Rate Law by norainihamzah9

4.3 Rate law Learning outcomes At the end of the chapter students will be able to:

define rate law determine reaction orders identify rate constant

RATE LAW The rate law is an expression for the reaction rate in terms of concentrations of chemical species involved in the reaction. For example,

Rate = k [A]m [B]n where : k is a rate constant, m is reaction order with respect to A, n is reaction order with respect to B, m+n is the overall reaction order.

The Rate Law For any general reaction occurring at a fixed temperature

bB â&#x2020;&#x2019; cC

Rate = k[A]m[B]n

The term k is the rate constant, which is specific for a given reaction at a given temperature. The exponents m and n are reaction orders and are determined by experiment.

The values of m and n are not necessarily related in any way to the coefficients a and b. 3

Individual and Overall Reaction Orders For the reaction 2NO(g) + 2H2(g) â&#x2020;&#x2019; N2(g) + 2H2O(g): The rate law is rate = k[NO]2[H2] The reaction is second order with respect to NO, first order with respect to H2 and third order overall.

Note that the reaction is first order with respect to H2 even though the coefficient for H2 in the balanced equation is 2. Reaction orders must be determined from experimental data and cannot be deduced from the balanced equation.

Reaction Orders A reaction has an individual order “with respect to” or “in” each reactant. For the simple reaction A → products: If the rate doubles when [A] doubles, the rate depends on [A]1 and the reaction is first order with respect to A. If the rate quadruples when [A] doubles, the rate depends on [A]2 and the reaction is second order with respect to [A]. If the rate does not change when [A] doubles, the rate does not depend on [A], and the reaction is zero order with respect to A.

Determining Reaction Order from Rate Laws PROBLEM:

For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.

(a) 2NO(g) + O2(g) (b) CH3CHO(g)

2NO2(g); rate = k[NO]2[O2] CH4(g) + CO(g); rate = k[CH3CHO]3/2

I3-(aq) + 2H2O(l); rate = k[H2O2][I-]

PLAN: Look at the rate law and not the coefficients of the chemical reaction. SOLUTION: (a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.

(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall. (c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+, while being 2nd order overall. Info taken from: https://goo.gl/images/pDRA21

Plots of reactant concentration, [A], vs. time for first-, second-, and zero-order reactions

7 Image taken from: https://goo.gl/images/myZLDZ

Plots of rate vs. reactant concentration, [A], for first-, second-, and zero-order reactions.

8 Image taken from: https://goo.gl/images/p5jXut

RATE CONSTANT 

The rate constant is independent of the concentrations of the chemical species involved in the reaction.



However, it depends on other factors such as temperature or ionic strength, e.g., k(T).



The units of the rate constant depend on the overall reaction order.



In general, the form of the rate law is NOT determined by the reaction stoichiometry, but is determined by the reaction mechanism. 9

UNITS OF THE RATE CONSTANT Example: second order Rate = k [A][B] or

UNITS OF THE RATE CONSTANT (CONT.) Assuming time in second(s). Order Zero First Second Third

Unit of rate constant moldm-3s-1( Ms-1) s-1 dm3mol-1s-1 (M-1s-1) dm6mol-2s-1 (M-2s-1)

RATE CONSTANTS AND THE ORDERS 2NO + O2 ï&#x201A;® 2NO2

Usually, the rate of a reaction is a function of the concentrations of reactants . example : Rate = k [O2] [NO]2 The orders of 1 and 2 for [O2] and [NO] respectively has been determined by experiment, NOT from the chemical equation. 12

RATES AS FUNCTIONS OF REACTANT CONCENTRATIONS In a general reaction, a A + b B + ï&#x201A;® products Rate = k[A]x[B]y If concentrations of B is kept constant, you can measure the reaction rate of A at various concentrations. 13

Determining Reaction Orders

For the general reaction A + 2B â&#x2020;&#x2019; C + D, the rate law will have the form Rate = k[A]m[B]n

To determine the values of m and n, we run a series of experiments in which one reactant concentration changes while the other is kept constant, and we measure the effect on the initial rate in each case.

Initial Rates for the Reaction between A and B Initial Rate Experiment (mol/LÂˇs)

Initial [A] (mol/L)

Initial [B] (mol/L)

1.75x10-3

2.50x10-2

3.00x10-2

3.50x10-3

5.00x10-2

3.00x10-2

3.50x10-3

2.50x10-2

6.00x10-2

7.00x10-3

5.00x10-2

6.00x10-2

[B] is kept constant for experiments 1 and 2, while [A] is doubled. Then [A] is kept constant while [B] is doubled. Info taken from: https://goo.gl/images/K6Ugn7

Finding m, the order with respect to A: We compare experiments 1 and 2, where [B] is kept constant but [A] doubles:

Rate 2 Rate 1

k[A] m [B] n 2 k[A] m 1

2 [B] n 1

3.50x10-3 mol/LÂˇs 1.75x10-3mol/LÂˇs

[A]2

m [A]1

[A]2

[A]1

5.00x10-2 mol/L

2.50x10-2 mol/L

Dividing, we get 2.00 = (2.00)m so m = 1

116

Finding n, the order with respect to B: We compare experiments 3 and 1, where [A] is kept constant but [B] doubles:

Rate 3 Rate 1

k[A] m [B] n 3 k[A] m 1

[B]

3.50x10-3 mol/LÂˇs 1.75x10-3mol/LÂˇs

3 n 1

[B]3 n [B]1

[B]3

[B]1

6.00x10-2 mol/L

3.00x10-2 mol/L

Dividing, we get 2.00 = (2.00)n so n = 1

Initial Rates for the Reaction between O2 and NO O2(g) + 2NO(g) → 2NO2(g)

Rate = k[O2]m[NO]n

Initial Reactant Concentrations (mol/L) Initial Rate Experiment (mol/L·s)

[O2]

[NO]

3.21x10-3

1.10x10-2

1.30x10-2

6.40x10-3

2.20x10-2

1.30x10-2

12.48x10-3

1.10x10-2

2.60x10-2

9.60x10-3

3.30x10-2

1.30x10-2

28.8x10-3

1.10x10-2

3.90x10-2

Info taken from: https://goo.gl/images/qEvR1j

Finding m, the order with respect to O2: We compare experiments 1 and 2, where [NO] is kept constant but [O2] doubles:

Rate 2

Rate 1

n k[O2] m [NO] 2 2 m

k[O2] 1 [NO]1

6.40x10-3 mol/LÂˇs 3.21x10-3mol/LÂˇs

[O2] m 2 = [O2]2 = m [O2]1 [O2]1 2.20x10-2 mol/L

1.10x10-2 mol/L

Dividing, we get 1.99 = (2.00)m or 2 = 2m, so m = 1 The reaction is first order with respect to O2. 19

Sometimes the exponent is not easy to find by inspection. In those cases, we solve for m with an equation of the form a = bm:

log a log 1.99 m= = = 0.993 log b log 2.00 This confirms that the reaction is first order with respect to O2.

Reaction orders may be positive integers, zero, negative integers, or fractions.

Finding n, the order with respect to NO: We compare experiments 1 and 3, where [O2] is kept constant but [NO] doubles:

Rate 3

Rate 1

[NO]3

[NO]1

12.8x10-3 mol/LÂˇs 2.60x10-2 mol/L = 3.21x10-3mol/LÂˇs 1.30x10-2 mol/L

Dividing, we get 3.99 = (2.00)n or 4 = 2n, so n = 2. Alternatively:

log a log 3.99 = = 2.00 log b log 2.00

The reaction is second order with respect to NO. The rate law is given by: rate = k[O2][NO]2 21

SAMPLE PROBLEM A reaction between substances A and B is represented stoichiometrically by A+B â&#x2020;&#x2019;C Observation on the rate of this reaction are obtained in three separate experiments as follows: What is the order with respect to each reactant, and what is the value of the rate constant?

[A]0, M

[B]0, M

1.0

Duration of experiment, Î&#x201D;t, h 0.50

Final concentration, [A]f, M 0.0975

0.100 0.100

2.0

0.50

0.0900

0.0500

1.0

2.00

0.0450

Determining Reaction Orders from Rate Data

PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these is NO2(g) + CO(g) â&#x2020;&#x2019; NO(g) + CO2(g) rate = k[NO2]m[CO]n Use the following data to determine the individual and overall reaction orders: Initial Rate Initial [NO2] Initial [CO] (mol/LÂˇs) Experiment (mol/L) (mol/L) 1

0.0050

0.10

0.080

0.40

0.10

0.0050

0.10

0.20

Info taken from: https://goo.gl/images/4hMp4p

PLAN: We need to solve the general rate law for m and for n and then add those orders to get the overall order. We proceed by taking the ratio of the rate laws for two experiments in which only the reactant in question changes concentration. SOLUTION: To calculate m, the order with respect to NO2, we compare experiments 1 and 2:

rate 2

rate 1

k [NO2]m2[CO]n2 [NO2] 2 = k [NO2]m1 [NO2] 1 [CO]n1

0.080 = 0.0050

0.40 0.10

16 = (4.0)m so m = 2

The reaction is second order in NO2.

To calculate n, the order with respect to CO, we compare experiments 1 and 3:

rate 3 rate 1

k [NO2]m3[CO]n3 [CO] 3 = m k [NO2] 1 [CO] 1 n [CO] 1

0.0050 = 0.0050

0.20 0.10

1.0 = (2.0)n so n =0

The reaction is zero order in CO.

rate = k[NO2]2[CO]0 or rate = k[NO2]2

EVALUATION OF ORDER BY EXPERIMENTS –INITIAL RATE METHOD Example : Derive the rate law for the reaction A + B + C  products from the following data : Expriment

[A]0 [B]0 [C]0 rate

1 0.100 0.100 0.100 0.100

2 0.200 0.100 0.100 0.800

3 0.200 0.300 0.100 7.200

4 0.100 0.100 0.400 0.400 26

EVALUATION OF ORDER BY EXPERIMENTS (CONT.) Assume: rate = k [A]x [B]y [C]z From experiment 1 and 2, we have: 0.800 k [0.2]x [0.1]y [ 0.1]z ----- = -----------------------0.100 k [0.1]x [0.1]y [0.1]z Thus, 8 = 2x; and x = ln8 / ln2 = 3 By similar procedures, we get y = 2 and z = 1. Thus, the rate law is: rate = k [A]3 [B]2 [C] overall order = 3+2+1 = 6

The following data refers to the following reaction: CH3COCH3 + I2 + H+ ï&#x201A;® CH3COCH2I + HI Expt. 1

3.0

[H+], [I2], Initial -1 M M rate, Ms 0.2 0.02 6.0 x 10-6

3.0

0.4

0.02 12.0 x 10-6

4.0

0.4

0.02 16.0 x 10-6

4.0

0.2

0.04 8.0 x 10-6

[CH3COCH3],

Based on the above information,     

Calculate the order with respect to CH3COCH3 , I2 and H+. Write the rate law for the reaction. What is the overall order of the reaction? Determine the rate constant,k with the correct unit. Calculate the rate of reaction if the concentration of propanone, iodine and hydrogen ion are each 0.5 mol dm-3. 29

SOLUTION I) Let the rate law equation be : rate = k [CH3COCH3]x[H+]y[I2]z For value of y; = Exp. 2 / exp 1 = 12 x 10-6 / 6 x 10-6 = (0.4)y / (0.2)y y = 1 For value of x; =Exp 3 / exp 2 = 16 x 10-6 / 12 x 10-6 = (4)x / (3)x x= 1 For value of z; = Exp 3 / exp 4 = 16 x 10-6 / 8 x 10-6 = (0.4)1(0.2)z / (0.2)1(0.04)z z = 0

SOLUTION ii) rate = k [CH3COCH3][H+] iii) overall order = 1+1 = 2 (second order) iv) consider exp 1; 6 x 10-6 = k (3.0)(0.2) ď &#x153;k = 1.0 x 10-5 s-1 ( wrong / no unit, give them no mark) v) Rate = (1.0 x 10-5) (0.5) ( 0.5) = 2.5 x 10-6 mol dm-3s-1 31

ZEROTH ORDER REACTION you will get a horizontalline, because rate = k[A]0 rate = k (a horizontal line)

rate

independent to [reactant]

[A] 32

FIRST ORDER REACTION For first order, rate = k[A] rate

ï&#x201A;¢

the plot is a straight line (linear)

[A]

SECOND ORDER REACTION ï&#x201A;¢

the plot is a branch of a prabola, because rate = k [A]2

Rate

[A]

DETERMINATION

RATE CONSTANTS

USING THE INTEGRATED RATE LAWS Example; first order integrated rate laws , ln [A]t = -kt + ln [A]0 y = mx + c

ln[A]t ln[A]o

you should plot ln [A]t vs. t and find a line to best fit your data. The slope of the line is -k. t 35

DETERMINATION OF ORDER BY LINEAR PLOT 

For zero order; [A]0



plot

– [A]t = kt

[A]t vs t , slope = -k

DETERMINATION OF ORDER BY LINEAR PLOT (CONT.)

[A]o /[A]t = kt Plot ln [A] vs t , slope = - k 

For first order; ln

DETERMINATION OF ORDER BY LINEAR PLOT (CONT.) 



For second order,

plot

1 1 --- - ---- = kt [A]t [A]o

1 --- vs t , slope = k [A]

Integrated rate laws and reaction order

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

ln[A]t = -kt + ln[A]0

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

Image taken from: https://goo.gl/images/neJiWA

Graphical determination of the reaction order for the decomposition of N2O5 19-Sep-17 Dr NORAINI HAMZAH

Image taken from: https://goo.gl/images/wW8fjL

Integrated Rate Laws [A] rate = -

t

first order rate equation

= k [A]

[A]t

= - kt

ln [A]t = -kt + ln [A]o

[A]0 [A] rate = -

t

= k [A]2

1 [A]t

second order rate equation -

1 [A]0

= kt

1 [A]t

= kt

1 [A]0

[A] rate = -

t

= k [A]0

zero order rate equation

[A]t - [A]0 = - kt

Determining Reaction Concentration at a Given Time PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87s-1, to two molecules of ethylene (C2H4). (a) If the initial C4H8 concentration is 2.00M, what is the concentration after 0.010 s? (b) What fraction of C4H8 has decomposed in this time? PLAN:

Find the [C4H8] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration.

SOLUTION: ln (a)

[C4H8]0 [C4H8]t

= kt

; ln

2.00

= (87s-1)(0.010s)

[C4H8]

[C4H8] = 0.83mol/L (b)

[C4H8]0 - [C4H8]t [C4H8]0

2.00M - 0.87M =

2.00M

Info taken from: https://goo.gl/images/RCFKaL

= 0.58

EXERCISE: 1.

The decomposition of dinitrogen pentoxide is a first-order reaction with a rate constant of 5.1x10-4s-1 at 450C. given: 2N2O5(g) â&#x2020;&#x2019; 4NO2(g) + O2(g) a) If the initial concentration of N2O5 was 0.25 M, what is the concentration after 3.2 min ? ( 0.227 M) b) How long will it take for the concentration of N2O5 to decrease from 0.25 M to 0.15 M ? (16.7 min) c) How long will it take to convert 62% of the starting material? (31.62min) 2. The conversion of cyclopropane to propane in the gas phase is a first-order reaction with a rate constant of 6.7 x 10-4s-1 at 5000C. Calculate the half-life of the reaction. (17.24 min) 3. Given : 2A â&#x2020;&#x2019;B The above reaction is first order in A with a rate constant of 2.8 x 10-2s-1 at 800C. How long will it take for A to decrease from 43 0.88 M to 0.14 M ? (65.65 seconds)

EXERCISE 1.

The recombination of iodine atoms to form molecular iodine in the gas phase : I(g) + I(g) â&#x2020;&#x2019; I2(g) follows second order kinetic and has the high rate constant of 7.0 x 10-9 M-1s-1at 230C. a) If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. b) Calculate the half-life of the reaction if the initial concentration of I is 0.60 M. 44

HALF LIVES , T1/2 ď&#x201A;˘ Definition

The half-life (t1/2) of a reaction is the time period required to reduce the reactant to half of its original value. Example: 8M t1/2

t1/2

HALF LIFE OF FIRST ORDER REACTION constant, independent of the initial concentration.  Thus , first half-life(t’1/2) = second half-life(t’’1/2 ).  The rate constant and half-life has the relationship: t1/2 = ln(2) = 0.693 k k 

First - order rate law : ln [A]o / [A]t = kt At t= t1/2, [A]t = ½ [A]o; subtituting ln ([A]o / ½[A]o) = kt1/2 or t1/2 = (ln 2)/k or t1/2 = 0.693/k

A plot of [N2O5] vs. time for three half-lives for a first-order process t1/2 =

ln 2 k

0.693

47 Image taken from: https://goo.gl/images/F1THdG

Determining the Half-Life of a First-Order Reaction PROBLEM:

Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via the following first-order reaction: CH2 ď &#x201E; H3C CH CH2 (g) (g) H2C CH2

The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? 0.693 PLAN: Use the half-life equation, t1/2 = , to find the half-life. k One-quarter of the initial value means two half-lives have passed. SOLUTION: (a)

t1/2 = 0.693/9.2s-1 = 0.075s

(b)

2 t1/2 = 2(0.075s) = 0.150s

Image taken from: https://goo.gl/images/BS1rN1

HALF LIVE FOR 2ND ORDER REACTIONS depends on the initial concentration, [A]o, and the rate constant k .  Thus second half-life (t’’1/2 ) = 2t’1/2  The rate constant and half-life has the relationship: t1/2 = 1 k [A]o 

HALF LIVE FOR ZERO ORDER REACTIONS 

The rate constant and half-life has the relationship:

t1/2 = [A]0 2k 

depends on the initial concentration, [A]o, and the rate constant k .

DETERMINATION OF ORDER BY HALF LIFE METHOD Plot [reactant] vs time

NOTE : If t’1/2 =t”1/2 , first order

[A]o

But if t”1/2 = 2t’1/2 , second order [A]o/2

[A]0/4

t’1/2

t”1/2

time

An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions

Zero Order

First Order

Second Order

Rate law

rate = k

rate = k [A]

rate = k [A]2

Units for k

mol/L*s

1/s

L/mol*s

Integrated rate law in straight-line form

[A]t = k t + [A]0

ln[A]t = -k t + ln[A]0

1/[A]t = k t + 1/[A]0

Plot for straight line

[A]t vs. t

ln[A]t vs. t

1/[A]t = t

Slope, y-intercept

k, [A]0

-k, ln[A]0

k, 1/[A]0

Half-life

[A]0/2k

ln 2/k

1/k [A]0 52

Image taken from: https://goo.gl/images/TVBLoR

EXAMPLE The rate constant for the decomposition of N2O5 to NO2 and O2 at 70C is 6.82 x 10-3 s-1. Suppose we start with 0.300 mol of N2O5(g) in a 0.500 L container.  



How many moles of N2O5 will remain after 1.5 min? How many minutes will it take for the quantity of N2O5 to drop to 0.030 mol? What is the half-life of N2O5 at 70C ?

SOLUTION Using;

ln [A]o /[A]t = kt

(1.5 min x 60 saat) = 90s

ln 0.300 mol = 6.82 x 10-3 s-1 x 90 s (N2O5)

0.300 ( N2O5 )

= 0.6138 = e0.6138 =1.847

(N2O5 ) = 0.162 mol 54

SOLUTION ln [A]o /[A]t = kt ln 0.300 = 6.82 x 10-3 s-1 x t 0.030 2.303 = 6.82 x 10-3 s-1 x t t

= 337.6 second = 5.6 min 55

SOLUTION For first order; t1/2 = 0.693 k = 0.639 6.82 x 10-3 s-1 = 101.6 seconds

SUMMARY: ď&#x201A;˘

Method to determine order of reaction: Initial rate method Half-life method Graph rate vs concentration (pattern) Linear plot (pattern)

The Effect of Temperature on Reaction Rate The Arrhenius Equation

Ea ď&#x20AC; RT k ď&#x20AC;˝ Ae

where k is the kinetic rate constant at T Ea is the activation energy R is the energy gas constant

ln k = ln A - Ea/RT

k2 k1

Ea = RT

T is the Kelvin temperature A is the collision frequency factor

1 T2

1 T1 58

Dependence of the rate constant on temperature

Image taken from: https://goo.gl/images/E3gGvT

Graphical determination of the activation energy

ln k = -Ea/R (1/T) + ln A

60 Image taken from: https://goo.gl/images/v4gmNe

Determining the Energy of Activation PROBLEM:

The decomposition of hydrogen iodide, 2HI(g)

H2(g) + I2(g)

has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea. PLAN:

Use the modification of the Arrhenius equation

to find Ea.

SOLUTION: ln

k2 k1

= -

Ea = - (8.314J/mol*K) ln

1 T1

1.10x10-5L/mol*s 9..51x10-9L/mol*s

Ea = - R ln

600K

Ea = 1.76x105 J/mol = 176 kJ/mol Info taken from: https://goo.gl/images/pqz84R

-1

1 500K 61

EXAMPLE: Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. N2O4 → 2NO2 If k = 2.5x103 s-1 at -5°C and k = 3.5x104 s-1 at 20°C,

what is the activation energy for the decomposition? 62

SOLUTION: Using ln (

) =k 2 (

E)a  R

1 1 T2 T1

for condition 1: k1 = 2.5x103 and T1 = -5°C = 268 K for condition 2: k2 = 3.5x104 and T2= 25°C = 298 K

SOLUTION(CONT.): 4 3.5x10 ln( )3 2.5x10

ln(14) 2.63

= =

(

Ea )  R

Ea  R

1 1 298 268

(0.00335 – 0.00373)

Ea  (– 0.00037) R

E = -7108  a R Ea

= 59 kJmol-1