kc=9.6 Find: [H2] Conceptual Plan: Balanced reaction2]→ equilibrium expression → [H At 575 K : Given: at equilibrium: [N

9.6=(0.128)2(0.110)(x)3 3 2 2 x=0.249 Check: Plug theKc=[NH]2[N] value for x back into the equilibrium expression and check the value.

Solution: [H]3 ( 0.110 ) ( 0.249 )3 9.6 = (0.128)2

=0.140[H2] M, Kc =0.0584 Find: [NH 2]=0.120 M, 3] Conceptual Plan: Balanced reaction → 3] equilibrium expression→[NH At 775 K : Given: at equilibrium: [N

[NH3]2[N2] [H2]3 x = 0.00439 0.0584= (x)2(0.120)(0.140)3 back into the equilibrium expression and check the value.

Check: Plug the value for x Solution: Kc=

2

3

0.0584 = ( 0.00439 ) ( 0.120 ) ( 0.140 )

16.40 At 25 °C : Given: 2]=0.0355 at equilibrium: M, [I2]=0.0388 [H M, and [HI]=0.922 M Find: Kc Conceptual Plan: Balanced reaction → equilibrium expression →Kc

2][I2]=(0.922) (0.0355)(0.0388)=617 Solution: Kc=[HI]2[H Check: The value is reasonable because the concentration of products is greater than the concentration of reactants.

2

[I2]=0.0455, [HI]=0.387, Kc=9.6 Find: [H2] Conceptual Plan: Balanced reaction2]→ equilibrium expression → [H At 340 °C : Given: at equilibrium:

2 I2[H ] [ ]9.6 = (0.387) (x) (0.0455) x = 0.343 Solution: Kc = [HI]2 Check: Plug the value for x back into the equilibrium expression and check the value.

2

( 0.387 )2 ( 0.343 ) ( 0.0455 )

9.6= Kc =0.0468, At 445 °C : Given: 2]=0.0485 at equilibrium: M, [I2] [H=50.2 Find: [HI] Conceptual Plan: Balanced reaction → equilibrium expression→[HI] (0.0468) ] [I2(x)2(0.0485) ] Solution: Kc = [HI]2 [H2 50.2= x = 0.338 Check: Plug the value for xback into the equilibrium expression and check the value.

2

50.2 = ( 0.338 ) ( 0.0485 ) ( 0.0468 ) 16.41 Given: P NO =125 torr; P Br =148 torr, Kp =28.4 Find: PNOBr 2

Conceptual Plan:torr → atm1 atm760 torr and then balanced reaction → equilibrium expression→P NOBr P Br = 148torr × 1 atm760torr =0.194_74 atm Solution: PNO= 125torr × 1 atm760torr =0.164_47 atm 2

Kp = P NOBrP NOP Br2 2

2

284.

=

(

)2

28.4 =

(

)2( )

2

0.164_47 x

(0.194_74)

x=0.386_788 atm=294 torr

Check: Plug the value for xback into the equilibrium expression and check the value.

0.386_788

0.164_47 2 (0.194_74)

16.42 Given: PSO2 =117 torr; P Cl2 =205 torr, Kp =2.91×10 3 Find: PSO2Cl2 Conceptual Plan: torr→atm

1 atm760 torr and then balanced reaction → equilibrium expression→P

Solution: PSO2 =117 torr× 1 atm760 torr =0.153_947 atm

SO2Cl2

P Cl2 = 205torr × 1 atm760torr =0.269_737 atm

Kp = P SO2P Cl2 P SO2Cl2

21.9×0=13

(0.153_947 )( ) 0.269_737

−5

( x)

x = 1.42_ 698 × 10 atm=0.0108 torr

Check: Plug the value for xback into the equilibrium expression and check the value.

2.91 × 103 =

(

0.153_947

) ) ( ) (1.42_698×10 −5

0.269_737

16.43 Given: PA, initial=1.12 atm; PA, eq=0.37 atm Find: Kp Conceptual Plan: 1. Prepare ICE table.

2. Calculate pressure change for the known value.

3. Calculate pressure changes for other reactants/products.

4. Determine equilibrium pressures.

5. Write the equilibrium expression and determine Kp. Solution: A(g) 2 B(g) ⇌

Initial

Change

Equil

PA

PB

1.12 atm

0.00

−x

+2x

0.37 atm

2x

( 1.5 ) ( 0.37 ) = 6. 0_ 81 = 6.1 PBKp= P BPA = atm)=1.5 x=1.12−0.37=0.75 atm2x=2(0.75 atm= 2

2

Check: The units (none) are correct. The magnitude is reasonable (6) since there are significant concentrations of both reactants and products.

16.44 Given:P A, initial =255 mmHg; PA, eq =55 mmHg Find: Kp 1 atm760 mmHg Conceptual Plan:mmHg → atm 1. Prepare ICE table.

2. Calculate pressure change for the known value.

3. Calculate pressure changes for other reactants/products.

4. Determine equilibrium pressures.

5. Write the equilibrium expression and determineKp .

Solution: 255 mmHg×1 atm760mmHg =0.335_526 atm; 55mmHg×

1 atm760mmHg =0.072_368 atm

2 A(g)

Initial

PA

PB

PC

0.335_526 atm

0.00

0.00

−2x

+x

+2x

0.072_368 atm

x

2x

Change

Equil

2x

=

0.335_526 atm−0.072_368 atm=0.263_158 atm=

P C;

B(g)+2 C(g) ⇌

x =0.131_579 atm= PBKp =

P PC P A B 2

2

= (0.131_579)

(0.263_158 ) (0.072_368 ) 2

Check: The units (none) are correct. The magnitude is reasonable (1.7) since there are higher concentrations of the products than the reactants.

16.45 Given: [Fe3+ ] initial =1.0×10 −3 M; [SCN]− initial =8.0×10 −4 M; [FeSCN2+] eq =1.7×10−4 M Find: Kc Conceptual Plan: 1. Prepare ICE table.

2. Calculate concentration change for the known value.

3. Calculate concentration changes for other reactants/products.

4. Determine equilibrium concentration.

5. Write the equilibrium expression and determine K c.

Solution: Fe3+(aq)+SCN−(aq) FeSCN3+(aq) ⇌[Fe3+]

[FeSCN3 + ]

[SCN−]

−3

8.0 × 10

−4

−1.7 × 10

−4

6.3 × 10

Initial

1.0 × 10

Change

−1.7 × 10

Equil

8.3 × 10

−4

0.00

−4

+1.7 × 10

−4

1.7 × 10

−4

−4

3.3 × 102 c = [FeSCN2+ ] [Fe3+)(]8.3×10−4 [SCN− )(6.3 × 10] =) =(1.7×10−4 −4

K

Check: The units (none) are correct. The magnitude is reasonable (330) since there are significant concentrations of both reactants and products.

2 eq =1.2×10 −2 M Find: K 16.46 Given: [SOCl] 2 2initial =0.020 M; [Cl] c Conceptual Plan: 1. Prepare ICE table.

2. Calculate concentration change for the known value.

3. Calculate concentration changes for other reactants/products.

4. Determine equilibrium concentration.

5. Write the equilibrium expression and determine Kc. Solution: SO2Cl2(g) 2g Cl SO()+ 2(g) ⇌

Initial

Change

Equil

[SO2Cl2]

[SO2]

[Cl2]

0.020

0.00

0.00

−2

−2

+1.2 × 10

−2

1.2 × 10

−1.2 × 10

+1.2 × 10

0.0080

1.2 × 10

−2

2 Kc=[SO][Cl]

2

−2

[SOCl] 2 2

=

1 (.2×10−2 ) (1.2×10 ) −2

( 0.0080 )

= 0.018

Check: The units (none) are correct. The magnitude is reasonable since there are significant concentrations of both reactants and products.

16.47 Given: 3.67 L flask, 0.763 g H2 initial, 96.9 g I 2initial, 90.4 g HI equilibrium Find: Kc Conceptual Plan: g → mol → Mn= gmolar mass M = nVand then 1. Prepare ICE table.

2. Calculate concentration change for the known value.

3. Calculate concentration changes for other reactants/products.

4. Determine equilibrium concentration.

5. Write the equilibrium expression and determineK . c 1 mol H2.016gH 0.3785 mol H2 3.67 L Solution: 0.763 gH× =0.103 M This is an initial concentration.

2 2 =0.378_5 mol H then 2 2 _ mol I 2 I 2=0.381_8 mol I 2 then0.3818 3.67 L 2 =0.104 M This is an initial concentration.

96.9g I2 × 1 mol I253.80g _ 3.67 L =0.193 M This is an equilibrium concentration.

90.4g HI× 1 mol HI127.91g HI =0.706_8 mol HI then 0.7068 mol HI H2(g) + I2(g)

Initial

Change

Equil

[H2]

[I2]

[HI]

0.103

0.104

0.00

− 0.0965

− 0.0965

+0.193

0.0065

0.0075

0.193

2HI(g)

⇌

Because HI gained 0.193 M, H2 and I2 had to lose 0.193/2=0.0965 M from the stoichiometry of the balanced reaction.

Kc = [HI]2[H2 ] [I2 ] = (0.193)2(0.0065) (0.0075) = 764 Check: The units (none) are correct. The magnitude is reasonable (764) since there are higher concentrations of the products than the reactants.

16.48 Given: 5.19 L flask, 26.9 g CO initial, 2.34 g H2 initial, 8.65 g CHOH equilibrium Find: Kc 3 Conceptual Plan:g → mol → Mn = gmolar mass M= nVand then 1. Prepare ICE table.

2. Calculate concentration change for the known value.

3. Calculate concentration changes for other reactants/products.

4. Determine equilibrium concentration.

5. Write the equilibrium expression and determine Kc. _ mol CO 0.9604 Solution: 26.9 gCO×1 mol CO28.01g CO=0.960_4 mol CO then

5.19 L

=0.185 M

2.34g H× 2

1 mol H2.016g H 2=1.16_1 mol H 2

2 then1.161_ mol H 2

5.19 L

CH3OH =0.270_0 mol CH 3OH then0.270_0 mol CH3OH 3 8.65 g CH3OH ×1 mol CHOH32.04g

=0.224 M 5.19 L

=0.0520 M

2

= 1. 7_ 399 = 1.7

CO(g) + 2H2(g)

Initial

Change

Equil

[CO]

[H2]

[CH3OH]

0.185

0.224

0.00

− 0.0520

− 0.104

+ 0.0520

0.133

0.120

0.0520

CH3OH(g) ⇌

Because CH3OH gained 0.0520 M, CO and H2 had to lose 0.0520 M and 0.104 M, respectively, owing to the stoichiometry of the balanced reaction.

133)(0.120)2 3 2 Kc=[CHOH][CO][H]2=

(0.0520)(1.

=27.2

Check: The units (none) are correct. The magnitude is reasonable (27) since there are higher concentrations of the products than the reactants.

The Reaction Quotient and Reaction Direction 16.49 Given: K=8.5×10−3

; [NH3]=0.0822 M; [H2S]=0.0822 M Find: whether solid will form or decompose c Conceptual Plan: Calculate Q→ compare Q→ and Kc Solution: Q= [NH][H 2S]= 3 Q=6.76×10−3 and K

(0.0822)(0.0822) = 6.76 × 10 −3

=8.5×10−3 so Q<Kc and the reaction will shift to the right; so more solid will decompose. c

16.50 Given: Kp =2.4×10−4 ; PH=0.041 atm; PS=0.037 atm;P 2 2 Find: whether the reaction is at equilibrium Conceptual Plan: Calculate Q→ compare Q and Kp (0.615)2 Solution: Q=P H2PP S2 H2S =(0.041)2(0.037) 2

H2S =0.615 atm

=1.6×10−4 Q=1.6×10−4 and K=2.4×10−4 p

2

Q < Kp, so the system is not at equilibrium, and the reaction will shift to the right.

16.51 Given: 5.14 g Ag2SO4, 1.5 L solution, Kc=1.1×10

−5

Find: whether more solid will dissolve

, g2SO4]→[Ag +][SO2− 4

Conceptual Plan: g AgSO→mol AgSO→[A 2 4 2 4

1 mol Ag2SO4

]→calculate Q and compare to Kc

311.80 g

[ ]=

2

Q = [ A g+ ][SO 4 2 − ]

mol Ag2SO4vol solution

(1 mol )Ag2SO4311.80gAg2SO4

Solution: 5.14 gAg2SO4

=0.0164_849 mol Ag 2SO4

0.0164_849 mol Ag2SO4

[Ag+]=2 [Ag

1.5 L solution

=0.011 M Ag2SO4

] = (0.022)2(0.011) = 5.3 × 10 ][SO ]=[Ag2SO4]=0.011 MQ=[Ag 2SO4]=2(0.011 M)=0.022 M; [SO2−4 + 2

4

2−

−6

Q < Kc, so the system is not at equilibrium and is an unsaturated solution. Therefore, if more solid is added, it will dissolve. p=6.7 at 298 K; 2.25 L flask; NO2=0.022 mol; N2O4=0.097 mol 16.52 Given: K Find: whether the reaction is at equilibrium Conceptual Plan: K→p p Solution: Kc =K(RT)Δn

Q = [ N 2O4 ] [ N O ]

2

=

Kc and mol→MK = (6.7)

(

[( )(

0.097 mol2.25 L

p= KRTΔn c( )

M=molV and then calculate Q and compare to Kc −1

) ]= 16_ 4 =450 )

0.08206L·atmmol·K (298K) 0.022 mol2.25 L

2

Q > Kc, so the reaction is not at equilibrium and will shift to the left.

Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant 16.53 a. Given: [A]=1.0 M, [B]=0.0, Kc=4.0; a=1, b=2 Find:[ A], [B] at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine [A] and [B].

Solution: A(g) B(g) ⇌

Initial

Change

Equil

[A]

[B]

1.0

0.00

−x

+x

1.0 − x

x

Q=[B][A]=01.0=0 Q<K Therefore, the reaction will proceed to the right by x.

Kc = [B] [A] = (x) (1.0−x) = 4.0; x = 0.80 [A]=1−0.80=0.20 M; [B]=0.80 M Check: Plug the values into the equilibrium expression:cK= 0.800.20 = 4.0 c=4.0; a=2, b=2 b. Given:[A]=1.0 M, [B]=0.0, K Find:[ A], [B] at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kc,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine[A]and[B].

Solution: 2A()g 2B()g ⇌

Initial

Change

Equil

[A]

[B]

1.0

0.00

−2x

+2x

1.0 − 2x

2x

2

Q = [ B ] 2 [ A ] = 01.0 =0 Q<K Therefore, the reaction will proceed to the right by x.

2 2 Kc = [ B ] [ A ] = ( 2x) 2(1.0−2x)2 = 4.0 ( 2 x ) 2 ( 1 . 0 − 2x) 2 =

√

√4.0 Take the square root of this expression. x=0.33

[A]=1−2(0.33)=0.34 M; [B]=2(0.33)=0.66 M 2 Check: Plug the values into the equilibrium expression:cK= ( 0.66 ) ( 0.34 ) 2 = 3.8. c. Given:

a=1, b=2 K Find: [A], [B] at equilibrium [A]=1.0 M, c=4.0; [B]=0.0,

Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine [A] and [B] .

Solution: A(g) 2B(g) ⇌

Initial

Change

Equil

[A]

[B]

1.0

0.00

−x

+2x

1.0 − x

2x

Q<K Therefore, the reaction will proceed to the right by x.

Q =[B]2[A]=01.0= 0 Kc = [ B ] 2 [ A ]=(2x)2(1.0−x)=4.0 4x2+4x−4=0 Solve using the quadratic equation, found in Appendix I. x =−1.6 or x=0.62; therefore, x=0.62.

[A]=1.0−0.62=0.38 M [B]=2x=2

(0.62)=1.2_ M 2

2

Check: Plug the values into the equilibrium expression: cK= ( 1.2 ) ( 0.38 ) = 3.8.

16.54 =2 a. Given: [A]=1.0 M, [B]=1.0 M, [C]=0.0, Kc=5.0; a=1, b=1, c Find: [A], [B], [C] at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine [A], [B], and [C].

Solution: A(g)+B(g) 2C(g) ⇌

Initial

Change

Equil

[A]

[B]

[C]

1.0

1.0

0.0

−x

−x

+2x

1.0 − x

1.0 − x

2x

Q=[C]2[A][B]=0(1.0)(1.0)=0 Q<K Therefore, the reaction will proceed to the right by x.

Kc=[C]2[A][B]=(2x)2(1.0−x)(1.0−x)=5.0; x=0.53 Solve by taking the square root of both sides.

[A]=[B]=1.0−0.53=0.47 M [C]=2x=2(0.53) M=1.1 M Check: Plug the values into the equilibrium expression:cK= (1.1)2(0.47) (0.47) = 5.4. b. Given:

c=5.0; a=1, [C]=0.0, b=1, c=1 [A]=1.0 M, [B]=1.0 M, K

Find: [A], [B], [C] at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine[A], [B],and[C]. g g Solution: A()+B() C()g ⇌

Initial

Change

Equil

[A]

[B]

[C]

1.0

1.0

0.0

−x

−x

+x

1.0 − x

1.0 − x

x

Q=[C][A][B]=0(1.0)(1.0)=0 Q<K Therefore, the reaction will proceed to the right by x.

Kc = [ C ] [ A ] [ B ] = ( x ) ( 1 . 0 − x ) ( 1 . 0 − x ) = 55.x02; − 1 1 x + 5 = 0 Solve by using the quadratic equation. x=1.6 or x=0.64 Therefore, x=0.64.

[A]=[B]=1.0−0.64=0.36 M [C]=x=0.64 M Check: Plug the values into the equilibrium expression:cK= (0.64) (0.36) (0.36) = 4.9. c. Given:

c=5.0; a=2, [C]=0.0, b=1, c=1 [A]=1.0 M, [B]=1.0 M, K

Find: [A], [B], [C] at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine[A], [B],and[C].

Solution: 2A()g + B() g C()g ⇌

Initial

Change

Equil

[A]

[B]

[C]

1.0

1.0

0.0

−2x

−x

+x

1.0 − 2x

1.0 − x

x

Q=[C][A]2[B]=0(1.0)(1.0)=0 Q<K Therefore, the reaction will proceed to the right by x.

2 Kc = [ C ] [ A ] [ B ] = ( x ) ( 1 . 0 − 2 x ) 2 ( 1 . 0 − x ) = 5 . 0 ; −20.x3+40.x2−26x+5=0 Solve by using successive approximations or a cubic equation calculator found on the Internet. x = 0.34 [A]=1−2x=0.32 M, [B]=1−x=0.66 M, [C]=x=0.34 M Check: Plug the values into the equilibrium expression:cK= (0.34) (0.32)2(0.66) = 5.0.

[N

[N2O4], [NO2] at equilibrium

2O4]=0.0500 M, [NO2]=0.0, Kc=0.513 16.55 Given: Find: Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kc,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium

[ 2]. [ 4]andNO

expression, and solve for x. DetermineNO 2 Solution: N2O4(g)

2 NO2(g)

⇌

Initial

Change

Equil

Q = [NO]22 [N2O4] = 00.0500 =0 2

[N2O4]

[NO2]

0.0500

0.00

−x

+2x

0.0500 −x

2x

Q<K Therefore, the reaction will proceed to the right by x.

4x2 + 0.513x − 0.02565 = 0

2 Kc= [ N O2 ] [N2O4] = (2x) (0.0500− x) = 0.513

√b2−4ac 2a

= − 0.513 ±√0.6735

2(4) = −0.513±√(0.513)2−4(4) ( −0.02565)

−b±

2(4)

x=−0.1667 or x=0.0385 Therefore, x=0.0385.

[N

2O4]=0.0500−0.0385=0.0115 M [NO2]=2x=2(0.0385)=0.0770 M

Check: Plug the values into the equilibrium expression:cK=(0.0770)20.0115 = 0.516.

2]=0.175 [COCl2]=0.0, Kc=255 [CO]=0.1500 M,M,[Cl [CO], [Cl

16.56 Given:

2], [COCl2] at equilibrium Find: Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and K c,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium

[ ] [ 2], and [COCl2].

expression, and solve for x. DetermineCO, Cl Solution:

CO(g) + Cl2(g)

Initial

[CO]

[Cl2]

[COCl2]

0.1500

0.175

0.0

−x

−x

+x

0.1500 − x

0.175 − x

Change

Equil

COCl2(g) ⇌

x

Q=[COCl2][CO][Cl2]=0(0.1500)(0.175)=0 Q<K Therefore, the reaction will proceed to the right by x. = 255 255x2 − 83.875x + 6.69375 = 0 K c= [COCl2] [CO] [Cl2] = x(0.1500−x) (0.175−x) −b ±

√b2−4ac 2a

2(255) = − ( −83.875) ±√( −83.875)2−4(255) (6.69375)

x=0.192_7 or x=0.136_22 Therefore, x=0.136_22.

[CO]=0.1500−0.136_22=0.013_78 M[Cl

2]=0.175−0.136_22=0.038_78 M[COCl2]=x=0.136_22 M

Check: Plug the values into the equilibrium expression:cK= (0.13622) (0.01378) (0.03878) = 255.

16.57 Given:

c=4.0×10 Find: CO2] [CO]=0.20 M, [CO2]=0.0, K [ at equilibrium 3

Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and K c,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium

[ ] [ 2], and [COCl2. ]

expression, and solve for x. DetermineCO, Cl Solution:

NiO(s) + CO(g)

Initial

[CO]

[CO2]

0.20

0.0

−x

+x

Change

0.20 −x

Equil

Ni(s) + CO2(g) ⇌

x

Q=[CO2][CO]=0(0.20)=0 Q<K Therefore, the reaction will proceed to the right by x.

Kc = [CO2] [CO] = x(0.20−x) = 4.0 × 103 4.010× (0.20 3 − x) = x x = 0.19_ 9 [CO2]=0.19_9 M Check: Because the equilibrium constant is so large, the reaction goes essentially to completion; therefore, it is reasonable that the concentration of the product is 0.19_9 M. 2O]=0.110 [CO]=0.110 M, M, [H[CO2]=0.0, [H2]=0.0, Kc=102 [CO], [H

16.58 Given:

2O], [CO2], [H2] at equilibrium Find: Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine[CO][Cl] , 2, and [COCl2].

Solution: CO(g) + H2O(g)

Initial

[CO]

[H2O]

[CO2]

[H2]

0.110

0.110

0.0

0.0

−x

−x

+x

+x

0.110 − x

0.110 − x

x

x

Change

Equil

CO2(g) + H2(g) ⇌

Q = [CO2 ] [ H 2 ] [ C O ] [ H 2 O ]= 0 ( 0.110 ) ( 0.110 ) =0 Q <K Therefore, the reaction will proceed to the right by. x Kc = [CO2][H2][CO][H2O] = (x) (x) (0.110−x) ( 0.110 − x) = 102

√

(

x)(x)(0.110−x)(0.110−x)

= √102

x0.110 − x = ± 10.0_ 995

±10.0_ 995

(0.110 − x) = x

x=0.100_09 or x=0.122_09 Therefore, x=0.100_09=0.100_.

[CO] = [H 2O]=0.110−0.100_=0.010=0.010 M[H2]=[CO2]=x=0.100=0.100_ M which is close to the true value; so the answers are valid.

K Check: Plug the values into the equilibrium expression:c=(0.100)(0.100)(0.010)(0.010)=101.2=1.0×102, , [C2H3O2]=0.0, Kc=1.8×10 [HCHO]=0.210 M, [HO+]=0.0 232] at equilibrium Find: [HCHO], [HO+], [CHO− 16.59 Given:

232

232

−

3

−5

2

Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium

[ ] [ 2], and [COCl2].

expression, and solve for x. DetermineCO, Cl Solution:

HO+ (aq) + C2H3O2 − (aq)

HC2H3O2(aq) + H2O(l) HC2H3O2

[H2O]

[HO+ ] 3

[C2H3O2 − ]

3

⇌

Initial

Change

0.210

0.0

0.0

−x

+x

+x

x

x

0.210 − x

Equil

232] Q=[HO+][CHO− 3

[HC2H3O2] =0(0.210)=0 Q<K Therefore, the reaction will proceed to the right by x. + − = (x) (x) (0.210−x) = 1.8 × 10 −5 Kc = [H3O ] [C2H3O2 ] [HC2H3O2] Assume that x is small compared to 0.210.

(1.8×10−5)

x2 = 0.210

x=0.00194 Check assumption: 0.001940.210×100=0.92%; assumption is valid.

[HO+ ]=[CHO− 232]=0.00194 M[HC2H3O2]=0.210−0.00194=0.208_1=0.208 M (0.208)

3

=1.81×10−5. The answer is the same to two significant figures with the true value, so the answers are valid.

K Check: Plug the values into the equilibrium expression:c=(0.00194)(0.00194) 16.60 Given:

[SO

−7

2Cl2]=0.175 M, [SO2]=0.0, [Cl2]=0.0, Kc=2.99×10

[SO

2Cl2], [SO2], [Cl2] at equilibrium Find: Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine[SOCl][SO]and[Cl] 2, 2.

2 2, Solution: SO2Cl2(g) [SO2Cl2]

[SO2]

[Cl2]

0.175

0.0

0.0

−x

+x

+x

x

x

Initial

Change

0.175 − x

Equil

SO2(g) + Cl2(g) ⇌

Q=[SO2][Cl2][SO2Cl2] =0(0.175)=0 Q<K Therefore, the reaction will proceed to the right by x. =2.99×10−7 Kc=[SO 2][Cl2] [SO2Cl2] = (x)(x)(0.175−x) Assume that x is small compared to 0.175.

(0.175 )

x2 = 0.299 × 10−7

x=2.28_7×10−4 Check assumption: 2.28_7×10−4

0.175

×100=0.13%; assumption is valid.

[SO 2]=[Cl2]=x=2.28_7×10−4=2.29×10−4 M[SOCl]=0.175−2.28_7×10−4 2 2

=0.174_8=0.175 M

Check: Plug the values into the equilibrium expression: Kc =

(2.28_7×10−4)(2.28_7×10−4)

=2.989×10−7=3.00×10−7

(0.175)

This is within 0.011×10−7 of the true value; therefore, the answers are valid. =1.11×10−4 Find: P 16.61 Given: PBr=755 torr, PCl=735 torr, PBrCl=0.0, K 2

p

2

BrCl at equilibrium

Conceptual Plan: Torr→atm and then prepare an ICE table, calculate Q, compare Q and Kp ,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine PBrCl. Solution: PBr=755 torr×1 atm760torr=0.993_4 atm P =735torr×1 atm760torr=0.967_1 atm Cl2

2

2 BrCl(g)

Br2(g) + Cl2(g) P Br2

P Cl2

P BrCl

0.9934

0.9671

0.0

−x

−x

+2x

0.9934 − x

0.9671 − x

2x

Initial

Change

Equil

Q=PBrClPBrP 2Cl 2 = 0( 0.9934)(0.9671)

=0

⇌

Q<K Therefore, the reaction will proceed to the right by x.

2

Kp = P BrClPBr2 PCl 2=

(2x)2(0.9934−x) (0.9671−x)

2

= 1.11 × 10

−4

Assume that x is small compared to 0.9934 and 0.9671.

(2x)2(0.9934) (0.9671) = 1.11 × 10 −4 4x2 = 1.066 × 10 −4 x=0.00516 atm=3.92 torrP

BrCl=2x=2(3.92 torr)=7.84 torr

Check: Plug the values into the equilibrium expression: × 10 −4 = 1.12 × 10. −4 = 1120 .

Kc = (2(0.00516) )2(0.9934−0.00516) (0.9671−0.00516) This is within 0.01×10−4 of the true value; therefore, the answers are valid.

16.62 Given: PCO=1344 torr, PHO=1766 torr, PCO=0.0, PH=0.0, Kp=0.0611 Find: P 2 2 2

CO ,2P H2 at equilibrium

Conceptual Plan: Torr→atm and then prepare an ICE table, calculate Q, compare Q and K the equilibrium expression, and solve for x. Determine P CO and P H2.

c,predict the direction of the reaction, represent the change with x

2

Solution: PCO=1344 torr×1 atm760torr=1.768_4 atm PHO=1766torr×1 atm760torr=2.323_7 atm 2

CO(g) + H2O(g)

Initial

P CO

P H2O

P CO2

P H2

1.768_ 4

2.323_ 7

0.0

0.0

−x

−x

+x

+x

1.7684 − x

2.3237 − x

x

x

Change

Equil

P P 0 ( 1.7684 ) ( 2.3237 ) =0 Q<K Therefore, the reaction will proceed to the right by x.

Q =PCOP 2H2COH2O= P P 2H2COH2O Kp =PCOP = ( x) ( )x (1.7684− )x (2.3237− x) = 0.0611

( ) (4.1092 −x )4.0921x + x = (0.25107−0.2500 x+0.0611 x2) x2 = 0.0611 2

2

CO2(g) + H2(g) ⇌

, sum the table, determine the equilibrium values, put the equilibrium values in

0.9389x2 + 0.2500x − 0.25107 = 0 −b ±

√b2−4ac

=−0.2500 ± √(0.2500)2−4(0.9389)(−0.25107)2(0.9389)

2a

x=0.4008 or −0.6671 so x=0.4008 PCO=P H2 =x=0.4008 atm=305 torr 2 Check: Plug the values into the equilibrium expression: Kp = (0.4008)2(1.7684−0.4008) (2.3237−0.4008) = 0.06109 = 0.061.

The answers are the same to two significant figures with the true value; therefore, the answers are valid.

16.63 c=1.0 Find:[ A], [B], [C] [A]=1.0 M, [B]=[C]=0.0, Kat equilibrium

a. Given:

Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine[A], [B],and[C].

Solution: A(g)

Initial

Change

[A]

[B]

[C]

1.0

0.0

0.0

−x

+x

+x

x

x

1.0 −x

Equil

B(g) + C(g) ⇌

Q = [ B ] [ C ] [ A ] = 0(1.0) =0 Q < K Therefore, the reaction will proceed to the right by. x Kc = [ B ] [ C ] [ A ] = (x) ( x) (1.0−x) = 1.0

(1.0)− x

x2 =1.0

x2 + x − 1 = 0 −b ±

2

√b − 4ac

= −1±√12−4 ) (−1 )(1 2(1)

2a

x=0.61_8 or x=−1.618 Therefore, x=0.61_8.

[B]=[C]=x=0.61_8=0.62 M[A]=1.0−0.61_8=0.382=0.38 M Check: Plug the values into the equilibrium expression: (0.38) Kc=(0.62)(0.62) =1.01=1.0, which is the equilibrium constant; so the values are correct. b. Given:

c=0.010 Find:[ A], [B],K [C] at equilibrium [A]=1.0 M, [B]=[C]=0.0,

Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine[A], [B],and[C].

Solution: A(g)

Initial

Change

[A]

[B]

[C]

1.0

0.0

0.0

−x

+x

+x

x

x

1.0 − x

Equil

Q = [ B ] [ C ] [A] Kc = [ B ] [ C ] [ A ] 2

x = 0.010

B(g)+C(g) ⇌

= 0(1.0) =0 Q<K Therefore, the reaction will proceed to the right by x. = (x) ( x) (1.0−x) = 0.010

(1.0 − x)

x2 + 0.010x − 0.010 = 0 − b ± b2 − 4ac

√

2a

2(1) = − (0.010) ±√(0.010)2−4(1) ( −0.010)

x=0.09512 or x=−0.1051 Therefore, x=0.095_12.

[B]=[C]=x=0.095_12=0.095 M[A]=1.0−0.095_12=0.90488=0.90 M Check: Plug the values into the equilibrium expression: (0.90) =0.01003=0.010, which is the equilibrium constant; so the values are correct.

Kc=(0.095)(0.095) c. Given:

Find: [A], [B], [C] at equilibrium [A]=1.0 M, [B]=[C]=0.0, K=1.0×10−5 c

Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kc, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the ] [C]. equilibrium expression, and solve for .xDetermineA [ ], [ B,and Solution: A(g)

Initial

Change

[A]

[B]

[C]

1.0

0.0

0.0

−x

+x

+x

x

x

1.0 − x

Equil

B(g)+C(g) ⇌

Q=[B][C][A] =0(1.0)=0 Q<K Therefore, the reaction will proceed to the right by x.

Kc = [ B ] [ C ] [ A ] = ( x ) ( x ) ( 1 . 0 − x ) = 1 . 0 × 1 0 − 5 Assume that x is small compared to 1.0.

(1.0×10−5)

x2 = 1.0

x=0.0031_6 Check assumption: 0.0031_61.0

×100%=0.32%; assumption is valid.

[B]=[C]=x=0.0031_6=0.0032 M[A]=1.0−0.0031_6=0.9968=1.0 M Check: Plug the values into the equilibrium expression: =1.024×10−5=1.0×10−5, which is the equilibrium constant; so the values are correct (1.0) Kc=(0.0032)(0.0032) 16.64 a. Given: PB=1.0 atm, PA=0.0 atm, Kp=1.0 Find: PB, PA at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kp. predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine P A and PB. Solution: A(g) PA

PB

2 B(g) ⇌

Initial

Change

Equil

1.0

+x

−2x

1.0 − 2x

x

BPA = 1.0

Q=P

0.0

2

0 Because there is no A, the reaction shifts to the left.

2

Kp

=P BPA 2

x2

(x)

=(1.0 −2 )

= 1.0

1.0 − 4x + 4x2 = 1.0x 4x2−5x+1.0=0

√b2−4ac 2a

= − ( −5) ±√( −5)2−4(4)2(4) (1)

−b±

x=0.25 or x=1.0x Therefore, x=0.25.

PA=x=0.25 atm; PB=1.0−2x=1.0−0.50=0.50 atm Check: Plug the values into the equilibrium expression: Kp= (0.50)2 (0.25)=1.0, which is the equilibrium constant. b. Given: −4 P B =1.0 atm, P A=0.0 atm, p=1.0×10 Find: PB, PA at equilibrium K Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kp,predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine P A and PB. Solution: A(g)

Initial

Change

Equil

PA

PB

0.0

1.0

+x

−2x

1.0 − 2x

x

Q = P BPA = (1.0)

2

0

Because there is no A, the reaction shifts to the left.

2

Kp = P BPA = (1.0−2x)

2 B(g) ⇌

2

(x)

= 1.0 × 10−4

2

(1.0 × 10−4)x

1.0 − 4x + 4x2 =

4x2 − 4.0001x + 1.0 = 0

√b2−4ac 2a

4(4)(1) = − ( −4.0001) ±√(−−4.0001)2

−b±

x=0.50_35 or x=0.49_65 Since 1−2x must be positive, x P A = x=0.49_65=0.50 atm; P

−4

2(4)

= − ( − 4.0001 ) ±√ 8.0 × 10

2(4)

= 0.49_ 65.

(0.49_65)=0.01 atm

B=1.0−2x=1.0−2

Check: Plug the values into the equilibrium expression:

(0.49_65)=9_.869×10−5=1×10−4, which is the equilibrium constant.

Kp = ( 0.007 ) 2 c. Given:

P B =1.0 atm, PA =0.0 atm, Kp =1.0×105 Find: PB, PA at equilibrium Conceptual Plan: Prepare an ICE table, calculate Q,compare Q and Kp, predict the direction of the reaction, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for .xDetermine P A and PB. Solution: A(g)

Initial

Change

Equil

PA

PB

0.0

1.0

+x

−2x

x

P

Q=P

2 B(g) ⇌

1.0 − 2x

=1.02 0 Because there is no A, the reaction shifts to the left.

2

x )2 Kp = P BPA = (1.0−2

(x)

= 1.0 × 105

2

Since Kp is so large, assume that 2x is small compared to 1.0.

1.0 x = 1.0 × 105 1.0 x=1.0×10−5 Check assumption: 2(1.0×10−5) ×100%=0.002%; assumption is valid.

atm; PB=1.0−2x=1.0−2(1.0×10−5)=0.99998 atm

P A=x = 1 . 0 × 1 0 − 5

Check: Plug the values into the equilibrium expression: K

p=(0.99998)2(1.0×10−5)=9.9996×104=1.0×105, which is the equilibrium constant.

Le Châtelier’s Principle 16.65 a. When A is added, the [A] will increase sharply and then decrease to a value that is higher than the initial value. The [B] will increase as the excess A reacts.

16.66 b. When the temperature of an exothermic reaction in increased, the reaction will shift to the left, so the [A] will increase and the [B] will decrease.

16.67 Given: CO(g)+Cl 2(g) ⇌COCl2(g) at equilibrium Find: What is the effect of each of the following? a. COCl2 is added to the reaction mixture. Adding COCl2 increases the concentration of COCl2 and causes the reaction to shift to the left. b. Cl2 is added to the reaction mixture. Adding Cl2 increases the concentration of Cl2 and causes the reaction to shift to the right. c. COCl2 is removed from the reaction mixture. Removing the COCl2 decreases the concentration of COCl2 and causes the reaction to shift to the right.

16.68 Given: 2 BrNO(g) ⇌2 NO(g)+Br2(g) at equilibrium Find: What is the effect of each of the following? a. NO is added to the reaction mixture. Adding NO increases the concentration of NO and causes the reaction to shift to the left. b. BrNO is added to the reaction mixture. Adding BrNO increases the concentration of BrNO and causes the reaction to shift to the right. c. Br2 is removed from the reaction mixture. Removing Br2 decreases the concentration of Br2 and causes the reaction to shift to the right.

16.69 Given: 2 KClO3(s) ⇌2 KCl(s)+3 O2(g) at equilibrium Find: What is the effect of each of the following? a. O2 is added to the reaction mixture. Adding the O2 increases the concentration of O2 and causes the reaction to shift to the left. b. KCL is removed from the reaction mixture. Removing KCL does not cause any change in the reaction. KCL is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. c. KClO3 is removed from the reaction mixture. Removing KClO3 does not cause any change in the reaction. KClO3 is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. d. O2 is removed from the reaction mixture. Removing O2 decreases the concentration of O2 and causes the reaction to shift to the right.

16.70 Given: C(s)+H 2O(g) ⇌CO(g)+H2 (g) Find: What is the effect of each of the following? a. C is removed from the reaction mixture. Removing C does not cause any change in the reaction. C is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. b. H2O is added to the reaction mixture. Adding H2O increases the concentration of H2O and causes the reaction to shift to the right.

c. CO is removed from the reaction mixture. Removing CO decreases the concentration of CO and causes the reaction to shift to the right. d. H2 is added to the reaction mixture. Adding H2 increases the concentration of H2 and causes the reaction to shift to the left.

16.71 a. Given: I2(g) ⇌2 I(g) at equilibrium Find: the effect of increasing the volume The chemical equation has 2 moles of gas on the right and 1 mole of gas on the left. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles). b. Given: 2 H2S(g) ⇌2 H2(g)+S2(g) Find: the effect of decreasing the volume The chemical equation has 3 moles of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of gas particles). c. Given: I2(g)+Cl2(g) ⇌2 ICl(g) Find: the effect of decreasing the volume The chemical equation has 2 moles of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure but causes no shift in the reaction because the moles are equal on both sides.

16.72 a. Given: CO(g)+H2O(g) ⇌CO2(g)+H2(g) Find: the effect of decreasing the volume The chemical equation has 2 moles of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure but causes no shift in the reaction because the moles of gas are equal on both sides. b. Given: PCl3(g)+Cl2(g) ⇌PCl5(g) Find: the effect of increasing the volume The chemical equation has 2 moles of gas on the left and 1 mole of gas on the right. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the left (toward the side with more moles of gas particles). c. Given: CaCO3(s) ⇌CaO(s)+CO2(g) Find: the effect of increasing the volume The chemical equation has 1 mole of gas on the right and 0 mole of gas on the left. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles).

16.73 Given: C(s)+CO 2(g) ⇌2 CO(g) is endothermic. Find: the effect of increasing the temperature Because the reaction is endothermic, we can think of the heat as a reactant. Increasing the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. This will cause an increase in the concentration of products and a decrease in the concentration of reactants; therefore, the value of K will increase.

Find: the effect of decreasing the temperature Because the reaction is endothermic, we can think of the heat as a reactant. Decreasing the temperature is equivalent to removing a reactant, causing the reaction to shift to the left. This will cause a decrease in the concentration of products and an increase in the concentration of reactants; therefore, the value of K will decrease.

16.74 Given: C6H12O6(s)+6 O2(g) ⇌6 CO2(g)+6 H2O(g) is exothermic.

Find: the effect of increasing the temperature Because the reaction is exothermic, we can think of the heat as a product. Increasing the temperature is equivalent to adding a product, causing the reaction to shift to the left. This will cause a decrease in the concentration of products and an increase in the concentration of reactant; therefore, the value of K will decrease.

Find: the effect of decreasing the temperature Because the reaction is exothermic, we can think of the heat as a product. Decreasing the temperature is equivalent to removing a product, causing the reaction to shift to the right. This will cause an increase in the concentration of products and a decrease in the concentration of reactants; therefore, the value of K will increase.

16.75 Given: C(s)+2 H 2(g) ⇌CH4(g) is exothermic. Find: Determine which will favor CH4. a. Adding more C to the reaction mixture does not favor CH4. Adding C does not cause any change in the reaction. C is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. b. Adding more H2 to the reaction mixture favors CH4. Adding H2 increases the concentration of H2, causing the reaction to shift to the right. c. Raising the temperature of the reaction mixture does not favor CH4. Because the reaction is exothermic, we can think of heat as a product. Raising the temperature is equivalent to adding a product, causing the reaction to shift to the left. d. Lowering the volume of the reaction mixture favors CH4. The chemical equation has 1 mole of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the right (toward the side with fewer moles of gas particles). e. Adding a catalyst to the reaction mixture does not favor CH4. A catalyst added to the reaction mixture only speeds up the reaction; it does not change the equilibrium concentration. f. Adding neon gas to the reaction mixture does not favor CH4. Adding an inert gas to a reaction mixture at a fixed volume has no effect on the equilibrium.

16.76 Given: C(s)+H 2O(g) ⇌CO(g)+H2(g) is endothermic. Find: Determine which will favor H2. a. Adding more C to the reaction mixture has no effect on the quantity of H2. Adding C does not cause any change in the reaction. C is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. b. Adding more H2O to the reaction mixture results in the formation of additional H2. Adding H2O increases the concentration of H2O and causes the reaction to shift to the right.

2. Because the reaction is endothermic, we can think of heat as a reactant. Raising the temperature is equivalent to adding a reactant, causing the reaction to c. Raising the temperature of the reaction mixture results in the formation of additional H shift to the right. d. Increasing the volume of the reaction mixture results in the formation of additional 2. The H chemical equation has 2 moles of gas on the right and 1 mole of gas on the left. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles). e. Adding a catalyst to the reaction mixture has no effect on the quantity of 2. A H catalyst added to the reaction mixture only speeds up the reaction; it does not change the equilibrium concentration. f. Adding an inert gas to the reaction mixture has no effect on the quantity of H2. Adding an inert gas to a reaction mixture at a fixed volume has no effect on the equilibrium.

Cumulative Problems 16.77 1, then 1/Kadd to equation 2. To find K a. To find the value of K for the new equation, combine the two given equations to yield the new equation. Reverse equation 1 and use HbO2 (aq)

HbCO(aq)

K1 = 1/1.8Hb(aq) + CO(aq)

Hb(aq) + 2O (aq)

K2=306

(K K1) for equation 3,2). use (1/

¯HbO2(aq)+CO( aq)

2 b. Given: O=20%, Find: the ratio [HbCO][HbO] 2 ⇌ CO=0.10% ⇌ Conceptual Plan: Determine the equilibrium expression and then determine [HbCO][HbO2].

(0.10=170(0.10)(20.0)=0.851.0 ) Solution: K= [HbCO] [O2] [HbO2] [CO] 170= [HbCO] (20.0) [HbO2] then [HbCO][HbO2] Because the ratio is almost 1:1, 0.10% CO will replace about 50% of the O2 in the blood. The CO blocks the uptake of O2 by the blood and is therefore highly toxic.

K3 = K1K2=

2(aq) HbCO(aq) + O

(1/1.8 )(306 ) =170

⇌

Find: [NO] in molecules/cm3 =298 K, N=78%, O 2=21%, K=4.1×10−31 2 p

16.78 Given:P =1 atm, T Conceptual Plan:

%vol→n→M and then Kp→Kc

PV = nRT n 1 L air

equilibrium expression, and solve for x . Determine

Kp = Kc (RT)

Δn

and then prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the

[NO] in mol/L

→ molecules/cm3.

6.022 × 1023 molecules

mole

1 L1000 mL mLcm3

Solution: Assume 1 L of air. nN2 = (1 atm)(0.78 L)

(

)

0.08206 L · atmmol · K (298 K) =0.031_90 mol

n O2 = (1 atm)(0.21 L)

(

)

0.08206 L · atmmol · K (298 K) =0.0085_9 mol[N 2]=0.031_90 mol/L

Initial

Change

Equil

[O2]

[NO]

0.031_ 9

0.0085_ 9

0.0

−x

−x

+2x

0.031_ 9 − x

0.0085_ 9 − x

2x

The reaction will proceed to the right by x.

2 (0.0319−x)(0.00859−x)=4.1×10−31 2 2 Kc=[NO] [N][O]=(2x)2 Assume that x is small compared to 0.00859 and 0.0319. x=5.3_0×10−18 Check assumption: 5.3_0×10−18 0.00858 ×100%=6.2×10 [NO] = 2x = 2

−14

Δn p = Kc (RT)

Kc = Kp (RT)

Δn

−31

= 4.1 × 10

((

)

0.08206 L · atmmol · K (298

2 NO(g)

N2(g)+O2(g) [N2]

[O2]=0.0085_9 mol/LK

⇌

%

(5.3_)0 × 10 −18

=1.0_6×10

−17

M

×L1000 mL×mLcm3 = 6.38×103molecules cm3 1.0_6×10−17molL ×6.022×1023 molecules mol = 6.4×103molecules cm3 Check: The answer is reasonable because the reaction has a small equilibrium constant; so you would not expect to produce much product.

The reaction to produce NO is endothermic, so we can think of heat as a reactant. Therefore, raising the temperature (as in an automobile engine) shifts the reaction to the right, producing more NO. 16.79 a. Given: 4.45 g CO2, 10.0 L, 1200 K, 2.00 g C, Kp=5.78 Find: total pressure 1 mol C12.01 g C and then determine limiting reactant and then mol CO2→ PCO 2. PV = nRT Prepare an ICE table, represent the Conceptual Plan: g CO2→mol CO2 and g C→mol C 1 mol CO244.01 g CO2 change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x.

Solution: CO2(g) + C(s) ⇌ 2 CO(g) 4.45 g CO×1 =0.101_1 mol CO2 2.00 g C× 1 mol12.01 g C =0.166_5 mol C 2 mol44.01 g CO2 Because the stoichiometry is 1:1, the CO 2 is the limiting reactant.

P CO= 2

(0.101_1 mol)

(0.08206)(1200 L · atm K) mol · K

10.0 L

=0.995_6 atm

2 CO(g)

CO2(g) + C(s)

Initial

Change

Equil

P CO2

P CO

0.996

0.0

−x

+2x

0.996 − x

2x

⇌

The reaction will proceed to the right by x. x2 P =5.78. Solve using the quadratic equation, found in Appendix I. Kp=P COCO=(2)(0.996−x) 2 2

x=0.678 atm

(0.678 atm)=1.35_6 atm

P CO=0.996 atm−0.678 atm=0.318 atm P CO = 2 2 P total=1.67 atm Check: Plug the values for the partial pressure into the equilibrium expression. (1.356)20.318=5.78, which is the value of the equilibrium constant. b. Given: 4.45 g CO2, 10.0 L, 1200 K, 0.50 g C, Kp=5.78 ,Find: total pressure Conceptual Plan:gCO 2→ molCO and g C→mol C 2 mol CO→ 2

PCO2 and mol C→mol CO→

1 mol CO244.01 g CO2

P CO

1 mol C12.01 g C

PV=nRT

and then determine limiting reactant and then

PV = nRT

Solution: CO2(g) + C(s) ⇌2 CO(g) 4.45 g CO2×1 mol44.01 g CO2=0.101_1 mol CO2 0.50 g C×1 mol12.01 g C=0.041_6 mol C Because the stoichiometry is 1:1, the C is the limiting reactant; therefore, the moles of CO formed will be determined from the reaction, not the equilibrium.

0.041_6 mol C×2 mol CO1 mol C=0.083_2 mol CO CO2(g)+C(s)

Initial

Change

0.1011

0.041_ 6

− 0.041_ 6

− 0.041_ 6

2 CO(g) ⇌

0.0

(0.041_ 6)

+2

0.059_ 5

Equil

( (

P CO2 = (0.059_5 mol)

0.08206 L · atm

mol · K

P CO=(0.083_2 mol) 0.08206 L · atm mol · K P total=0.58_6+0.81_9=1.405=1.41 atm

0

0.083_ 2

) )

(1200 K)

10.0 L

=0.58_6 atm

(1200 K)

10.0 L

=0.81_9 atm

Check: The pressure is less than the equilibrium pressure, which is reasonable because the C was the limiting reactant.

16.80 Given: at equilibrium: 0.13 mol H2, 0.13 mol CO, 0.43 mol H2O, then react all H2 Find: CO at new equilibrium Conceptual Plan: Equilibrium values→K c and then new value of H2O. Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x.

Solution: H2O(g)+C(s) ⇌H2(g)+CO(g) Equil 0.43 [H2O] (0.43) 0.13 Kc=[H2][CO] =(0.13)(0.13) 0.13 =0.039_3

2(g)+O2(g)→2 H2O(g) So 0.13 mol H2 produces 0.13 mol H2O, giving new initial conditions of 0.56 mol H2O(0.13+0.43), 0

2H

H2O(g) + C(s)

Initial

Change

0.56

0

0.13

−x

+x

+x

0.56 − x

Equil

H2(g) + CO(g) ⇌

0.13 + x

x

=0.039_3 so x2+0.1693x−0.0220=0 [H2O] =(x)(0.13+x) (0.56−x) Kc=[H2][CO] Solve using the quadratic equation, found in Appendix I. x=0.086 [CO]=0.13+0.086=0.216 M=0.22 M therefore, 0.22 mol CO. Check: Determine equilibrium concentration and plug into the equilibrium expression.

[H2O] (0.47) =(0.086)(0.22) =0.040. This is within 0.001 of the equilibrium value, so the answer is reasonable.

Kc=[H2][CO] 16.81 Given: V=10.0 L, T=650 K, 1.0 g MgO, P CO=0.026 atm, K =0.0260 p 2 Find: mass MgCO3 when volume is 0.100 L Conceptual Plan: P(10.0 L)→P(0.100 L). Prepare an ICE table, represent the change with x moles CO2, the limiting reactant, and the mass of MgCO3 formed.

, sum the table, determine the equilibrium value, put the equilibrium values in the equilibrium expression, and solve for x . Then determine

P1V1 = P2V2

PV =nRT

Solution: P1V1= P2V2

x=2.60 atm (0.0260 )(10.0 atm L)= (x)( 0.100 L) MgCO3(s) 2.60

Initial

MgO3(s) + CO2(g) ⇌

−x

Change

2.60 − x

Equil

Kp = P CO2 =0.026=2.60−x x=2.57_9 atm

CO=(2.57_9 atm)10.0 L(0.08206 K_)5 L · atm mol · K=)0(.645803 mol CO2 1.0 g MgO×1 mol40.31 g MgO=0.024_8 mol MgO

n

2

Therefore, MgO is the limiting reactant and produces 0.0248 mol MgCO 3 0.0248 mol MgCO×84.32 g MgCO31 mol MgCO 3=2.09 g MgCO3 3

Check: The units (g) are correct. The magnitude (2) is reasonable since the pressure and volume are small. Much less than 1 mole is anticipated.

16.82 Given: at equilibrium: PI=0.21 atm, P=0.23 atm I 2 Find: P of each gas when the volume is compressed to half the initial volume Conceptual Plan: P at equilibrium to equilibrium constant, determine P at new volume,P1V1=P2V2 prepare an ICE table, represent the change with x , sum the table, determine the equilibrium value, put the equilibrium values in the equilibrium expression, and solve for. x

⇌2 I(g) K=PP=(0.23)2 Solution: I(g) II 2 p

(0.21)

= 0.25_ 2

P 1V1 = P2V2

2: (0.21 atm)(1)=(x atm)(0.5) x=0.42 atm

For I

For I: (0.23 atm)(1)=(y atm)(0.5) y=0.46 atm 2 I(g)

I2(g)

Initial

0.42

0.46

+x

−2x

0.42 + x

0.46 − 2x

Change

Equil

⇌

Q = (0.46)2 (0.42) = 0.50 Q >K, so the reaction shifts to the left.

Kp=(0.46−2x)2(0.42+x)=0.25_2 x=0.056_7 Solve using the quadratic equation, found in Appendix I. P

0.477=0.48 atm (I2)=0.42+0.057=

P

(I)=0.46−2(0.057)=0.346=0.35 atm

2

Kp = (0.35) (0.48) =0.25_5=0.26. The value is within one significant figure of the original Kp, so the answer is reasonable.

Check: Plug the equilibrium values into the equilibrium expression: 16.83 Given: CH()+Cl() is exothermic Find: Which of the following will maximize CHCl? 2 4 g 2 g CHCl() 2g 24

24

2

a. Increasing the reaction volume will not maximize 24 CHCl.2 The chemical equation has 1 mole of gas on the right and 2 moles of gas on the left. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to ⇌ shift to the left (toward the side with more moles of gas particles). b. Removing CHCl 2H4Cl2.

Removing the CHCl2 will4 decrease the concentration of CHCl 2and 4will 2 4as 2 it forms will maximize CHCl. 2 42 2 2 cause the reaction to shift to the right, producing more C more C c. Lowering the reaction temperature will maximize2CHCl. 4 2 The reaction is exothermic, so we can think of heat as a product. Lowering the temperature will cause the reaction to shift to the right, producing2H4Cl2.

Cl increases the concentration of Cl, so d. Adding Cl2 will maximize CHCl. 2 4 Adding 2 2 2 the reaction shifts to the right, which will produce more CHCl. 2 42 16.84 Given: CH()+I() Find: CHI() is endothermic Which of the following will maximize CHCl 2 4 g2 g 242g 24 2 a. Decreasing the reaction volume will maximize 242 CHI. The chemical equation has 1 mole of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to ⇌ the right (toward the side with fewer moles of gas particles). b. Removing I2 from the reaction mixture will not maximize 2CHI. 2.I 42 Removing I 2decreases the concentration of I, 2 and the reaction will shift to the left to produce more more C c. Raising the temperature of the reaction will maximize2CHI. 42 The reaction is endothermic, so we can think of heat as a reactant. Raising the temperature will cause the reaction to shift to the right, producing2H4I2. d. Adding CH 2 4to the reaction mixture will maximize CHI. 2 42Adding CH2will 4 increase the concentration of CH2and 4 cause the reaction to shift to the right to produce more C2H4I2.

P at equilibrium?

16.85 Given: reaction 1 at equilibrium:P H2 =0.958 atm, PI2 =0.877 atm, PHI PH2 = PI2 = 0.621 atm, P HI =0.101 atm Find: Is reaction 2 at equilibrium? If not, what is theHI =0.020 atm; reaction 2: Conceptual Plan: Use equilibrium partial pressures to determineK p. UseKpto determine whether reaction 2 is at equilibrium. Prepare an ICE table, calculate Q, compare Q and Kp, predict the direction of the reaction, represent the change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. DetermineK HI. Solution: H2(g) + I2(g) 2HI(g) Reaction 1:

⇌

Equil

P H2

P I2

KHI

0.958

0.877

0.020

2

P I2 = ( 0.020 ) ( 0.958 ) ( 0.877 ) = 4.76_ 10 × 10−4 Kp = P HIPH2 2

2

P I2 = ( 0.101 ) ( 0.621 ) ( 0.621 ) =0.0264 Q>K, so the reaction shifts to the left.

Q = P HIPH2 2

H2(g) + I2(g) P H2

P I2

P HI

0.621

0.621

0.101

+x

+x

−2x

0.621 + x

0.621 + x

0.101 − 2x

Reaction 2:

Initial

Change

Equil

2 HI(g)

⇌

2

4 Kp = P HIP H2 PI2 = (0.101−2x) ( 0.621 +x) (0.621+x) = 4.76_ 10 × −10 2

(0.101−2x)2 ( 0.621 + x) ( 0.621 + x) =

√

4.76_ 10 × 10−4

√

(0.101−2x) (0.621+x) = 2.18_ 2 × 10−2 x = 0.04325 = 0.0433 P H=P I =0. 621+ x =0.621+0.0433=0.664 atm 2 2 P HI=0.101−2x=0.101−2(0.0433)=0.0144 atm Check: Plug the values into the equilibrium expression: Kp= (0.0144)2 (0.664)2 = 4.703 × 10 −4 = 4.70 × ;10 this−4 value is close to the original equilibrium constant.

16.86 Given: reaction 1 initial: HS=0.500 M, SO=0.500 M, HO at2 equilibrium =0.0011 M; reaction 2: HS=0.250 M, SO=0.325 2M 2 2 2

Find: [H2O] at equilibrium in reaction 2

Conceptual Plan: Prepare an ICE table. Determine the equilibrium concentrations in reaction 1 and determineK c. UseKcto determine the equilibrium concentrations for reaction 2. Prepare an ICE table, calculate Q, compare Q and Kc,

2O].values, put the equilibr predict the direction of the reaction, represent the change with x, sum the table, determine the equilibrium Solution: 2 H2S(g) + SO2(g) Reaction 1:

Initial

Change

Equil

⇌

2 g 3 sS()+2 HO() 2S] [H

[SO2]

[S]

[H2O]

0.500

0.500

−2x

−x

+2x

0.500 − 2x

0.500 − 2x

0.0011

constant

0

2x=0.0011, x=5.5×10−4 −4 [H2S]=0.500−0.0011=0.4989 [SO 2]=0.500−5.5×10 =0.49945 2 .0011)2 −6 Kc =[HO]2[HS]2[SO] 2 2 2 = (0 = 9.73_ 3 × 10

2 H2S(g) + SO2

Reaction 2:

Initial

[H2S]

[SO2]

0.250

0.325

[S]

constant

[H2O] 0

⇌

3 S(s)+2 H 2O(g)

−2x

−x

+2x

0.250 − 2x

0.325 − x

2x

Change

Equil

Reaction shifts to the right.

2 2 [SO]=(2x)2 (0.250−2x)2(0.325−x)=9.73_3×10−6 2 2 Kc = [H2O] [HS] Assume that 2x and x are small, respectively, compared to 0.250 and 0.325.

4x2 = 1.97_ 70 × 10−7 x2 = 4.94_ 25 × 10−8 x=2.22_3×10−4 Check assumption: 2.22×10−4 0.250 =8.88×10−4×100%=0.089%; assumption is valid.

[HO]=2x=2 2

M (2.22×10−4)=4.44×10−4 (2.22×10−4 )=0.250 M

[H2S] = 0.250 − 2

[SO]2= 0.325 − 2.22 × 10−4

=0.325 M Check: Plug the values into the equilibrium expression:

c=(4.44×10−1)2(0.250)2(0.325)=9.71×10−6; this value is close to the original equilibrium constant.

K

=5.3×10−5 Find: mass in g of NH2 and % yield Given: 200.0 L container; 1.27 kg N; 2 0.310 kg H;2 725 K; K p 16.87 1000 g kg gmolar mass molvol and then prepare an ICE table. Represent the change with x, sum the table, determine the equilibrium values, put the K=K(RT)Δn Conceptual Plan: K→ Kc and then kg→g→mol→M p c p equilibrium values in the equilibrium expression, and solve for

g and then determine x. Determine [NH M→mol→ 3]. Then

M×vol mol×molar mass

theoretical yieldNH→%yield.

3 determine limiting reactant % yield = actual yieldtheoretical yield× 100% −2 L · atmmol · K Solution: Kp=Kc (RT) Δn Kc=Kp (RT) Δn = 5.3×10−5 (725 K)=

( (0.08206 ) ) 0.18_ 76

nN =1.27 kg N2×1000 gkg 2

× 1 mol N228.02 g N2 =45.3_25 mol N 2

[N2] =

45.3_25 mol200.0 L

=0.226_63 M

nH2 =0.310 kg H2×1000 gkg

× 1 mol H22.016 g H2 =153_.77 mol H 2

[H2] = 153_.77 mol200.0 L

=0.768_85 M N2(g)+3 H2(g)

[N2]

[H2]

[NH3]

0.2266

0.7689

0.0

−x

+3x

+2x

0.2266 − x

0.7689 − 3x

2x

Reaction 1:

Initial

Change

Equil

2 NH3(g) ⇌

Reaction shifts to the right.

3 2 2 Kc=[NH]2[N][H]3=(2x)2

(0.2266− x )(0.7689−3) x3 =0.1876 Assume that x is small compared to 0.2268 and 3x is small compared to 0.7689.

(2x)2(0.2266) (0.7689)3 = 0.1876 x = 0.06951 Check assumptions: 0.069510.2268×100%=30.6, which is not valid, and 3(0.06951) 0.7689 Use method of successive substitution to solve for x. This yields x=0.0460.

×100%=27.1%.

3]=2x=2(0.0460)=0.0920 M

[NH

Check: Plug the values into the equilibrium expression: 2 Kc = (0.0920) (0.2266−0.0460) ( (0.7689−3(0.0460) )3 = 0.1866; this value is close to the original equilibrium constant.

0.0920 mol NH3 L ×200.0 L×17.03 g NH3mol NH3=31_3.8 g=3.1×10 g Determine grams NH: 3 Determine the theoretical yield and the limiting reactant: 1 mol N228.02g N2 2 mol NH31mol N2 × 17.03 g NH3mol NH 3 =154_4 g NH3 1.27 kg N2× 1000 gkg × × 0.310 kg H2×1000 gkg ×1 mol H22.016g H2×2 mol NH33mol ×17.03 H2 g NH3mol NH3 =1745 g NH3

2

3 g NH3.

N2 produces the least amount of NH3; therefore, it is the limiting reactant, and the theoretical yield is 1.54×10 % yield=3.1×102 g1.54×103 g×100=20.%

16.88 Given: V=85.0 L, 22.3 kg CH 4, 55.4 kg CO2, T =825 K, p=4.5×10 K

2

Find: g H2 at equilibrium

Conceptual Plan:kg → g → mol → P and 1000 gkg gmolar mass PV=nRT then prepare an ICE table. Represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium P P H. and then determine 2 Then →mol→g theoretical yieldH→%yield.

2 determine limiting reactant % yield =actual yieldtheoretical yield×100% x expression, and solve for.Determine

Solution: nCH=22.3 kg×1000 gkg 4

PV=nRT mol×molar mass

×1 mol16.04 g=139_0.3 mol

(

)

85.0L

=110_7 atm

(

)

85.0L

=100_3 atm

P CH =nRTV × (139_0 mol) 0.08206 L · atmmol · K (825 K) 4 nCO =55.4 kg× 1000 g kg ×1 mol44.01 g=125_9 mol 2 P CO2 = nRTV × (125_9 mol) 0.08206 L · atmmol · K

(825 K)

CH4(g) + CO2(g)

Initial

P CH4

P CO2

P CO

P H2

1107

1003

0.00

0.00

−x

−x

+2x

+2x

1107 − x

1003 − x

2x

2x

Change

Equil

Kp=P

2

COP H2P CH4P CO2 =(2x)2(2x)2 2

(

110_7−x

2 CO(g)+2 H 2(g)

⇌

) ( )=4.5×10 2

1003−x

Assume that x is small compared to 1003 and 110¯7 x4 = 3.13 × 107; x = 74.8 Check assumptions: 74.81003×100%=7.46% and 74.81107×100%=6.78%; assumptions are not valid.

Use method of successive substitutions, which yields x=72.0.

Check: Plug into equilibrium expression: Kp=P

P P COP H2CH2CO2 2

(

=(2(72.0))2(2(72.0))2110_7−72.0

2

2.

Determine grams of H

)(

1003−72.0

)=4.46×10 2

2 =4.5×10, which is the equilibrium constant.

H=2(72.0)=144 atm; n =PVRT=(144 ) atm)(85.0 L)(0.08206L · atmmol · K(82=5 K1)80_.8 mol

P 2

180_.8 mol×2.016 g H2

H2 mol =364.5 g=365 g H

Determine % yield.

1394 mol CH4×2 mol H2mol CH4=2788 mol H2

2

1259 mol CO2×2 mol H2mol CO2=2518 mol H2 CO2 is the limiting reactant.

180_.8 mol H2 2518 mol H2×100%=7.18% yield atm; 16.89 Given: at equilibrium: PCO=0.30 atm; P Cl=0.10 2

P COCl2 =0.60 atm, add 0.40 atm Cl2

Find: PCO when system returns to equilibrium Conceptual Plan: Use equilibrium partial pressures to determine Kp. For the new conditions, prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine PCO. Solution: CO(g)+Cl2(g) COCl2(g) ⇌

Condition 1:

P CO

P Cl2

P COCl2

0.30

0.10

0.60

( 0.60 ) ( 0.30 ) ( 0.10 ) =20 PCOPCl Kp = P COCl 2 2 =

CO(g) + Cl2(g) P CO

P Cl2

P COCl2

0.30

0.10 + 0.40

0.60

−x

−x

+x

0.30 − x

0.50 − x

0.60 + x

Condition 2:

Initial

Change

Equil

COCl2(g) ⇌

Reaction shifts to the right because the concentration of2Cl was increased.

2 =(0.60+x)(0.30−x)(0.50−x) =20 Kp=PCOClPCOPCl 2 20x2 − 17x + 2.4 = 0

√b2−4ac 2a

−b±

2(20) = − ( −17) ±√( −17)2−4(20) (2.4)

x=0.67 or 0.18 so x=0.18 P CO=0.30−0.18=0.12 atm; PCl=0.50−0.18=0.32; P 2 COCl2=0.60+0.18=0.78 atm Check: Plug the values into the equilibrium expression: (0.78) (0.12) (0.32) Kp= =20.3=20.; this is the same as the original equilibrium constant. atm; PO=1.00 atm; at equilibrium, P total =3.75 atm, T=27 °C Find: Kc 16.90 Given: PSO=3.00 2 2 Conceptual Plan: Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, use the total pressure, and solve for x . Determine partial pressure of each at equilibrium. Determine n Kp → Kc.Kp=Kc 2(g)(RT)Δ + O2(g) Solution: 2 SO

2 SO3(g)

⇌

P SO2

P O2

P SO2

3.00

1.00

0.00

−2x

−x

+2x

3.00 − 2x

1.00 − x

2x

Initial

Change

Equil

3

P total = P SO2 + P O2 + P SO

(10 ) (

3.75 = 3.00 − 2x + . 0 −

(

)

x +2x

x = 0.25 P SO2 = 3.00−2(0.25) =2.50 atm; PO2 = 1.00−0.25 p=PSOP P =(0.50)2(2.50)2(0.75) =0.053_3=0.053 SO2O2 K 3 2

)

( )

=0.75 atm; PSO3=20.25

=0.50 atm

2

p to Kbe less than 1.

Check: The value of the pressure of SO3 is small compared to the pressures of SO2 and O2; therefore, you would expect

((

Kp =K c(RT)Δn 0.053_ 3 = Kc 0.08206 L · atmmol · K c=1.3_12=1.3 K

−1

)(27+273 K))

16.91 Given: Kp=0.76; Ptotal at equilibrium =1.0 atm Find: Pinitial CCl4 Conceptual Plan: Prepare an ICE table, represent thePCCl with A and the change with x , sum the table, determine the equilibrium values, use the total pressure, and solve for A in terms of x. Determine partial pressure of each at 4 equilibrium, use the equilibrium expression to determine x, and determine A. Solution: CCl4(g) C(s)+2 Cl2(g) ⇌

A

Initial

Equil

0.00

constant

+2x

A− x

2x

x =1.0− x A P total = P CCl4 + PCl2 1.0= A −+2

Kp = P Cl2PCCl4 =

P Cl4

−x

Change

P CCl4 = (A −x)=

PC

P CCl4

x

(1.0−x)−x=1.0−2x; PCl 2

(2x)2(1.0−2x)

=2x

= 0.76

2

2 4x+1.52−0.76=0 Solve the quadratic equation. x=0.285 or −0.665 so x=0.28_5 A=1.0−x=1.0−0.285=0.715=0.72 atm Check: Plug the values into the equilibrium expression: 2 2 =0.756=0.76; the original equilibrium is constant.

Kp=PClPCCl 4 = (2x)(A−x)=(2(0.285))(0.715−0.285) 2 2

16.92 Given: K=3.0; SO 2=2.4 mol initial; SO3=1.2 mol equilibrium Find: mol NO2 initial Conceptual Plan: Assume 1.0 L, prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, and determine the initial values.

Solution: SO2(g)+NO2(g) ⇌ SO3(g) + NO(g)

Initial

Change

Equil

[SO2]

[NO2]

[SO3]

[NO]

2.4

y

0.0

0.0

−x

−x

+x

+x

1.2

y − 1.2

1.2

1.2

x=1.2 Fill in the table.

K = [SO3] [NO] [SO2] [NO2] 3.0 = (1.2) (1.2) (1.2) (y−1.2) y = 1.6 mol NO2 initial=1.6 mol Check: The initial amount 1.6 mol is greater than the amount lost.

16.93 Given: V=0.654 L, T=1000 K, K p =3.9×10−2 Find: mass CaO as equilibrium →n Conceptual Plan: Kp→PCO→n PV=nRT stoichiometry g=n(molar mass) (CO2) (CaO) →g 2 Solution: Because CaCO3 and CaO are solids, they are not included in the equilibrium expression.

; n=PVRT=(3.9×10−2 atm)(0.654 L)(0.08206L · atmmol · K)(100=0 K3).1_08×10−4 mol CO

Kp= PCO2 = 3.9 × 10−2

2

3.1_08×10−4 mol CO

2×1 mol CaO1 mol CO2×56.08 g CaO1 mol CaO=0.0174 g=0.0174 g CaO Check: The small value of K would give a small amount of products, so we would not expect a large mass of CaO to form.

16.94 Given: at equilibrium: N2O4, P=0.28 atm, NO2, P=1.1 atm; T=350 K Find: equilibrium pressures when volume doubles →K pand then P when volume doubles. Then prepare an ICE table, represent the change with x , sum the table, and determine the equilibrium values.

Conceptual Plan: P (N2O 4), P (NO2) 2

P (1.1) Solution: Kp=P NO2N2O4=

0.28

=4.3_21

2

When the volume is doubled, the partial pressure of each gas decreases by half.

2 NO2(g)

N2O4(g) P N2O2

P NO2

0.28/2

1.1/2

−x

+2x

0.14 − x

0.55 + 2x

Initial

Change

Equil

⇌

The reaction shifts to the side with more moles—to the right.

(0 1 )−x=4.3_21 Kp = P NO2PN2O4 = (0.55+2x)2 .4 2

4x2 + 6.521x − 0.3024 = 0 − ( b ) ± b2

2(4) = − (6.521) ±√(6.521)2−4(4) (0.3024)

2a

√ −4ac

x =0.0451 or −1.675 so x=0.0451

( )

P NO=0.14−0.0451=0.0949 atm=0.095 atm; P

NO2 =0.55+20.0451

24

=0.6402=0.64 atm

this is the same as the original equilibrium constant.

K Check: Plug the values into the equilibrium expression:p=(0.6402)2(0.0949)=4.32=4.3; 16.95 Given: Kp=3.10, initial PCO=215 torr, PCl=245 torr Find: mole fraction COCl 2 2

Conceptual Plan: P in torr→P1 atm760 torr in atm. Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, use the total pressure, and solve for mole fraction. P COCl 2 P total

(

( )=0.322_4 atm )=0.282_9 atm

Solution: P CO =(215 torr) 1 atm760 torr

P Cl2 =(245 torr) 1 atm760 torr

CO(g) + Cl2(g) 0.2829

0.3224

0

−x

−x

+x

0.282_ 9 − x

0.322_ 4 − x

Initial

Change

Equil

Q<K, so the reaction shifts to the right.

(

)(

0.322_4−x Kp=PCOClP2COPCl=x0.282_9−x 2

COCl2(g) ⇌

x

)=3.10 so 3.10x2−2.87634x+0.2827=0 Solve using the quadratic equation, foun under like conditions so the mole fraction

can be determined using the pressure values.

PCO=0.282_9−0.111_7=0.171_2 PCl=0.322_4−0.111_7=0.210_7 PCOCl=0.111_7 mole2 fraction COCl2=PCOClP2CO +PCl+P2 2 Check: Plug the equilibrium pressures into the equilibrium expression:p= K 0.111_7

COCl=0.111_70.171_2+0.210_7+0.111_=70.226_3=0.226 2

(0.171_2)(0.210_7 )=3.0966=3.10, which is the equilibrium constant; so the an

torr Find: % mass H at equilibrium 16.96 Given: Kp =1.60×10−3 , T=700 K, 1.55 L, PHO=145 2 2 Conceptual Plan: P(torr)→P(atm).1 atm760 torr Prepare an ICE table, represent the change with x, sum the table, and determine the equilibrium values. →

P

n

→

Solution: PHO=(145 torr) 2

→

mass for each

(

mass %

PV=nRT

28.01 g CO1 mol CO 2.016 g H21 mol H2 18.02 g H2O1 mol H2O mass H2total mass×100

)=0.1908 atm

1 atm760 torr

H2O(g) + C(s) 0.190_ 8

Initial

constant

−x

Change

0.190_ 8 − x

Equil

Q <K reaction shifts to the right.

(

0.190_ 8− x

2 Kp= P COPH2 PHO = (x)(x)

P CO=PH=0.0166_8 atm P 2

0

+x

+x

x

x

x=0.0166_8 or −0.01828 so x = 0.0166_ 8

H2O =0.190_8−0.0166_8=0.174_1 atm

(0.0166_8 ) ( )( atm )=(470.50.00_ K1)×10 ) ( )(0.08206 =)4(.7600_ ( L · atm K9)8×10−3 mol 1.55 L

nCO = nH2 = 2O nH =

)=1.60x10−3

0

CO(g) + H2 (g) ⇌

0.0174_1 atm

(4.50_ 1 × 10

−4

0.08206 L · atm

1.55 L

mol · K

−4

mol

mol · K

mol CO)28.01 g CO1 mol CO =0.0126_1 g CO; (4.50_1×10

18.02 g HO1 2 mol HO2 (4.69_ 8 × 10 − 3 mol HO) 2

=0.0846_6 g H 2O

9.07_4×10− 4 g H2

_

−4

mol H)22.016 g H21 mol H2

= 9.07_ 4 × 10

−4

g H2

(0.0981_774 )100%=9.07_4×10−4 gg H2g (0.0126_1 g+9.07_4×10−4 )g+0.08466 (0.0166_8 ) ( ) (0.174_1 ) ×100%=0.924_2%=0.924% H

×

Kp = Check: Plug the equilibrium values into the equilibrium expression:

0.0166_8

Challenge Problems 16.97 a. Given: P NO =522 torr, P O2 =421 torr; at equilibrium, P total =748 torr Find: Kp

2

= 1.60 × 10 − 3, which is the equilibrium constant. Thus, the answer is reasonable.

Conceptual Plan: Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, use the total pressure, and solve for x . torr → atm → Kp g Solution: 2 NO(g)+O2(g) 2 NO2() ⇌

P NO

P O2

P NO2

522 torr

421 torr

0.00

−2x

−x

+2x

522 − 2x

421 − x

2x

Initial

Change

Equil

(

NO2

)

P total = PNO +2 PO +P 748 = 522 − 2x + 421 − x + 2x x =195 torr P

NO=(522−2(195))=132 torr

(

( )

)

torr P P O=421−195=226 2

NO2 =2195=390 torr

P NO=132 torr×1 atm760 torr=0.173_7 atm;

PO=226 torr× 2

1 atm760 torr =0.297_4 atm

P NO=390 torr×1 atm760 torr=0.513_2 atm 2

2

( )2( )2

Kp = P NO2P NOPO = 0. 513_2 2

2

(0.297_4)= 29.35 = 29.3

_ 0.1737

b. Given: PNO=255 torr, P O =185 torr, Kp =29.3 Find: equilibriumPNO 2

2

Conceptual Plan:torr →atmatm760 torrand then prepare an ICE table. Represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x .

Determine PNO.2 Solution: PNO=255 torr× 1 atm760 torr =0.335_5 atm P O2 =185 torr× 1 atm760 torr =0.243_4 atm 2 NO(g)+O 2(g) P NO

P O2

P NO2

0.3355

0.2434

0.00

−2x

−x

+2x

0.3355 − 2x

0.2434 − x

2x

Initial

Change

Equil

2

2 NO2(g) ⇌

2

Kp = P NO2P NOPO2 = (2x) (0.3355−2x) ( 0.2434 − x) = 29.3 2

2

− 117.2x3 + 63.847x2 −12.867x+0.80282=0. Solve using successive approximations or a cubic equation calculator found on the Internet.

2

x = 0.111_3 P NO = 2x = 2

(0. 1 _ ) =0.222_6 atm

113

NO = (0.3355 − 2(0.1113)) ( ) = 0.1129 PO = 0.2434 − 0.1113

P 0.222_6 atm×760 torr1 atm

= 0.1321

2

=169.2 torr=169 torr p

(0.222_6 )2(0.1129)2(0.1321)

=29.428=29.4; this is within 0.1 of the original equilibrium constant.

Check: Plug the values into the equilibrium expression: K = =0.355 Find: Ptotalat equilibrium 16.98 Given: 2.75 L, 950 K, 0.100 mol SO,20.100 mol O; 2 Kp = nRTand then prepare an ICE table. Represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, solve for x , and then Conceptual Plan:n→PPV determine P for each reactant and product→ P total.

P total = P SO2 + P O2 + P SO3 Solution: P = nRTV

P O2 = P SO2 = (0.100 mol)

( 0.08206 ) L · atmmol · K

(950 K)

2.75 L

=2.83_5 atm 2 SO2(g) + O2(g)

P SO2

P O2

P SO3

2.835

2.835

0.00

−2x

−x

+2x

2.835 − 2x

2.835 − x

2x

Initial

Change

Equil

2

2 SO 3(g) ⇌

2

Kp = P SO3P SO2P O2 = (2x) (2.835−2x) ( 2.835 − x) = 0.355 2

2

−1.42x3+4.051x2−14.266x+8.089=0. Solve using successive approximations or a cubic equation calculator found on the Internet. x=0.6628=0.663 and the other two roots are imaginary numbers.

PSO=2x=2(0.663)=1.32_6=1.33 atm 3

)

P SO=(2.836−2x)=(2.836−2(0.663) =1.51_0=1.51 atm 2 P O=(2.836−x)=(2.836−0.663)=2.17_3=2.17 atm 2 P total=1.33+1.51+2.17=5.01 atm Check: Plug the values into the equilibrium expression: K p=PSOP3 2

P SO2O

2

=(1.33)2(1.51)2 (21) .7

=0.3575

2

This is within 99% of the significant figures of the original equilibrium constant.

16.99 Given: PNOClat equilibrium=115 torr; Kp =0.27, T =700 K Find: initial pressure NO, Cl2 atm760 torr Conceptual Plan:torr→atm and then prepare an ICE table. Represent the change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and determine initial pressure.

1 atm760torr 115torr× =0.151_3 atm Solution:

2 NOCl(g)

2 NO(g)+Cl 2(g)

Initial

P NO

P Cl2

P NOCl

A

A

0.00

Let A = initial pressure of NO and Cl2.

⇌

−2x

−x

+2x

A −2x

A− x

0.151

A − 0.151

A − 0.0756

Change

Equil

2

( (A − 0.151) = 0.151)2 KpP NOClPNOPCl 2 2

(A − 0.0756)

2x = 0.151;

x = 0.0756

= 0.27

2

0.27A3−0.1019A2+0.01231A−0.023266=0. Solve using successive approximations or a cubic equation calculator found on the Internet.

A = 0.566 PNO=PCl=A=0.566 atm=430 torr 2 Check: Plug the values into the equilibrium expression: 2 2 (0.151) (0.566−0.151) (0.566−0.0756) =0.2699=0.27; KpP NOClPNOPCl 2= 2 2

this is the same as the original equilibrium constant.

1=1×104, K reaction 2=0.10 16.100 Given: P NO=1 atm, K reaction p p 2 4

Find: Which component will haveP >0.2 atm?

Then prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x.

Conceptual Plan: Use reaction 2 to determine PNO. 2 Solution: 2 NO(g)

N2O4(g) ⇌

Initial

Change

Equil

Kp = PN 2O 4P NO= ( 1.00 − x)

(2x) 2

2

P NO

P N2O4

0.00

1.00

+2x

−x

+2x

1.00 − x

= 0.10

0.40x2 + x − 1.00 = 0 − b ± b2

2(0.40) = −1±√12−4(0.40) ( −1.00)

2a

√ −4ac

x = 0.7655 = 0.77 PNO2 = 2x = 1.54 Because Kp for reaction 1 is so large, essentially all of the materials are products; so Pthe NO and P O2 in reaction 1 will be negligible.

16.101 Given: P=0.750 atm, density =0.520 g/L, T=377°C Find: Kc Conceptual Plan: Prepare an ICE table, represent the P CCl with A and the change with x, sum the table, determine the equilibrium values, use the total pressure, and solve for A in terms of x . Determine partial pressure of each at 4

equilibrium in terms of x, use the density to determine the apparent molar mass, and use d = PM RT the mole fraction (in terms of P) and the molar mass of each gas to determine x. χA = PAPtotal Solution: 2 NO2(g)

2NO(g) + O2(g) ⇌

Initial

Change

Equil

P NO2

P NO

P O2

A

0.00

0.00

−2x

+2x

+x

A − 2x

2x

x

P total=PNO+P NO+PO2 0.750=A−2x+2x+x A=0.750 −x 2 P NO2 =(A−2x)=(0.750−x)−2x=(0.750−3x); P NO=(2);x P O =2 x d=PMRT

=34.71 g/mol M=dRTP=(0.520 gL)(0.08206 L·atmmol·K)(610K) 0.750atm

P P NOtotalM P M = χNOM2 NO+χ NOMNO+χOMO= PNO2totalMNO+ NO+ P O2P totalMO 2 2

2

2

2

NO+P M2 NOMNO+POM2 PtotalM = PNO O2 2 (0.750)(34.7) = (0.750 − 3x)(46.0) + 2x(30.0) + x(32.0) x = 0.184 Kp=P

2

NOP O2 P NO

2

= (2x)2(x)(0.750−3x)

= (2(0.184))2

( 0.184 ) ( 0.750 − 3 ( 0.184 ) 2) = 0.635_ 6

2

Kp = Kc (RT)Δn

0.635_6=Kc

) )1

((

0.08206 L·atmmol ·K(610K)

Kc = 1.27 × 10−2 Check: The units (none) are correct. The magnitude (0.01) is reasonable since there are higher concentrations of the products than the reactants.

16.102 Given: reaction 1: Kc=7.75; reaction 2: Kc=4.00, [N2O5] initial =4.00 M, [O] 2 equil =4.50 M Find: concentration of other species at equilibrium Conceptual Plan: Combine reaction 1 and reaction 2 to get the overall reaction; then prepare an ICE table. Represent the change with x,sum the table, determine the equilibrium values, use the total pressure, and solve for x . Use x to determine equilibrium concentration. + 2 O2(g) K = K1K2= 31.0 N2O(g) ¯N2O5(g) Solution: N 2O5(g) N2O3(g) + O2(g) K1 = 7.75N2O3(g) N2O(g) + O2(g) K2 = 4.00 Initial

⇌

Change

Equil

] 2 [NO] 25 2 Kc= [ N2 O ] [ O =31 .0=(x)(4.50)2 x=2.41¯95=2.42 [N2O5]=4.00−x=4.00−2.42=1.58 M 2O]=x=2.42 M [N2]=4.50 [O UseMreaction 1 or reaction 2 and

4.00

0.00⇌

0.00

−x

+x

+2x

4.00 − x

(4.00−x)

x

31.0(4.00−x)=20.2_5x

4.50

124=51.2¯5x

solve for [N 2O3] represented as y.

Kc = [N2O3] [O2][N2O5] = 7.75 = (y) (4.50) (1.58) y = 2.72 [N2O3]=y=2.72 M Check: Plug the values into any of the equilibrium expressions. For example, for the overall reaction: [NO] (1.58) Kc=[NO][O]2 2 2 25 = (2.42)(4.50)2 =31.0, the equilibrium constant.

For example, for reaction 2: [N2O3] (2.72) = (2.42) (4.50) Kc = [N2O] [O2] = 4.00 16.103 Given: P total=3.0 atm, mole fraction O2=0.12, T=600 K Find: Kp

⇌

Conceptual Plan:

mole fraction→P

O

→

2

→

P SO 2

P SO 3

→

2P SO2 P O

mole fraction= P O2P total

K p

Kp = P SO2P O2 P SO3

P SO3 = P total − P SO2 − P O2

2

2

2

)(3.0)=0.36 atm(0.36 O=(0.12 PSO=2P atm)=0.72 atm

Solution: P 2

O=2 2

2

P SO3 = 3.0 − 0.36 − 072 = 1. 9_2 K =p P

P

SO2O2

P SO3

= (0.72)2(0.36) (1.9 2)2 _

−2

=0.050_6=5.1×10

2

2

Check: The units (none) are correct. The magnitude (0.05) is reasonable since there are higher concentrations of the products than the reactants.

Conceptual Problems 16.104 The equilibrium constant is very small; therefore, at equilibrium, the concentration of products will be small compared to the concentration of reactants. In reaction mixture d, the initial concentration of reactants is large and there are no products.

Because only a small amount of products will be formed, the approximation that x is small will most likely apply. In reaction mixture b, there is initially no reactant and only products. To reach equilibrium, most of the product would have to go to reactants, and the x is small approximation would not apply. In reaction mixture a, there is a smaller amount of reactants than in reaction mixture d. Because there are fewer initial reactants, the percentage needed to reach equilibrium is higher and the x is small approximation is less likely to apply. In reaction mixture c, the initial concentrations of A and B are equal. The reaction will have to lose a large amount of product to reach equilibrium, and the x is small approximation will not apply.

16.105 Yes, the direction will depend on the volume. If the initial moles of A and B are equal, the initial concentrations of A and B are equal regardless of the volume. Because K=[B]2[A] =1, if the volume is such that the [A]=[B]<1.0, then c

Q < K and the reaction goes to the right to reach equilibrium. However, if the volume is such that the [A]=[B]>1.0, then Q>K and the reaction goes to the left to reach equilibrium.

16.106 Kp=0.50 means that P(products)<P(reactants). If the reactants and products are in their standard states, then P of each reactant and product=1.0; so Q>K. To reach equilibrium, the reaction will have to shift to the left.

16.107 An examination of the data shows that whenAP=1.0, P B =1.0; therefore, K p =P BPA = b

b

a

a

=1.0. Therefore, the value of the numerator and denominator must be equal. We see from the data that PB= PA , so P B = PA. Because

√

2

the stoichiometric coefficients become exponents in the equilibrium expression, a=1 and b=2.

16.108 When the concentration of A is increased, the rate of the reaction in the forward direction is increased. This will in- crease the amount of product, B. The rate of reaction in the reverse direction will increase slightly as the concentration of product increases, but not to the same extent as the forward reaction. According to Le Châtelier’s principle, increasing the amount of reactant will increase both the reactant and product at equilibrium. This is consistent with the rate of reaction increasing to a greater extent in the forward direction than in the reverse direction.

16.109 Looking at the plot, at equilibrium [A]=0.59 M and [B]=0.78 M. 2 B(g), K=[B]2[A]=(0.78)20.59

Since A(g)

=1.0. This value is reasonable, since the concentrations of the reactant and product are similar.

⇌

Questions for Group Work

/(K K.)12

16.110 To find the value of K for the new equation, combine the three given equations to yield the new equation. Use equation a as is, reverse and halve equation b, and reverse and halve equation c. To find K for new reaction a. b use c K .

C(s) + O2(g)

⇌ CO(g)2

K

=

1.363×1069

H2O(g) ⇌ H2(g)+12O2(g)

K

=

1K.= b

11.389×1080=8.484_942×10−41

CO() 2g

√ √

CO()+ g

12 O() g 2

⇌

K =

1Kc. =

11.477×10 90 =

√ √

16.111 a. K=[NH]23 [N2 ][H]32 b. Adding H2 will increase the denominator and not change the numerator. c. Since the denominator is larger, Q will be less than K. d. Since Q<K, the denominator needs to be decreased—so the reaction will shift to the right or in the forward direction. e. Yes, the reaction will shift to return Q to K. 16.112 For the first reaction, A ⇌ B, K=[B][A]. For the second reaction, C ⇌ 2 D, K=[D]2[C]≠[D][C]. Since K is the same as the ratio of products to reactants for the first reaction, it is independent of the starting concentrations.

Since K is not the same as the ratio of products to reactants for the second reaction, the ratio of products to reactants is depen- dent on the starting concentrations. The equilibrium constant value is independent of the starting concentrations.

(2) (1) =−1.3×104±√(1.3×104)2− (4) (1) ( −2.6×103) 16.113 a. x2/(0.2−x)=1.3×104→x2=1.3×104(0.2−x)→x2=2.6×103−1.3×104x→ x2+1.3×104x−2.6×103=0x=−b±√b2−4ac The x is small approximation is not valid.

2a

2a b. x2/(0.2−x)=1.3→x2=1.3(0.2−x)→x2=2.6−1.3x→x2+1.3x−0.26=0x=−b±√b2−4ac =−1.3±√(1.3)2−(4)(1)(−0.26)

(2)(1)

=−

=−1.4_7614 or 0.1_76136.

The x is small approximation is not valid.

2

2

2a = c. x/(0.2−x)=1.3×10−4→x2=1.3×10−4(0.2−x)→x2=2.6×10−5−1.3×10−4x→x2+1.3×10−4x−2.6×10−5=0x=−b±√b−4ac

−1.3×10

−4

−4 2

(2) (1) √(1.3×10 ) − (4) (1) ( −2.6×10−5)

±

The x is small approximation is valid.

2

2a = −1.3×10 d. x2/(0.01−x)=1.3×10−4→x2=1.3×10−4(0.01−x)→x2=1.3×10−6−1.3×10−4x→x2+1.3×10−4x−1.3×10−6=0x=−b±√b−4ac

−4

−4 2

(2) (1) √(1.3×10 ) − (4) (1) ( −1.3×10−6)

±

The x is small approximation is not valid.

In order for the x is small approximation to be valid in the expression x2/(y−x)=z, y/z> ∼1000. So y needs to be much larger than z.

16.114a. When a concentration of a reactant is increased, the denominator of Q increases and the reaction must shift to the right. b. When a solid product is added, the value of Q will not change (as long as the volume is not changed) because solids do not appear in the equilibrium expression. c. When the volume is decreased, the reaction will shift to the side of the reaction that has the smaller number of moles of gas. If the number of moles of gas is the same on both sides of the reaction, the reaction will not shift as the volume is changed. d. Heat can be considered a reactant in an endothermic reaction and a product in an exothermic reaction. As the temperature is increased, an endothermic reaction will shift to the right. As the temperature is increased, an exothermic reaction will shift to the left.

Data Interpretation and Analysis 16.115

a. CaSiO3(s) + CO2(g) ⇌ CaCO3(s)+SiO2(s) (System 1) KP=1PCO 2= 191 atm =0.011 MgSiO3(s)+CaCO3(s)+SiO2(s) ⇌ CaMgSi2O6(s)+CO2(g) (System 2) KP=PCO=91 atm=91 2 = 2.9 2.0 PCO2 PCO2 == 800 100atm atm b. For System 1 at 740 K log PCO2 CO2 = For System 2 at 740 K log P c. The partial pressure of carbon dioxide in part a (91 atm) is a little lower than the midpoint partial pressure of carbon dioxide of System 1 in part b, so System 1 is most likely responsible for the carbon dioxide on the surface of Venus. d. For supercritical carbon dioxide to form, the partial pressure of2CO needs to be between 80 atm and 7000 atm and the temperature needs to be above 300 K. In the past on Venus, the partial pressure of CO 2 could have been above 100 atm, and the temperature could have been above 300 K. Yes, supercritical fluid CO2 could have been present on Venus.

17 Acids and Bases Review Questions 17.1 The pain of heartburn is caused by hydrochloric acid that is excreted in the stomach to kill microorganisms and activate enzymes that break down food. The hydrochloric acid can sometimes back up out of the stomach into the esophagus, a phenomenon known as acid reflux. When hydrochloric acid comes in contact with the lining of the esophagus, the H+ ions irritate the esophageal tissues, resulting in a burning sensation. The simplest way to relieve mild heartburn is to swallow repeatedly. Saliva contains bicarbonate ion (HCO− 3) that acts as a base and, when swallowed, neutralizes some of the acid in the esophagus. Heartburn can also be treated with antacids such as Tums, milk of magnesia, and Mylanta. Because these overthe-counter medications contain more base than does saliva, they do a better job of neutralizing the esophageal acid.

17.2 Acids have the following general properties: a sour taste, the ability to dissolve many metals, the ability to turn blue litmus paper red, and the ability to neutralize bases.

Bases have the following general properties: a bitter taste, a slippery feel, the ability to turn red litmus paper blue, and the ability to neutralize acids.

17.3 A carboxylic acid is an organic acid that contains the following group of atoms: O H─O─C─. Carboxylic acids are often found in substances derived from living organisms. Examples include citric acid, malic acid, and acetic acid.

17.4 An Arrhenius acid is a substance that produces H+ in aqueous solution. ∥ An Arrhenius base is a substance that produces OH− in aqueous solution.

17.5 The hydronium ion is HO+.

In water, H+ions always associate with H2O molecules to form hydronium ions and other associated species with the general formula H(H2O) .

3 n

+

17.6 A Brønsted–Lowry acid is a proton (H+) donor.

A Brønsted–Lowry base is a proton (H+) acceptor.

17.7 According to Huheey, “The differences between the various acid–base concepts are not concerned with which is right, but which is most convenient to use in a particular situation.” There is no single correct definition; we use the definition that is best for a particular situation.

17.8 Amphoteric substances can act as acids or bases depending on the circumstances. Some amphoteric substances2O areand H NH3.

17.9 A conjugate acid–base pair consists of two substances related to each other by the transfer of a proton. In the reaction: + and H2O/OH− .

NH(aq)+HO(l) NH 4(aq)+OH−(aq), the conjugate pai rs are NH3/NH4 3 2 +

⇌ 17.10 A strong acid completely ionizes in solution. An example is HCl. A weak acid only partially ionizes in solution. An example is HF. 2SO4.

17.11 A diprotic acid contains two ionizable protons. An example is H 3PO4.

A triprotic acid contains three ionizable protons. An example is H 17.12 An acid ionization constant (K a) is the equilibrium constant for the ionization reaction of a weak acid. The smaller the acid ionization constant, the weaker the acid. + − and has the ion product of water Kw=[H3O][OH]. + − + H2O(l)⇌H 3O(aq)+OH(aq) 17.13 The autoionization of water is 2O(l) H + −14 ⇌H+(aq)+OH−(aq), and the ion product of water is K w=[H][OH].

At−25 °C,Kw=1.0×10 .

It can also be written HO(l) 2 − + − times the con centration of OH− is always 1.0×10 −14 ; therefore, if [H3O]+increases, [OH] decreases 17.14 Because Kw =1.0×10−14, the concentration of HO+ and if the [H3O] decreases, [OH] increases.

3

17.15 We define pH as pH=−log[H 3O+ ]. An acidic solution has a pH<7, a basic solution has a pH>7, and a neutral solution has a pH=7.

17.16 We define pH as pOH=−log[OH−]. An acidic solution has pOH>7, a basic solution has pOH<7, and a neutral solution has pOH=7. + + 17.17 We can neglect the contribution of the autoionization of water to the 3OH concentration in a solution of a strong or weak acid because it is negligible compared to the concentration of H3O ions generated from the dissociation of the acid. Even the weakest of the weak acids has an ionization constant that is four orders of magnitude larger than that of water.

17.18 The x is small approximation is used when the equilibrium constant is relatively small and the initial concentration is relatively large. We assume that x is small compared to the initial concentration. If the assumption is valid, x is <5% of the initial concentration.

17.19 The percent ionization is defined as percent ionization=concentration of ionized acidinitial concentration of acid×100. The percent ionization of a weak acid decreases with the increasing concentration of the acid. +and suppresses the formation of additional H3O + by the ionization of the weak acid. Because of Le Châtelier’s principle, the formation of H3O + 17.20 The complete ionization of the strong acid produces a significant concentration 3O of H by the

strong acid causes the weak acid to ionize even less than it would in the absence of the strong acid. − ⇌BH+( aq)+OH(aq) 17.21 B(aq)+HO() 2 l − 17.22 An anion that is the conjugate base of a weak acid is itself a weak base. A−(aq)+H 2O()l HA(aq)+OH(aq) An anion that is the conjugate base of a strong acid is pH-neutral.

17.23 At 25 °C, the product of K a for an acid and Kb for its conjugate base is Ka×Kb=Kw=1.0×10 −14 . ⇌ 3+ 17.24 A cation that is the conjugate acid of a weak base such as NH+ is aeak w acid. Small highly charged metal cations such as Al and Fe3+ form weakly acidic solutions.

4 + + 17.25 For most polyprotic acids, aK1 is much larger thanKa.2 Therefore, the amount of HO 3 contributed by the first ionization step is much larger than that contributed by the second or third ionization step. In addition, the production 3 of HO by the

first step inhibits additional production of3O H + by the second step because of Le Châtelier’s principle.

17.26 For a weak diprotic acid, [X2−]=Ka.2 This result is general for all diprotic acids in which the x is small approximation is valid.

17.27 For an H─Y binary acid, the factors affecting the ease with which this hydrogen will be donated are the polarity of the bond and the strength of the bond.

17.28 The factors affecting the ease with which the hydrogen will be donated (and therefore acidic) are the electronegativity of the element Y and the number of oxygen atoms attached to the element Y. 17.29 A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.

17.30 A Lewis acid has an empty orbital (or can rearrange electrons to create an empty orbital) that can accept an electron pair. A Lewis base has a lone pair of electrons that it can donate to the Lewis acid.

Problems by Topic The Nature and Definitions of Acids and Bases 17.31

( )+NO HNO3(aq) → H +aq

acid a.

3(aq) −

+ NH4 +(aq ) → H (aq) + NH3(aq)

b. acid c. bas d. e acid

KOH(aq) → K+(aq) + OH−(aq)

+ H C2 H 3O2(a)q→ H (aq) + C2H3O 2(aq) −

17.32 a. base

NaOH(aq)→Na+(aq)+OH−(aq)

b. acid c. acid

HSO(aq)→2H+(aq)+SO2− 2 4

4

(aq)

HBr(aq)→H+(aq)+Br−(aq) Sr(OH)(aq)→Sr2+ (aq)+2OH−(aq)

d. base

2

17.33 +

− a. Because HCO 2 3don ates a proton to H2O, it is the acid. After H2CO3 donates the proton, it becomes HCO3, the conjugate base. Because H2O accepts a proton, it is the base. After H2O accepts the proton, it becomes H3O, the conjugate acid. − b. Because HO donatesthe proton, it becomes OH, the conjugate base. Because NH3 accepts a proton, it is the base. After NH3 accepts the proton, it becomes+ NH4, the conjugate acid.

2 donates a proton to NH, 3 it is the acid. After HO 2 +

− c. Because HNO3 donates a proton to HO, it is the acid. After HNO tes the proton, it becomes NO3, the conjugate base. Because H2O accepts a proton, it is the base. After H2O accepts the proton, it becomes H3O, the conjugate acid.

2 3 dona − + d. Because HO it is the acid. After HO 2 donates a proton to CHN, 55 2 donates the proton, it becomes OH, the conjugate base. Because C5H5N accepts a proton, it is the base. After C5H5N accepts the proton, it becomes C5H5NH, the conjugate acid.

17.34 a. Because HI donates a proton to 2HO, it is the acid. After HI donates the proton, it becomes I−, the conjugate base. Because HO accepts a proton, it is the base. 2After HO accepts the proton, it becomes HO+ , the conjugate acid.

2 3 − + b. Because HO 2 donates a proton to CH3NH2, it is the acid. After H2O donates the proton, it becomes OH, the conjugate base. Because CH3NH2 accepts a proton, it is the base. After CH3NH2 accepts the proton, it becomes CH3NH3, the conjugate acid. − 2− accepts a proton, it is the base. After CO3 2− accepts the proton, it becomes HCO3, − the conjugate c. Because HO donates a proton to CO2−, it is the acid. After HO dona tes the proton, it becomes OH, the conjugate base. Because CO3 2 3 2 acid. + − d. Because HBr donates a proton to 2HO, it is the acid. After HBr donates the proton, it becomes Br, the conjugate base. Because H2O accepts a proton, it is the base. After H2O accepts the proton, it becomes H3O, the conjugate acid.

17.35 Adding H+ to the formula of the base will generate the conjugate acid.

4(aq)+H2O(l)→HO+( a. HSO− HSO− aq)+SO2− (aq) 4

b. HClO4 c. H2CO3

3

4

HClO4(aq)+H2O(l) ⇌HO+( aq)+ClO−4(aq) H2CO3(aq)+H2O(l) ⇌HO+( aq)+HCO−3(aq)

NH+ d. NH+ 4(aq)+H2O(l) ⇌HO+ (aq)+NH3(aq) 4 3 17.36 Removing H+ to the formula of the acid will generate the conjugate base. a. HSO− b. Br− 4 c. I− d. CHO−2

)+HO+ H2SO4(aq) + H2O(l)⇌HSO−( (aq) 4 aq 3 HBr(aq)+HO(l)→Br− (aq)+HO+3 2 2 HI(aq)+HO(l)→I−

(aq) + H3O+(aq)

)+HO+ (aq) HCHO2(aq)+H2O(l) ⇌CHO−( 2 aq 3 + 17.37 HPO−(aq)+HO(l) H3O(aq)+HPO2−( aq) HPO−( 2 4 2 4 2 4 aq)+HO() 2 l ⇌

HPO( 4aq )+OH− (aq) 3 ⇌

17.38 HCO−(aq)+HS− (aq) ⇌H2S(aq)+CO 3 2 − (aq) 3

HCO− 3 (aq) + HS−(aq) H2CO3(aq)+S2−(aq) ⇌

Acid Strength andKa 17.39 a. HNO3 is a strong acid. b. HCl is a strong acid. c. HBr is a strong acid. d. H2SO3 is a weak acid. H2SO3(aq)+H2O()l K=[HO+3 ][HSO3]− [H2SO3] a

17.40

− H3O(+aq )+HSO3(aq)

⇌

+aq − a. HF is a weak acid. HF(aq)+H2O(l)⇌H 3O( )+F(aq) K=[HO+][F−] [HF] 3 a

− H3O(+aq )+CHO2(aq)

b. HCHO2 is a weak acid. HCHO2(aq)+H2O()l K=[HO+][CHO−] 3 2

[HCHO2]

a

⇌

c. H2SO4 is a strong acid. d. H2CO3 is a weak acid. H2CO3(aq)+H 2O()l ][HCO3]− 3 Ka1 =[HO+

− H3O(+aq )+HCO3(aq)

[H2CO3]

⇌ 17.41 The strength of an acid is determined by how much it ionizes and how much undissociated acid is left over. The larger the fraction of the acid that ionized, the stronger the acid. a. contains no HA, 10 H+, and 10 A− b. contains 3 HA, 7 H+, and 7 A− c. contains 9 HA, 1 H+, and 1 A− So solution a > solution b > solution c.

17.42 (b) HB The strength of an acid is determined by how much it ionizes and how much undissociated acid is left over. The larger the fraction of the acid that ionized, the stronger the acid. a. HA contains 9 HA, 1 H+ and 1 A− b. HB contains no HB, 10 H+, and 10 B− So solution HB is a stronger acid than HA. −10 17.43 HCl is a strong acid, K(HF)=6.8×10−4 , K a(HClO)=2.9×10 −8K , a(HC6H5O)=1.3×10 . a

+ id and the greater the [H3O].

The larger the value of K, a the stronger the ac The order of decreasing [HO+] is HCl>HF>HClO>HC6H5O. 3

+ 17.44 HBr is a strong acid, K(HCHO)=1.8 ×10 −4 , HI is a strong acid, Ka(HCN)=4.9×10 −10 . The larger the value of Ka, the stronger the acid and the greater the [H3O ]. a 2 The order of decreasing [HO+] is 0.20 M HI=0.20 M HBr>0.20 M HCHO2>0.20 M HCN. 3

17.45 a. CN− is a stronger base than is ClO−2 .

CN− is the conjugate base of HCN a=4.9×10 (K b. ClO− is a stronger base than is ClO−4 .

−10

); ClO2 − is the conjugate base of HClO2K(a=1.1×10

−2

); the weaker the acid, the stronger the conjugate base.

ClO− is the conjugate base of HClO (a weak acid); ClO− is the conjugate base of HClO4 (a strong acid); the weaker the acid, the stronger the conjugate base.

4

c. HO is a stronger base than is Br−.

2 HO is the conjugate base of HO+ ; Br − is the conjugate base of HBr (a strong acid); the weaker the acid, the stronger the conjugate base.

2

3

17.46 a. ClO− is a stronger base than is F−.

F− is the conjugate base of HF (K a=6.8×10 b. F− is a stronger base than is Cl−.

−4

); ClO − is the conjugate base of HClO (K a=2.9×10

−8

); HClO is the weaker acid, so ClO− will be the stronger base; the weaker the acid, the stronger the conjugate base.

F− is the conjugate base of HF (a weak acid); Cl− is the conjugate base of HCl (a strong acid); the weaker the acid, the stronger the conjugate base. − c. NO− 2 is a stronger base than is NO3.

NO− is the conjugate base of HNO2 (a weak acid); NO3− is the conjugate base of HNO3 (a strong acid); the weaker the acid, the stronger the conjugate base.

2

Autoionization of Water and pH 17.47 a. Given: Kw=1.0×10−14, [H3O+]=1.5×10−3 M Find: [OH−] and classify solution + − Conceptual Plan:[HO]→[OH] Kw =1.0×10−14=[H 3 Solution: Kw= 1.0 × 10−14 = (1.5 × 10−3)[OH− ]

3O+ ][OH − ]

[OH−]=6.7×10−12 M [OH−] < [H3O+], so the solution is acidic. −7 Check: The units (M) are correct. The magnitude (7×10−12 M) makes sense because [H+]>10 M. ] =3.5×10−9 M Find: [OH−] and classify solution b. Given: Kw =1.0×10−14, [HO+ 3 + − K w =1.0×10−14=[H Conceptual Plan:[HO]→[OH] 3 Solution: − Kw= 1.0 × 10−14 = (3.5 × )10−9 [OH ]

+

3O ][ OH−]

[OH−]=2.9×10−6 M [H O+] < [OH− ], so the solution is basic.

3

Check: The units (M) are correct. The magnitude (3×10−6 M) makes sense because [H+]<10−7 M. c. Given: Kw =1.0×10−14, [HO+]=8.3×10−11 M Find: [OH]−and classify solution 3 + − K w =1.0×10−14=[HO+][OH− Conceptual Plan:[HO]→[OH] 3 3 Solution: Kw= 1.0 × 10−14 = (8.3 × 10−11)[OH− ]

]

[OH−]=1.2×10−4 M [H O+] < [OH− ], so the solution is basic.

3

Check: The units (M) are correct. The magnitude (1×10−4 M) makes sense because [H+]<10−7 M. 17.48 [HO+]=8.2×10−9 M a. Given: K=1.0×10−14, w 3

Find: [OH−] and classify solution

+ − K w =1.0×10−14=[HO+][OH− Conceptual Plan:[HO]→[OH] 3 3 − Solution: Kw = 1 . 0 × 1 0 − 1 4 = ( 8 . 2 ×M)[OH] 10−9

]

[OH−]=1.2×10−6 M [H O+] < [OH− ], so solution is basic 3

Check: The units (M) are correct. The magnitude of the answer (1×10−6 M) makes sense because3[HO+]<10−7M. b. Given: Kw=1.0 ×10−14, [HO+ ]=3.1×10 −3 M Find: [OH]−and classify solution 3

+

−

Conceptual Plan:[HO]→[OH] Kw =1.0×10−14=[HO+][OH− 3 3 Solution: Kw=1.0 × 10−14=(3.1×10−3 M)[OH−]

]

[OH−]=3.2×10−12 M [H O+]> [OH−], so solution is acidic 3

Check: The units (M) are correct. The magnitude of the answer (3×10−12 M) makes sense because3[HO+]>10−7M. M Find: [OH]−and classify solution c. Given: Kw=1.0× 10−14, [HO+]=5.8×10−11 3 + Conceptual Plan:[HO]→[OH] 3 Solution:

−

Kw =1.0×10−14=[H

3O+ ][OH − ]

Kw= 1.0 × 10−14 = (5.8 × 10−11 M)[OH]− [OH−]=1.7×10−4 M [ H O +]<[O H −], so solution is basic 3

] < 10 Check: The units (M) are correct. The magnitude of the answer (2×10−4 M) makes sense because3[HO+

−7

M.

17.49 a. Given: Kw= 1.0×10−14, [OH− ]=5.2×10

−3 M

Find: [HO+3] and classify solution

+ − Conceptual Plan:[OH]→[HO] Kw =1.0×10−14=[H 3 Solution: K = 1 . 0 × 1 0 − 1 4 =3 [ H](5.2 O +× 10 −) 3 w [H3 O+] = 1.9 × 10−12 M

+

−

3O ][OH ]

[H O+] < [OH− ], so the solution is basic.

3

Check: The units (M) are correct. The magnitude (2×10−12 M) makes sense because [OH−]>10−7 M. M Find: [HO+3] and classify solution b. Given: Kw =1.0×10−14, [OH−]=8.9×10−9 − Conceptual Plan:[OH]→[HO] 3 + K w =1.0×10−14=[HO+][OH− 3 Solution: Kw= 1.0 × 10−14 = 3[H O+](8.9 × ) 10−9 M [HO+]=1.1×10−6 3

[OH−]<[HO+ 3

]

], so the solution is acidic.

Check: The units (M) are correct. The magnitude (1×10−6 M) makes sense because [OH−]<10−7 M. M Find: [HO+3] and classify solution c. Given: Kw =1.0×10−14, [OH−]=2.4×10−2 − Conceptual Plan:[OH]→[HO] 3 + K w =1.0×10−14=[HO+][OH− 3 Solution: Kw= 1.0 × 10−14 = 3[H O+](2.4 × ) 10−2 M [H3 O+] = 4.2 × 10−13

]

[H3 O+] < [OH− ], so the solution is basic.

Check: The units (M) are correct. The magnitude (4×10−13 M) makes sense because [OH−]>10−7 M. 17.50 a. Given: Kw =1.0×10−14, [OH−]=7.6×10−8

M Find: [HO+3] and classify solution

− + K w =1.0×10−14=[HO+][OH− Conceptual Plan:[OH]→[HO] 3 3 Solution: Kw= 1.0 × 10−14 = 3[H O+](7.6 × M) 10−8 M [HO+]=1.3×10−7 3

]

[H3 O+] > [OH− ], so solution is acidic Check: The units (M) are correct. The magnitude of the answer (1×10−7 M) makes sense because [OH−]<10−7 M. M Find: [HO+3] and classify solution b. Given: Kw =1.0×10−14, [OH−]=5.3×10−3 − Conceptual Plan:[OH]→[HO] 3 + K w =1.0×10−14=[HO+][OH− 3 Solution: Kw= 1.0 × 10−14 = 3[H O+](5.3 × M) 10−3 M [H3 O+] = 1.9 × 10−12

]

[H3 O+] < [OH− ], so solution is basic Check: The units (M) are correct. The magnitude of the answer (2×10−12 M) makes sense because [OH−]>10−7 M. M Find: [HO+3] and classify solution c. Given: Kw =1.0×10−14, [OH−]=1.4×10−13 −

Conceptual Plan:[OH]→[HO] 3 + K w =1.0×10−14=[HO+][OH− 3 Solution: Kw= 1.0 × 10−14 = 3[H O+](1.4 × 10−13 M) M [HO+]=7.1×10−2 3

]

[H3 O+] > [OH− ], so solution is acidic Check: The units (M) are correct. The magnitude of the answer (7×10−2 M) makes sense because [OH−]<10−7 M. 17.51 a. Given: [HO+]=1.7×10−8 3

M Find: pH and pOH

+ pH=−log[HO+ Conceptual Plan:[HO]→pH→pOH ] pH+pOH=14 3 3 Solution: pH=−log(1.7×10−8)=7.77 pOH=14.00−7.77=6.23 pH>7, so the solution is basic.

Check: The units (none and none) are correct. When [H+]<10−7 M the pH>7, pOH<7.

b. Given: [HO+]= 3

1.0×10 −7 M Find: pH and pOH

+ pH=−log[HO+ ] pH+pOH=14 Conceptual Plan:[HO]→pH→pOH 3 3 Solution: pH=−log(1.0×10−7)=7.00 pOH=14.00−7.00=7.00 pH=7, so the solution is neutral.

Check: The units (none and none) are correct. When [H+]=1×10−7 M the pH=7=pOH. c. Given: [HO+]= 2.2×10 −6 M Find: pH and pOH 3

+ pH=−log[HO+ Conceptual Plan:[HO]→pH→pOH ] pH+pOH=14 3 3 pOH=14.00−5.66=8.34 Solution: pH=−log(2.2×10−6)=5.66 pH<7, so the solution is acidic.

Check: The units (none and none) are correct. When [H+]<10−7 M the pH<7, pOH>7.

17.52 a. Given: [HO+]= 3

3.7×10

−3

M Find: pH and pOH

+

3 pH=−log[HO+ ] pH + pOH = 14 Conceptual Plan:[HO]→pH→pOH 3 Solution: pH = − log(3.7 × 10−3) = 2.43 pOH = 14 − pH = 14.00 − 2.43 = 11.57 pH<7, so the solution is acidic Check: The units (none and none) are correct. When 3[HO+]>10−7 M the pH<7, pOH is >7. b. Given: [HO+]=7.0×10−7 M Find: pH and pOH 3

pH=−log[H 3O+] pH + pOH = 14 Solution: pH = − log(7.0 × 10−7) = 6.15 pOH = 14 − pH = 14.00 − 6.15 = 7.85 pH<7, so the solution is acidic + Conceptual Plan:[HO]→pH→pOH 3

Check: The units (none and none) are correct. When 3[HO+]>10−7 M the pH<7, pOH is >7. c. Given: [HO+]=1.8×10−10 3

M Find: pH and pOH

+ pH=−log[HO+ 3 Conceptual Plan:[HO]→pH→pOH ] pH + pOH = 14 3 Solution: pH = − log(1.8 × 10−10) = 9.74 pOH = 14 − pH = 14.00 − 9.74 = 4.26 pH>7, so the solution is basic Check: The units (none and none) are correct. When 3[HO+]<10−7 M the pH>7, pOH is <7.

17.53 a. Given: pH=8.55

[OH− ] Find: [HO+], 3

+ − pH=−log[HO+] K Conceptual Plan:pH→[HO]→[OH] 3 3 3 8.5 Solution: pH=−log[HO+] 5 = − l o g [ H 3+O ]

w

=1.0×10−14=[HO+][OH− 3

]

− 8.55 −9 ] 3 −8.55 = log[H3O+ ] 10 − 8.55 =10 log[HO+ 3 10 =[H 3O+ ] [HO+ ] = 2 . 8 × 1 0 MKw =1×10

−14

= (2.8 × 10

− 9)[OH−] [OH−] = 3.6 × 10 −6

M

M and [OH−]>10−7 Check: The units (M and M) are correct. When pH>7, [HO+]<10−7 3 + Find: [HO], 3 [OH]

b. Given: pH=11.23

3

] Kw =1.0×10−14=[HO+][OH− 3

] pH=−log[HO+ 3

Conceptual Plan:pH→[HO+]→[OH− 3 Solution: pH=−log[HO

+ ] 11.23=−log[H

−11.23 = log[H3O+] 10

−11.23

+

−7

+ Find: [HO], 3 [OH]

+

log[H +3O ] −

10 −11.23 = [H3O+]

M and [OH]>10

−14

] w=1.0×10 ] pH=−log[H 3O+K

3

3

= 10

−7

−12 [HO+ 3 ] = 5 . 9 × 1 0 MKw =1×10

−14

= (5.9 × 10

−12)[OH−] [OH−] = 1.7 × 10 −3

M

M.

−

Conceptual Plan:pH→[HO+]→[OH− Solution: pH=−log[HO

]

3O+ ]

Check: The units (M and M) are correct. When pH>7, [H3O]<10 c. Given: pH=2.87

M.

−

=[H3O + ][OH − ]

3O+ ]

]2.87=−log[H

− 2.87

−2.87 = log[H3O+] 10

3O+ ]10 − 2.87

=1 0log[H

−3 3 =[H 3O+ ] [HO+ ] = 1 . 3 × 1 0 MKw =1×10

−14

−12 = (1.3 × 10 − 3)[OH−][OH−] = 7.7 × 10 M

+ −7 −7 M. Check: The units (M and M) are correct. When pH<7, [H3 O ] > 1 0 M and [OH−]<10

17.54 +

a. Given: pH=5.55

Find: [HO] 3 and [OH]

−

] pH=−log[H 3O+ ]

Conceptual Plan:pH→[HO+]→[OH− 3

Kw = 1.0 × 10 −14 = [H3 O+][OH−]

Solution:

+ 5 . 5 5 = − log[H3O 3 ] pH = − log[H O+] + 3 O+] 10−5.55= 10 log −5.55 = log[H [H3O ]10

−5.55 [ H O + ]=2.8×10 − 6 MKw =10×10 = [H3O+] 3 −7 Check: The units (M and M) are correct. When pH<7,[H3 O+ ] > 1 0 M and [OH−]<10−7 M. + Find: [HO] 3 and [OH]

b. Given: pH=9.01

= (2.8 × 10 − 6 M)[OH−][OH−]=3.5×10

−9

M

−

] pH=−log[H 3O+ ]

Conceptual Plan:pH→[HO+]→[OH− 3 Solution: pH = − log[H 3 O+]

−14

Kw = 1.0 × 10

−14 = [H O+][OH− ] 3

9 . 0 1 = − l o g [ H 3+O ]

+ −9.01 [ H O +]=9.8×10 − 1 0 MKw = 1.0 × 10 −14 = (9.8 × 10 −10 M)[OH−][OH−]=1.0×10 −9.01 = log[H3O+] 10−9.01 = 10log [H3O ]10 = [H3O+] 3 Check: The units (M and M) are correct. When pH>7,[H3 O+ ] < 1 0− 7 M and [OH−]>10−7 M. +

−

+

−

−5

M

c. Given: pH=4.66 Find: [H 3O ] and [OH ] Conceptual Plan: pH→[H 3O]→[OH] pH=−log[H3O] Solution: 3 O+] pH = − log[H + −4.66 = log[H 3O ]

+

Kw = 1.0 × 10

−14 = [H O+][OH− ] 3

4 . 6 6 = − l o g [ H 3+O ] −4.66

log

+

−4.66

−5

+

3 = 10 [H3O ]10 = [ H 3 O ] [HO+ ] = 2 . 2 × 1 0 MKw = 1.0 × 10 −7 −7 M. Check: The units (M and M) are correct. When pH<7,[H3 O+ ] > 1 0 M and [OH−]<10 −14 17.55 pH=−log [H 3O+K ] w=1.0×10 =[H3O + ][OH − ]

10

+

−

[H3O ]

pH

[OH ]

1.4 × 10

−11

3.15

3.7 × 10 −9

2.7 × 10

−6

8.43

−12

−3

7.1 × 10

8×10

6.3 × 10

−4

1×10

−4

1.6 × 10

11.1

−11

3.20

−14

= (2.2 × 10

− 5 M)[OH−][OH−]=4.6×10

−10

M

Acidic or Basic

acidic

basic

basic

acidic

− −14 −9 −6 −9 − 11.1 0 7 . 1 × 1−4 0 = 1 . 4 × 1−11 [H3 O+ ] = 1 0−3.15 = 7 . 1 × 1−40 [ O H− ] = 1 . 0 × 1−14 =8×10 0 [ O H ] = 1 . 0 × 1 0 3 . 7 × 1 0 = 2 . 7 × 1 0 pH = − log(3.7 × 10 ) = 8.43[H3O+ ] = 10 Check: When pH<7,[H+]>10−7 M and [OH−]<10−7 M. When pH>7,[H+]<10−7 M and [OH−]>10−7 M. 17.56 pH=−log[HO+] K =1.0×10−14=[HO+][OH− ] 3

w

−14

[OH−] = 1.0 × 10

8×10

−12

=1×10

−3

[H3O

3

[H3O+ ]

pH

[OH−] −12

2.46

2.6 × 10 − 8

3.8 × 10 − 7

7.58

1.8 × 10 − 9

5.6 × 10 − 6

8.74

7.1 × 10 − 8

1.4 × 10 − 7

7.15

3.5 × 10

−12

−3

2.9 × 10

Acidic or Basic acidic

basic

basic

basic

[ O H− ] = 1 . 0 × 1 0 − 1 4 3 . 5 × 1 0 − 3 = 2 . 9 × 1 0 − 1 2 p H = − l o g ( 3 . 5 × 1 03− 3 ) = 2 . 4 6 [ H O + ] = 1 . 0 × 1 0 − 1 4 3 . 8 × 1 0 − 7 = 2 . 6 × 1 0 − 8 p H = − l o g ( 2 . 6 × 1 0 − 8 ) = 7 . 5 8 [ O=H− 1 .−06 ×pH 1 0=− − 1 4log(1.8 1 . 8 ××110 0 −−)99= 8.74[H 5.6] ×= 10 Check: When pH<7,[H+]>10−7 M and [OH−]<10−7 M. When pH>7,[H+]<10−7 M and [OH−]>10−7 M. 17.57 Given: Kw=2.4×10

−14

+ at 37 °C Find: [H3O], pH −

+ ]→pHK w=[H3O][OH] pH=−log[H3O] Solution: H2O(l) + H2O(l)⇌H +3O(aq)+OH−(aq)

Conceptual Plan: K→[HO+ w 3

+

+ − Kw=[H3O][OH] − For pure water, [H 3 O+ ] = [ O H ] = pH = − log[H + −7 3O ] = − log(1.5 × 10 ) = 6.82

−14 =1.5×10 −7 √Kw=√2.4×10

+ Check: The value of Kw increased, indicating more products formed; so the H3O increases and the pH decreases from the values at 25 °C. w indicates that more products are formed as the temperature increases. According to Le Châtelier, this means that the heat is a reactant. Therefore, the autoionization of water is endothermic.

17.58 The increasing value of K

17.59

a. Given: [H +3O]=0.044 M Find: pH Conceptual Plan: [HO+]→pH

+

Conceptual Plan: [HO+]→pH 3

+

Conceptual Plan: [HO+]→pH p 3

+

pH=−log[H3O 3 Solution: pH=−log(0.044)=1.35_7=1.36 + b. Given: [H 3O]=0.045 M Find: pH pH=−log[H3O Solution: pH=−log(0.045)=1.34_7=1.35 + c. Given: [H 3O]=0.046 M Find: pH

]

]

H = − log[H3O ] Solution: pH=−log(0.046)=1.33_7=1.34 If the pH of the solution did not carry as many decimal places as the significant digits to the right of the decimal point, you would not be able to distinguish a difference in the pH of solutions (b) and (c). Also, solution (a) would have a calculated pH that was too high.

17.60 a. Given: pH=2.50 Find: [H 3O+ ]

Conceptual Plan: pH→[HO+ Solution: pH=−log[HO+3

b. Given: pH=2.51

] pH=−log[H 3O+ ]] + ] 2.50 = − log[H3O

3

] pH=−log[HO+ 3

Conceptual Plan:pH→[HO+3 +

Solution: pH=−log[HO 3

c. Given: pH=2.52

−2.50 = log[H3O+ ]

10

− 2.50

=10

log[HO+ ] 3

10

−2.51 = log[H3O+ ]

10

− 2.51

=10

log[HO+ ] 3

10

−2.52 = log[H O +]

+ 10 −2.52 = 10log[H +3O ]10 −2.52 = [H O +][ H3O]=3.0_2×10

− 2.50

=[H 3O+ ]

−3 −3 [HO+ 3 ] = 3 . 1 _ 6 × 1 0= 3.2 × 10 M

− 2.51

=[H 3O+ ]

−3 −3 [HO+ 3 ] = 3 . 0 _ 9 × 1 0= 3.1 × 10 M

+ Find: [HO] 3

]

2.51=−log[H 3O +]

]

+ Find: [HO] 3

] pH=−log[HO+ 3

Conceptual Plan:pH→[HO+3 +

Solution: pH=−log[HO 3

]

2.52=−log[H 3O +]

]

3

3

−3

=3.0×10

−3

M

The pH values have two digits to the right of the decimal point. This gives two significant figures in the concentration3 of HO+ . If you do not carry out the concentration to two significant figures, all of the concentrations will appear to be the same.

Acid Solutions 17.61 + − a. Given: 0.25 M HCl (strong acid) Find: [HO], 3 [OH], pH

pH=−log[HO+] ] [HCl]→[HO+] 3 3

Conceptual Plan:[HCl]→[HO+]→pH and then [HO+]→[OH− 3 3

[HO+][OH−]=1.0×10−14 3

+ Solution: 0.25 M HCl=0.25 M HO pH=−log(0.25)=0.60 3

− [OH]=1.0×10−14/0.25 M=4.0×10−14 Check: HCl is a strong acid with a relatively high concentration, so we expect the pH to be low and the [OH−] to be small. + − b. Given: 0.015 M HNO3 (strong acid) Find: [HO], 3 [OH], pH

Conceptual Plan:[HNO]→[HO+]→pHand then [HO+]→[OH− 3

3

pH=−log[HO+] 3 ] [HNO]→[HO+] 3 3

3

[HO+][OH−]=1.0×10−14 3

M HO+ pH=−log(0.015)=1.82 Solution: 0.015 M HNO=0.015 3 3 − [OH]=1.0×10−14/0.015 M=6.7×10−13 − to be small.

Check: HNO is a strong acid, so we expect the pH to be low and the [OH] 3

− 3 (strong acid) Find: [HO],+[OH], pH c. Given: 0.052 M HBr and 0.015 M HNO 3

pH=−log[HO+] 3 3 ] [HBr]+[HNO]→[HO+]

Conceptual Plan:HBr]+HNO]→[HO+]→pHand then [HO+]→[OH− [ [ 3 3 3 + Solution: 0.052 M HBr=0.052 M HO and 0.020 M HNO=0.020 M HO 3

3

3

[HO+][OH−]=1.0×10−14 3

+

3

Total H3O+ =0.072 M+0.020 M=0.072 M pH=−log(0.072)=1.14 [OH−]=1.0×10−14/0.072 M=1.4×10−13 − [OH] to be small.

Check: HBr and HNO3 are both strong acids that completely dissociate. This gives a relatively high concentration, so we expect the pH to be low and the

by mass, d. Given: HNO=0.655% 3 Conceptual Plan:

+ − Find: [HO],[OH], pH 3

dsolution =1.01 g/mL

%100

% mass HNO 3→g HNO3→mol HNO3and then g soln→mLsoln→Lsoln→MHNO3 [H + [ H N O3 ] → [3H +O ] p H = − l o g 3O ]

M HNO→M HO+→pHandthen[HO+]→[OH−] 3 3 3 Solution: 0.655 g HNO3 100 g soln

×1 mol HNO63.02 g HNO 3 ×1.01g solnmL soln 3

×1000mL soln

0.105 M HNO3= 0.105 M H3O +

1.01 g solnmL soln

mol HNO363.02 g HNO3

1000 mL soln

Ls

[H3O+][OH−] = 1.0 × 10−14 L soln

=0.105 M HNO3

− pH=−log(0.105)=0.979[OH]=1.00×10

−14

/0.105 M=9.52×10

−14

Check: HNO3 is a strong acid that completely dissociates. This gives a relatively high concentration, so we expect the pH to be low and the [OH−] to be small.

17.62 a. Given: 0.048 M HI (strong acid) Find: pH ] [HI]→[H 3O+]pH = − log[H3O+ ]

Conceptual Plan:[HI]→[HO+]→pHand then[HO+]→[OH− 3

3

[H3O+][OH −] = 1.0 × 10−14

Solution: 0.048 M HI=0.048 M HO +

pH=−log(0.048)=1.32 Check: HI is a strong acid with a relatively high concentration, so we expect the pH to be low. b. Given: 0.0895 M HClO4 (strong acid) Find: pH 3

3O+ ]

+ [HClO]→[HO] 4 3pH=−log[H

Conceptual Plan:[HClO]→[HO+]→pH 4 3 +

Solution: 0.0895 M HClO=0.0895 M HO

pH=−log(0.0895)=1.048 Check: HClO4is a strong acid, so we expect the pH to be low. c. Given: 0.045 M HClO 4and 0.048 M HCl (strong acids) Find: pH Conceptual Plan:[HCl]+HClO]→[HO+ + [ 3O ] pH=−log[H3O 3 ] → pH [HCl]+[HClO]→[H 4 4 + 0.048 M HCl=0.048 M H Solution: 0.045 M HClO=0.045 M HO and 3O+ 4

3

4

+

]

3

+

Total HO =0.045 M+0.048 M=0.093 M pH=−log(0.093)=1.03 3 Check: HClO4and HCl are both strong acids that completely dissociate. This gives a relatively high concentration, so we expect the pH to be low. d. Given: HCl=1.09% by mass, dsolution =1.01 g/mL Find: pH Conceptual Plan: %100 1 mol HCl36.46 g HCl % mass HCl→gHCl→molHCland thengsoln→mLsoln→Lsoln→MHCl

1.01 g solnmL soln

[HCl]→[HO]3 + pH=−log[HO]3 + 3 Solution: 1.09 g HCl100 g soln×1 mol HCl36.46 g HCl×1.01 g solnmL soln M HCl→MHO+→pH

×1000 mL soln 1 L soln =0.301_9 M=0.302 M HCl + 0.302 M HCl=0.302 M HO pH=−log(0.302)=0.520 3 Check: HCl is a strong acid that completely dissociates. This gives a relatively high concentration, so we expect the pH to be low.

17.63 a. Given: pH=1.25, 0.250 L

Find: g HI

Conceptual Plan:pH→[HO+

3

+ Solution: [HO]=10 3

b. Given: pH=1.75, 0.250 L

3

3

→

molHI

→ 1L

[HO+]→[HI] mol= MV g HI pH=−log[HO+] 3 3 ×0.250 L×127.91 g HI1 mol HI =1.8 g HI

g=127.91 g HI1 mol HI

]

→

[HI]→molHI→g HI

g=127.91 g HImol HI [HO+]→[HI] mol= MV 3 ×0.250 L×127.91 g HI1 mol HI =0.57 g HI

pH=−log[HO+] 3 L

Find: g HI

Conceptual Plan:pH→[HO+

3

3

[HI]

−1.75 =0.017_8 M=[HI]0.017_8 mol HI

c. Given: pH=2.85, 0.250 L + Solution: [HO]=10

→

Find: g HI

Conceptual Plan:pH→[HO+ + Solution: [HO]=10

]

−1.25 =0.056 M=[HI]0.056 mol HI

]

→

[HI]

→

molHI→g HI L

−2.85 =0.0014 M=[HI]0.0014 mol HI

pH=−log[HO+] 3

[HO+]→[HI] 3

×0.250 L×127.91 g HI mol HI

mol= MV =0.045 g HI

g=127.91 g HImol HI

Check: The units () g are correct. The magnitudes of the answers make sense since the volume is small (<1 L). The higher the pH, the lower the amount of HI. 17.64 a. Given: pH=2.50, 0.500 L

Find: g HClO4

+ + Conceptual Plan:pH→[HO+3 ] → HClO4] [ → mol HClO4 → g HClO4pH=−log[H 3O][H3O]→[HClO4] Solution: − 2.50

[H3 O+ ] = 1 0

=0.0031_6 M=[HClO 4]0.0031_6 mol HClO4

mol=MV

L

g=100.46 g HClO41 mol HClO4

×0.500 L×100.46 g HClO4

1 mol HClO4

b. Given: pH=1.50, 0.500 L Find: g HClO4 Conceptual Plan: pH→[HO+ ]→HClO4]→molHClO4→g HClO4 [ 3

3 pH=−log[HO+][

H3O+ ] → [HClO4]

mol= MV

g=100.46 g HClO4 1 mol HClO4

=0.16 g HClO 4

1000 mL soln 1 L soln

mol HClL so

Solution: [H3 O+] = 10− 1.50 =0.031_6 M=[HClO 4]0.031_6 mol HClO4

L

×0.500 L×100.46 g HClO4

mol HClO4

=1.6 g HClO 4

c. Given: pH=0.50, 0.500 L Find: g HClO 4 Conceptual Plan: pH→[HO+ 3

pH=−log[HO+][ 3

]→[HClO4]→mol HClO4→g HClO4

H3O+ ] → [HClO4]

g=100.46 g HClO4 1 mol HClO4

mol= MV

Solution: [HO+]=10−0.50=0.31_6 M=[HClO]0.31_6 mol HClO 3

4

L

4

×0.500 L×100.46 g HClO41 mol HClO4

=16 g HClO4

4.

Check: The units (g) are correct. The magnitudes of the answers make sense since the volume is small (<1 L). The higher the pH, the lower the amount of HClO 17.65 Given: 121 mL HCl,25.0 °C, 0.855 atm, 1.5 L solution Find: pH Conceptual Plan: vol HCl→mol HCl→[HCl]→[HO+

PV=nRT

]→pH

[HC]=[H 3O]+pH=−log[H3O]

M=mol HClL soln

+

(1 L1000 mL)(0.08206·L )((25=.00+.0207432.125_)K84)9 atm mol mol·K

Solution: n=(0.855 atm)(121 mL) [HCl]=0.00422_849 mol

3

+

1.5 L

=0.0028_20 M=[H 3O ] pH=−log(0.0028_90)=2.55 Check: The units (none) are correct. The magnitude of the answer makes sense since HCl is an acid and we will have a low pH. Find: vol soln 17.66 Given: 36.0% HCl, d soln=1.179 g/mL; 10.00 L, pH 2.10 Conceptual Plan: %100

% mass HCl→g HCl→mol HCl and then g soln→mL soln→L soln→M HCl

[HO+]=10−2.10=0.0079_433 MVM=VM

1 mol HCl36.46 g HCl + [H3O]=10

[HCl]→[HO+]

3O]+and the M1, V2, M2→V 1 Solution: 36.0 g HCl100 g soln×1 mol HCl36.46 g HCl×1.179 g solnmL soln M HCl→M HO+ and then pH→[H 3

3

×1000 mL soln

1 L soln

1.179 g soln1 mL soln

−pH

1000 L soln1 L soln

mol HClL soln

V1M1=V2M2

=11.6_4 M HCl

V(11.6_4 M)=(10.00 L)(0.0079_433 M) V1=0.0068_241 L=6.8 mL 1 Check: The units (mL) are correct. The magnitude of the answer (7 mL) makes sense since we are starting with a very high concentration of HCl (a strong acid). + Find: 17.67 Given: 0.200 M benzoic acid,Ka =6.5×10−5 3 [HO], pH ] Conceptual Plan: Write a balanced reaction. Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x . Determine [HO+ 3 and pH. + − Solution: HC7H5O2(aq)+H2O(l) H3O (aq)+C 7H5O2 (aq) 3

1

1

2

2

⇌

Initial

Change

0.200

0.0

0.0

‒x

+x

+x

x

x

0.200‒ x

Equil

= (x)(x)(0.200 − x) = 6.5 × 10−5 [ 3O+][C7 5O2 H ]− [HC7H5O2] Ka = H Assume that x is small compared to 0.200. x2=(6.5×10−5)(0.200); x=3.6×10−3 M=[H 3O+ ] Check assumption: 3.6×10−3 0.100 pH = − log(3.6 × 10−3) = 2.44

×100%=1.8%; assumption is valid.

Check: The units (M and none) are correct. The magnitude of the pH(3) makes sense since we expect a pH>−log(0.200)=0.700 when we have a weak acid. + 17.68 Given: 0.100 M formic acid,Ka =1.8×10 −4 Find: 3 [HO], pH Conceptual Plan: Write a balanced reaction. Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x . Determine [HO+ ] 3 and pH. HO+(aq)+CHO− (aq) Solution: HCH2O(aq)+H2O(l) 3 2 ⇌

Initial

Change

0.0

0.0

‒x

+x

+x

x

x

0.100‒ x

Equil

K = [H O+][CH2 O−] a

0.100

3

[HCH2O]

= (x)(x)(0.100 − x) = 1.8 × 10−4

Assume that x is small compared to 0.100. x2=(1.8×10−4)(0.100); x=4.2×10−3 M=[H

3O+ ]

Check assumption: 4.2×10−3 0.100 ×100%=4.0%; assumption is valid. pH = − log(4.2 × 10−3) = 2.37 Check: The units (M and none) are correct. The magnitude of the pH(2) makes sense since we expect a pH>−log(0.100)=1.000 when we have a weak acid.

17.69 −4 a. Given: 0.500 M HNO2, a=4.6×10 K Find: pH Conceptual Plan: Write a balanced reaction. Prepare an ICE table, represent the change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x . Determine ] and pH. [HO+ 3 Solution: HNO2(aq ) + H2O(l ) H3O+aq ( ) + NO2 −(aq)

Initial

⇌

0.500

0.0

0.0

‒x

+x

+x

x

x

Change

0.500‒ x

Equil

[HNO2] =(x)(x)(0.500−x)=4.6×10−4 Ka =[HO+][NO− 3 2 ] Assume that x is small compared to 0.500. x2=(4.6×10−4)(0.500); x=0.015 M=[HO+3 ] Check assumption: 0.0150.500×100%=3.0%; assumption is valid. pH = − log(0.015) = 1.82 Check: The units (none) are correct. The magnitude of the answer (2) makes sense since we expect a pH>−log(0.500)=0.301 when we have a weak acid. b. Given: 0.100 M HNO, 2K a =4.6×10 −4 Find: pH Conceptual Plan: Write a balanced reaction. Prepare an ICE table, represent the change with x , sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x . Determine [H 3O +] and pH. l HNO2aq ( ) + H O( H3O +aq ( ) + NO ( )2 − aq 2 ) Solution: Initial

⇌

0.100

0.0

0.0

‒x

+x

+x

x

x

Change

Equil

[HNO2] =(x)(x)(0.100−x) Ka =[HO+][NO− 3 2 ] Assume that x is small compared to 0.100. x2=(4.6×10−4)(0.100) x=0.0068 M=[H 3O+ ]

0.100‒ x

−4

= 4.6 × 10