Introduction to chemical engineering thermodynamics 8th edition smith solutions manual 1

Solution Manual for Introduction to Chemical Engineering

Thermodynamics 8th Edition Smith

Ness Abbott Swihart 1259696529 9781259696527

Download full solution manual at: https://testbankpack.com/p/solution-manual-for-introduction-to-chemicalengineering-thermodynamics-8th-edition-smith-ness-abbott-swihart-12596965299781259696527/

4.1.(7th edition prob. 4.1)

If we have set up a spreadsheet toevaluate the heat capacity integral like the example spreadsheet fromthe lecture notes, we can use it to evaluate the heat capacity integral in each case, and simplyenter the heat capacity parametersand read out the value of the integral:

(a) If 10 mol of SO is heated from 200 to1100C, the heat requirement will be 10*R*ICPH,where ICPH is the heat capacity integral Using our spreadsheettoevaluate ICPH, we get

(b) Similarly, if we heat 12 moles of propane from 250 to 1200 C, the heat requirement is 12*R*ICPH Putting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives

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4.2 (7th edition Prob. 4.2)

If we have set up a spreadsheet toevaluate the heat capacity integral like the example spreadsheet fromthe lecture notes, we can use it to evaluate the heat capacity integral in each case, and simply vary the final temperature until we get the desired Q

(a) If 800 kJ is added to10 molof ethylene, then Q = H = 800000 J/10 mol= 80000 Jmol-1 So, H/R = 80000 J mol-1 /8.314 J mol-1 K-1 = 9622 K, sotheintegral of C /R from 200C (473K) to the final temperature should be 9622 K Using our spreadsheettoevaluate this, we get

So,

(b) Similarly, if we add2500 kJ to15 molof 1-butene,we have H/R = 2500000 J/ (15 mol*8.314 J mol1 K-1 )= 20045 K Putting the heat capacity coefficients for 1-butene into the spreadsheet and trying final temperatures until we get this value, we get

(c) Thisis the same, butin English units, in whichwe can use R = 1.986 Btu lbmol

4.3 (New)

Tocompute the outlet temperature, use

But first the heat capacity at 373.15 K must be determine for eachcompound using the Table C.1 and Eqn 4.5. This can easily be done in a spreadsheet

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4.4 (7th edition Prob. 4.3)

Lookingin table A.2, we are reminded that the gas constant is R = 0.7302 ft3 atm (lb mol)-1 R-1 in the sort of units usedin this problem, sowe canconvert the volumetric flow rate to amolar flow rate using the ideal gas law(airat atmospheric conditions is very nearly an ideal gas) n

n = 0.59 lbmol s -1

To find the heat needed to heat the air at constant pressure from 122F to 932 F,we need to integrate the heat capacity over that temperature range

932 F

Q =

nC pdT

122 F

The ideal gas heat capacity for air is given in table C.1 as

Cp = R( A + BT + DT 2 )

T must be in Kelvins for use in this expression, so we convert 122

to 323.15

Note that becausethe heat capacity integral comesout with units of K, while we are using R with units of R, we havetomultiply by1.8 R/K

4.5 (7th edition Prob. 4.4) Howmuch heat is required when 10,000 kgof CaCO is heated at atmospheric pressure from50°C to 880°C? Thenumber of moles of CaCO is 107 g/ 100.09 g mol-1 = 99910.08 mol CaCO

4.6. (7th edition prob. 4.5)

For consistencywith the problem statement,we rewrite Eq. (4.8) as:

. DefineCPam asthe value of CP evaluated at the arithmetic mean temperature Tam. Then: Where

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Define δ as the difference between the two heat capacities:

This readily reduces to: Making the = T2/T re e nswer.

��2/��1 = ��2 ��1 4.7.

For consistency with the problem statement, we rewrite Eq. (4.8) as

C = A + B T  + + D ( p ) 1 ( 1) 2 T 2

subst ion τ ields the quir d a 1

As in the precedingp

Whence,

Define δ as the difference between the two heat cap

This readily reduces to:

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Making the substitution τ =

T2

��1 � � 1

4.8. (7th edition prob. 4.7)

Given  P  T = T 2 R/C p Solve for Cp 2 3    P3 

4.9 (7th edition prob. 4.8)

Except for the noble gases [Fig (4.1)], CP increases with increasing T Therefore, the estimate is likely to be low

4.10 (7th edition Prob. 4.9)

(a) In this part, we use equation 4.14 reduced temperature atanother reduced temperature. The criticaltemperature for n-pentane (from appendix

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B) is 469.7 K So, 25C corresponds to areduced temperature of Tr1 = 298.15/469 7 = 0.6348. The normal boiling point for n-pentane(also fromappendix B) is 309.2 K, which correspondstoa reduced temperature

Tr2 = 309.2/469.7 = 0.6583. So, using equation 4.14, we have

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This is essentially exact agreement with the handbook value (it differs by0.03%)

(b) In this part, we estimate the heat of vaporization at the normal boiling point without using the information on the heat of vaporization at 25C, using equation 4.13 also use the criticalpressure, Pc = 33.70 bar for n-pentane. Then, using equation 4.13:

Dividing by the molecular weight of n-pentane (72.15 g/mol)gives Hn = 358.6 J g . This is just 0.4% higher than the handbook value, and it only required knowledge of the critical properties and the normal boiling point temperature.

4.11. (Like7th edition 4.10, but indifferent units, solutions below are for7th edition version and

We want toevaluate the latent heat of vaporization using the Clapeyron equation:

(a) At -16C, we seethat the vapor volume is 0.12551 m 3kg-1and the liquid volume is 0.000743 m

= 0.124767 m 3kg-1 . Now,we can estimate the slope of the vapor pressurecurve fromthe values at -18, -16, and -14 C, whichare 1.446,1.573, and 1.708 bar, respectively. So, we could estimate the slope as

(b) At 0C, we seethat the vapor volume is 0.069309 m 3kg-1 and the liquid volume is 0.000722 m 3kg-

, so

V = 0.068587 m 3kg-1 . Now,we can estimate the slope of the vapor pressurecurve fromthe values at -2, 0, and2 C, which are2.722,2.928, and 3.146 bar, respectively. So, we could estimate the slope

H = (273 15 + (0))K*0 068587m3 kg

(c) At 12C, we seethat the vapor volume is 0.046332

andthe liquid volume is 0.000797

, so V = 0.045535 m 3kg-1 . Now,we can estimate the slope of the vapor pressurecurve fromthe values at 10, 12, and 14 C, which are 4.146,4.43, and 4.732 bar,respectively. So,we could estimate the slope as

H = (273 15 + (12))K*0 045535m

(d) At 26

we seethat the vapor volume is 0.029998 m

andthe liquid volume is

kg-1 Now,we can estimate the slope of the vapor pressurecurve fromthe values at 24, 26, and 28 C, which are 6.458,6.854, and 7.269 bar, respectively So, we could estimate the slope as

= 0.029167 m

(e) At 40C, we seethat the vapor volume is 0.019966 m 3kg-1 andthe liquid volume is 0.000872 m

kg-

= 0.019094 m 3kg-1 Now,we can estimate the slope of the vapor pressurecurve fromthe values at 35, 40,and 45 C, which are 8.87, 10 166 and 11 599 bar, respectively. So, we could estimate the slope as

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(a) Equation 4 14 is used to estimate the latent heat at a temperature of interest from the (known) latent heat at some other temperature, and may be written as:

For chloroform, the critical temperature is 536 4 K and the normal boiling point is 334.3 K At 273

This is 0.8% lower than the handbook value. Similarlyfor

and

This is 4.1% below the handbook value. Similarly

(also known ascarbon tetrachloride!), T

so

This is 0.5% below the handbook value.

(b) Equation 4 13 is used to estimate the heat of vaporization at the normal boiling point without knowing the heat of vaporization at any temperature. It can be written as

where P

, the criticalpressure, must be given in bar For chloroform, we get

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4.13 (7th edition Prob. 4.12)

Again, wewant to usethe Clapeyron equation to findthe latent heat of vaporization from the vapor pressure curve and molar volumes.

First, we should find the normal boiling point bysetting Psat = 101.3 kPa. Taking the derivative of the Antoine equation for the vapor pressuregives

or,using the numbers for benzene

For T = 353.2 K and Psat = 101.325 kPa this gives

Now,we can use generalized correlations to find the liquid and vapor volumes. The critical properties of benzene are Tc = 562.2 K, Pc = 48.98 bar,  = 0.210, Zc = 0.271, and Vc = 259 cm mol-1

For the volume of the saturated liquid, we can usethe Rackett equation, which gives

Since the reduced pressure is only 1.013/48 98 = 0.0207, we expect thatthevapor will not be far from ideal, even though the reduced temperature is only 353.2/562 2 = 0.6282. We can get the compressibility fromthe Pitzer correlation as in the previous homework:

Z =1 (0 8049 + 0 210 1 0730)

So, V = ZRT/P = 0.9661*83.145*353.2/1 =

So,

= 28370 -- 82.6 = 28288 cm

mol-1 . Finally,putting this all together, we have

So, our final answer is H = 31.15 kJ/mol.

7th edition Prob. 4 12, chooseethylbenzene

We want to usethe Clapeyron equation to find the latent heat of vaporization from the vapor pressurecurve and molar volumes.

First, we should find the normal boiling point bysetting Psat = 101.3 kPa. That gives

101.325) = 13.9726

T = 3259 93 212.3 = 136.2

C 13.9726 ln (101.325)

Reassuringly, this agrees with the value given in Table B.1exactly (tothe precision given).

Taking the derivative of the Antoine equation for the vapor pressuregives

or,using the numbers for ethylbenzene

3259 93P sat =

(T + 212.3)2

For T = 136.2 °Cand Psat = 101.325 kPa this gives

3259.93 101.325 = = 2 720 kPa K

(136 2 + 212 3)2

Now,we can use generalized correlations to find the liquid and vapor volumes The critical properties of ethylbenzene are Tc = 617 2 K, Pc = 36.06 bar,  = 0.303, Zc = 0.263, and Vc = 374 cm mol-1 .

For the volume of the saturated liquid, we can usethe Rackett equation, which gives

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Because the reduced pressure is only 1.013/36.06 = 0.0281, we expect that the vapor will not be far from ideal, even though the reduced temperature is only 353.2/562 2 = 0.6632. We can get the compressibility fromthe Pitzer correlation: B

So,

mol

. Finally, putting this all together, we have

So, our final answer is H = 35.7 kJ/mol This agrees with the value in the table to within a few tenths of a percent -- essentially perfect agreement

7th edition Prob. 4.12: Dothis for toluene and phenol.

Here, we want touse the Clapeyron equation tofind the latent heat of vaporization fromthe vapor pressure curve and molar volumes. The Clapeyron equation is:

We will use it, with the known vapor pressure curve and volume change tocompute the enthalpy of vaporization.

(a)

Toluene

First, we should find the normal boiling point by setting Psat = 101.3 kPa. Using the Antoine parameters for toluene this gives

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Reassuringly, this agrees with the value given in Table B.2essentially exactly.

Taking the derivative of the Antoine equation for the vapor pressuregives

or,using the numbers for toluene

For T = 110.6 Cand Psat = 101.325 kPa this gives

Now,we can use generalized correlations to find the liquid and vapor volumes. The critical properties of toluene are

For the volume of the saturated liquid, we can usethe Rackett equation, which gives

Since the reduced pressure is only 1.013/41.06 = 0.0247, we expect thatthevapor will not be far from ideal, even though the reduced temperature is only 383.8/591 8 = 0.6485. We can get the compressibility fromthe Pitzer correlation as in the previous homework:

So, V = 30297 -- 117.7 = 30180 cm 3 mol-1 . Finally, putting this all together, we have

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So, ourfinal answer is H = 33.3 kJ/mol This agrees with the value in table B 2 to within about 0.4%, which is about the precision of our calculations -- essentially exact agreement

(b)Now, we repeat the whole thing forphenol -- just forpractice:

First, we should find the normal boiling point by setting Psat = 101.3 kPa. Using the Antoine parameters for phenol, this gives

(101 325) = 14 4387

T = 175.4 + 3507.8 = 181.8 C 14 4387 ln (101 325

Reassuringly, this agrees with the value given in Table B.2essentially exactly.

Taking the derivative of the Antoine equation for the vapor pressuregives

or,using the numbers for phenol

3507 8Psat = (T +175.4)2

For T = 181.8 Cand Psat = 101.325 kPa this gives

sat 3507 8101 325 = = 2 786 kPa K

(181.8 +175.4)2

Now,we can use generalized correlations to find the liquid and vapor volumes The critical properties of phenol are T = 694.3 K, P = 61.3 bar,  = 0.444, Z = 0.243, and V = 229 cm 3 mol-1

For the volume of the saturated liquid, we can usethe Rackett equation, which gives

Since the reduced pressure is only 1.013/61 3 = 0.01653, we expect thatthevapor will not be far from ideal, even though the reduced temperature is only 455/694.3 = 0.6553 We can get the compressibility fromthe Pitzer correlation as in the previous homework:

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putting this all together, we

So, our final answer is H = 45.88 kJ/mol This agrees with the value in table B.2 to within about 0.7%, which is about the precision of our calculations -- essentially exact agreement

4.14 (7th edition Prob. 4.13)

Using the equation above, solve for the pressure

Differentiate the above equation and determine dP/dT

Using the Clapeyron equation, solve for V

Then using Eq. 3.37, Solve for B

4.15 (7th edition Prob. 4.14)

Tofind the total heatexchanger duty, we must consider three process steps: (1)heating of the subcooled liquid from300 K toits saturation temperature at 3bar, (2) vaporization at the saturation temperature, and (3) heating of the vapor from the saturation temperature to500 K

For the first step, we do a heat capacity integral from300 K to368.0 K, using the liquid phase heat capacity:

For the second step, we need the heat of vaporization at 3bar and 368.0 K In table B.2, we find the heat of vaporization at the normal boiling point (337.9 K) is 35.21 kJ/mol Tofind the heat of vaporization at 368 K, we can apply the Watson equation (p.134 of SVNA) This gives:

Finally, for the thirdstep, we have:

Comparing these, we seethat the vaporization step has the largest contribution tothe overall heat requirement Adding them gives Q = H = 62.64 kJ/mol/ If the total methanol flow is 100kmol/hr = 27.78 mol/s, then the total heatexchanger duty is 62.64*27.78 = 1740.13 kW.

4.16. (New)

Need togather several things for this proble, the boilingpoint, the latent heat of vaporization, and the heat capacities for eachcompound. Using the equation

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4.17 (7th edit )

ionProb.4.15

Estimate ΔHvusingRiedel ) (414)

equation(413 andWatsoncor ection

Assumethethrottlingprocessisadiabaticandisenthalpic.

4.18 (7th edition Prob. 4.16)

In each case, we can look upthe heat of formation of the compound as a gas in table C.4, where we find -1 -1

Hf , (acetylene(g)) = 227480 J mol , Hf(1,3-butadiene(g)) = 109240 J mol , Hf(ethylbenzene(g)) = 29920Jmol-1 , H (n-hexane(g)) = -166920 Jmol-1 , H(styrene(g)) = 147360 Jmol-1 . The heatof formation f f of the liquid is equal to the heat of formation of the gas minus the heat of vaporization at 25C If we only know the critical properties and normal boiling point of each species, we could estimate this by first using equation 4.12to estimate the heat of vaporization at the normal boiling point of each species, andthen using equation 4.13 toestimate the heat of vaporization at 25C We could, if we like, combine the equations to get

This is getting a bit cumbersome totype intothe oldcalculator, so let's evaluate it in a spreadsheet Doing so gives:

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In the final column above, we have subtracted the latent heat of vaporization at 25 C from the heat of formation of each species in the gas phase toget the heat of formation of each species in the liquid phase

4.19 (7th edition Prob. 4.17)

1st law: dQ = dU dW = CV dT + P dV (A)

Ideal gas: P V = R T and P dV + V dP = R dT

Whence V dP = R dT P dV (B)

Since P V = const then P V 1 dV = V dP

fromwhich V dP = P dV

Combines with (B) toyield:

Combineswith(A)togive:

Whichreducesto: Or

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Since CP is linear in T, the meanheatcapacity isthe value of CP at the arithmetic meantemperature Thus

With , and integrating (C):

4.20 (7th edition Prob. 4.18)

The heat of combustion of 6CHOH(g) is the heat of reaction for

+ 12*(241.82)-6*(-200.66) = -4059 kJ mol-1

The heatof combustion of CH (g) (assumed to be1-hexene, though it doesn't really say and there aremany

other possible CH isomers) is given bythe heat of reaction for

(g)+ 9O → 6CO (g)+ 6H O(g)

+ 6*(-241.82)(-41 95) = -3770 kJ mol-1

The difference between these two heats of combustion is the heat of reaction for the condensation reaction given in the problem statement That is, subtracting the second combustion reaction above from the first gives

6CHOH(g) → CH (g)+ 6H O(g)

3 6 12 2

For which the heat of reaction is -4059 - (-3770) = -289 kJ/mol

The heat of combustion of ethylene at 25C (with water vapor product) is the heat of reaction for

CH (g)+3 O(g) → 2CO (g)+ 2H O(g)

Thisis given by Hrxn = 2*Hf(CO2(g) ) + 2*Hf(H

(g) ) = 2*(-393.51) + 2*(-241.82)(52.51) = -1323 kJmol-1 = -1.323106 J mol-1 Theonly difference in each case is how much excess O and have tobe heated up bythe heat released byreaction.

(a) The products (per mole of ethylene burned) will be 2 moles of CO + 2 moles of H O + 3*(79/21) =

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this mixture from 298 K tosome final temperature and vary the final temperature until we get H/R =

1.323106 J mol-1/8.314 J mol-1 K-1 = 159120 K. Doing so, usingour handy-dandy heat capacity integrating spreadsheet gives 2533 K

(b) Similarly, here the products are be 2 moles of CO + 2 moles of H O + 0.25*3 = 0.75 moles of O +

3*1.25*(79/21) = 14.1 moles of N , and integrating the heat capacity of this mixture toget get H/R = 159120 K gives a final temperature of 2198 K

(c) and (d) are the same as (a) and (b), butwith different numbers that lead to final temperatures of 1951 and 1609 K, respectively

(e) Here, we can add the heat required to heat the air from 298K to 773K to the heat of reaction. We can imagine a path in whichwe coolthe air from 773K to 298K (removing heat H = 309424 J per mol

6

CH burned), carry out thereactionat298K (releasing the standard heatof reaction, Hrxn = -1.32310 )

and then heat the products with both the heat removed to cool the air to298K and the heat of reaction

6 (Htot = 1 66310 J per molCH burned), so the total Htot /R is 196330 K Both the heatremoved to cool theairfrom773to 298K andthe heatrequired to heatthe reactionproducts to thefinaltemperature are again computed using the heat capacity integrating spreadsheet This gives a final temperature of 2282 K. Todownload the spreadsheet I used tosolve this problem, click here.

NOTE: A problem with this homework problem is that several of the final temperatures areabove 2000 K, whereas the heat capacity expressions used (from table C.1 in the book) are only valid up to 2000 K. So, the results for parts (a), (b),and (e) are probably all incorrect (they are the correct answers to the homeworkproblem, but probably are not good values for the actual adiabaic flame temperature of ethylene)

4.21 (7th edition Prob. 4.19)

C2H4 + 3O2 = 2CO2 + 2H2O(g)

ΔH298=[2 ( 241818) + 2 ( 393509) 52510] = 1323164 J/mol

Parts (a) - (d) can be worked exactly asExample 4.7. However, with Mathcad capableof doing the iteration, it is simpler toproceed differently.

Index the product specieswith the numbers:

1= oxygen

2= carbon dioxide

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4= nitrogen

For the products

The integral is given by Eq (4.8) Moreover, byan energy balance,

Parts(b),(c),and(d)areworkedthesameway,theonlychangebeinginthe numbersofmolesofproducts.

(e)50%

airpreheatedto500degC.Forthisprocess,

For4.5/0.21=21.429molesofair:

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Given

(f)theoreticalamountofpureoxygen,

Starting with a guess and solving iteratively yields

4.22 (similar to 7th edition Prob. 4.20)

4.22

(a) the heat of combustion of methane gas in the heat of reaction for

4 + + 2O2 (g) → CO2 (g)+ 2 H2O (l)

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(b)

(c)The heat of combustion of ethylene gas is the heat of reaction for

2*(-285.83)-(-52

(d) The heat of combustion of propane gas is the heat of reaction for

(e) The heat of combustion of propylene, gas is the heat of reaction for

= 6*( -393.51) + 6*(-285.83)-2*(-19.710) = -4036.62 kJ mol-1

(f) The heat of combustion of n-butane , gas is the heat of reaction for

= 4*( -393.51) + 5*(-285.83)-(-125.790) = -2877.4 kJ mol-1 .

(g) The heat of combustion of 1-butene, gas is the heat of reaction for

= 4*( -393.51) + 4*(-285.83)-(-0.540)

(h) The heat of combustion of ethylene oxide, gas is the heat of reaction for

= 4*( -393.51) + 4*(-285.83)-(-0.540)

(i) The heat of combustion of acetaldehyde gas is the heat of reaction for

= 2*( -393.51) + 2*(-285.83)-(-166.190) = -1192.49 kJ

(j) The heat of combustion of methanol gas is the heat of reaction for

= 2*( -393.51) + 4*(-285.83)-2*(-200.660) = -1529.02 kJ mol-1 .

(k) The heat of combustion of ethanol gas is the heat of reaction for

2*( -393.51) + 4*(-285.83)-(-235.100) = -1695.24 kJ mol

4.23 (7th edition Prob. 4.21)

2(g))

H = 4*(90.25) + 6*(-241.82) -- 4*(-46.110) -- 5*(0)= -905.5 kJ mol-1

(c) Hrxn = 2*Hf(HNO3(l)) + Hf(NO(g)) -- 3*Hf(NO2(g))- Hf(H2O(l))

Hrxn = 2*(-174.1) + (90.25) -- 3*(33.18) - (-285.830) = -71.66 kJ mol .

(d) Hrxn = Hf(CaO(s) ) + Hf(C2H2(g)) -- Hf(CaC2(s)) - Hf(H2O(l))

Hrxn = -635.090 + 227 480 + 59.800 + 285 830 = -61.98 kJ mol

(e) Hrxn = 2*Hf(NaOH(s)) + Hf(H2(g)) - 2*Hf(Na(s)) -- 2*Hf(H2O(g))

Hrxn = Hf(CH3OH(g)) + Hf(H2O(g)) - Hf(CO2(g)) -- 3*Hf(H2(g)) -1 Hrxn = -200660 - 241818 - -393509 -- 3*0 = -48969 J mol

(k) Hrxn = Hf(HCHO(g)) + Hf(H2O(g)) - Hf(CH3OH(g)) -- ½*Hf(O2(g)) -1 Hrxn = -108570 - 241818 - -200660 -- ½*0 = -149728 J mol

(l) Hrxn = 2*Hf(H2O(g))+ 2*Hf(SO2(g)) - 2*Hf(H2S(g)) -- 3*Hf(O2(g)) -1

Hrxn = 2*-241818 -2*296830 +2*20630 -- 3*0 = -1036036 Jmol

(m) Hrxn = 3*Hf(H2(g)) + Hf(SO2(g)) - Hf(H2S(g)) -- 2*Hf(H2O(g)) -1

Hrxn = 3*0 + -296830 + 20630 -- 2*-241818 = 207436 J mol

(n) Hrxn = 2*Hf(NO(g))- Hf(N2(g))-- Hf(O2(g)) -1

Hrxn = 2*90250 -- 0-- 0= 180500 Jmol

(o) Hrxn = Hf(CaO(s)) + Hf(CO2(g)) -- Hf(CaCO3(s))

Hrxn = -635090 -- 393509 + 1206920 =178321 Jmol

(p) Hrxn =Hf(H2SO4(l))- Hf(SO3(g)) -- Hf(H2O(l)) -1 Hrxn = -813989 + 395720 + 285830 = -132439 Jmol

(q) Hrxn = Hf(C2H5OH(l)) - Hf(C2H4(g)) -- Hf(H2O(l))

Hrxn = -277690 -- 52510 + 285830 = -44370 J mol

(r) Hrxn = Hf(C2H5OH(g)) - Hf(CH3CHO(g)) -- Hf(H2(g)) -1 Hrxn = -235100 +166190 -- 0= -68910 Jmol

(s) Hrxn = Hf(CH3COOH(l)) + Hf(H2O(l)) - Hf(C2H5OH(l)) -- Hf(O2(g)) -1 Hrxn = -484500 -285830 + 277690 -- 0= -492640 Jmol

(t) Hrxn = Hf(CH2:CHCH:CH2(g)) + Hf(H2(g))- Hf(C2H5CH:CH2(g))

Hrxn = 109240 + 0+ 540 = 109780 Jmol

(u) Hrxn = Hf(CH2:CHCH:CH2(g)) + 2*Hf(H2(g)) - Hf(C4H10(g))

Hrxn = 109240 + 0+ 125790 = 235030 J mol

(v) Hrxn = Hf(CH2:CHCH:CH2(g)) + Hf(H2(g))- Hf(C2H5CH:CH2(g))- Hf(O2(g))-

Hrxn = 109240 +(-241818)-(-540) = -132038 Jmol

(w) Hrxn = 4*Hf(NH3(g)) -6* Hf(NO(g)) -- 6*Hf(H2O(g))-5*Hf(N2(g))-

4.24 (7th edition Prob. 4.22)

(a) A heat of reaction at a temperature other than standard temperature is given by the heat of reaction at standard temperature plus the integral of the difference in heat capacity between products and reactants:

A handy spreadsheet for evaluating the integral wasprovided in the lecture notes. Putting in the parameters for

Multiplying the result by R gives -

Cp The heat of reaction at 298.15 K is just twice the heat of formation of ammonia, because this is the formation reaction for ammonia, written with 2NH H298 = -46110 J/mol *2 = -92.22 kJ/mol. Adding on the

C

gives H600°C = -109.80 kJ/mol for the reaction as written.

(b) Repeating this for the reaction 4NH + 5O O gives

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Multiplying the result by R gives 4.729 kJ/mol for the Cp. The heat of reaction at 298.15 K is given by:

H298 H298(H2 H298(NO) -- H298(NH3)

H298 = 6*(-241.818) + 4*90.250 -- 4*(-46.11) = -905.47 kJ/mol. 500°C = -900.74 kJ/mol.

(f) A heat of reaction at a temperature other than standard temperature is given by the heat of reaction at standard temperature plus the integral of the difference in heat capacity between products and reactants:

A handy spreadsheet for evaluating this was provided in the lecture notes. Putting in the parameters for the reaction

6NO (g)+ 8NH (g) → 7N (g)+ 12 H O (g)

gives:

The heat of reaction at650C is -2716 kJ/mol Thisis only slightly different from the heat of reaction at standard conditions. It just happens that the average heat capacity of the reactants and products is about the sameover this temperature range.

4.25 (7th edition Prob. 4.23)

ThisisasimpleapplicationofacombinationofEqs.(4.19)&(4.20)with evaluatedparameters.Ineachcasethevalueof ΔH0298 iscalculatedinPb.

4.23.Thevaluesof ΔA, ΔB, ΔCand ΔDaregivenforallcasesexceptfor Parts(e),(g),(h),(k),and(z)intheprecedingtable.Thosemissingareas follows:

4.26 (New)

Use table C 5 tofind the standard enthalpies.

(a)Reactants = -1262 2 kJ/mol + -3627.9 kJ/mol = -4890.1 kJ/mol

Products = -2274.6 kJ/mol + -2638.5 kJ/mol = -4913.1 kJ/mol

Products -- Reactants = -4913.1 J/mol -- (-4881.4) J/mol = -23 J/mol

(b) Reactants = -2274 6 kJ/mol

Products = -2265.9kJ/mol

Products -- Reactants = 8.7 kJ/mol

(c)Reactants = -5893.8 kJ/mol

Products = -5956.8 kJ/mol

Products -- Reactants = -63 kJ/mol

(d) Reactants = -9144.4 J/mol

Products = -9083.66J/mol

Products -- Reactants = 60.74 kJ/mol

(e) Reactants= -9144 4 J/mol

Products = -9200.66 J/mol

Products -- Reactants = -56.26 kJ/mol

(f) Reactants = -9144.4 kJ/mol

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Products = -9231.66 kJ/mol

Products -- Reactants = -87.26 J/mol

(g) Reactants = -75.5 kJ/mol

Products = -571.66 kJ/mol

Products -- Reactants = -496.16 kJ/mol

(h) Reactants = -3941 1 kJ/mol

Products = -3913.73 kJ/mol

Products -- Reactants = -27.37 kJ/mol

(i) Reactants = -7957 kJ/mol

Products = -8399.12 kJ/mol

Products -- Reactants = --441.42 kJ/mol

(j) Reactants = -9141.6 kJ/mol

Products = -9231.66 kJ/mol

Products -- Reactants = -90.06 kJ/mol

(k) Reactants = -9137.4 kJ/mol

Products = -9231.66 kJ/mol

Products -- Reactants = -94.26 kJ/mol

(l) Reactants = -4703.8 kJ/mol

Products = -4707.2 kJ/mol

Products -- Reactants = - 3.4 kJ/mol 4.27

(New)

First determine how many moles of ethanol we have, MWetoh = 46.068 g/mol.

= 0.217 mol.

For the secondpart, write the equation

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C2H5OH + NAD+ → (2CO2 + 2H2O) + NADH + H

H f 298 =(2(-393509 J mol 1) + 2(-285830 J mol 1) + (-31.9 J mol

H = -0.217 mol × H f 298 = 0.217 mol × (-1358421.6 J mol 1) = -294777.4872 J

This reaction gives off heat as it proceeds in the reaction. rob. 4.24)

4.28(7th editionP

T = 288.71 K P= 1atm

Thehigherheatingvalueisthenegativeoftheheatofcombustionwithwaterasliquid product.

Calculatemethanestandardheatofcombustionwithwaterasliquidproduct:

CH4+2O2 → CO2+2H2O

Standard of ⋅

Assumingmethaneisanidealgasatst condit

ts Formation: andard ions:

4.29 (7th edition Prob. 4.25)

Calculatemethanestandardheatofcombustionwithwaterasliquidproduct

StandardHeatsofFormation:CH4+2O2 >CO2+2H2O

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