Chapter 13: Kinetics

Renee Y. Becker

Valencia Community College

Introduction 1. Chemical kinetics is the study of reaction rates

2. For a chemical reaction to be useful it must occur at a reasonable rate

3. It is important to be able to control the rate of reaction

4. Factors that influence rate

a) Concentration of reactants (molarity)

b) Nature of reaction, process by which the reaction takes place

c) Temperature

d) Reaction mechanism (rate determining step)

e) Catalyst

Reaction rate

Reaction rate- Positive quantity that expresses how the conc. of a reactant or product changes with time

Gen eq. N2O5(g) 2NO2(g) + ½ O2(g)

1. [ ] = concentration, molarity, mole/L

2. [N2O5] decreases with time

3. [NO2] increases with time

4. [O2] increases with time

5. Because of coefficients the concentration of reactant and products does not change at the same rate

6. When 1 mole of N2O5 is decomposed, 2 moles of NO2 and ½

mole of O2 is produced

N2O5(g) 2NO2(g) + ½ O2(g)

-[N2O5] = [NO2] = [O2]

2 ½ coef. From gen eq

 

-[N2O5] because it’s concentration decreases, other

positive because they increase

 

Rate of reaction can be defined by dividing by the change in time, t

 

rate = - [N2O5] = [NO2] = [O2]

t 2t ½ t

Reaction Rates

Generic Formula: aA + bB cC + dD

rate = - [A] = - [B] = [D] = [C]

at bt dt ct

Reaction Rate and Concentration The higher the conc. of starting reactant the more rapidly a reaction

takes place

1. Reactions occur as the result of collisions between reactant molecules

2. The higher the concentration of molecules, the greater the # of collisions in unit time and a faster reaction

3. As the reactants are consumed the concentration decreases, collisions decrease, reaction rate decreases

4. Reaction rate decreases with time and eventually = 0, all reactants consumed

5. Instantaneous rate, rate at a particular time

6.  Initial rate at t = 0

Rate Expression and Rate Constant

Rate expression / rate law:

rate = k[A]

where k = rate constant, varies w/ nature and temp.

[A] = concentration of A

Order of rxn involving a single reactant General equation rate expressionA prod rate = k[A]m

where m=order of reactionm=0 zero orderm=1 first orderm=2 second order

m, can’t be deduced from the coef. of the balanced eq.

Must be determined experimentally!

Order of rxn involving a single reactant

Rate of decomposition of species A measured at 2 different conc., 1 & 2

rate2 = k[A2]m rate1 = k[A1]

m

By dividing we can solve for m, to find the order of the reaction

Rate2 = [A2]m

Rate1 [A1]m (Rate2/Rate1) = ([A2]/[A1])m

 

Example 1

CH3CHO(g) CH4(g) + CO(g)

[CH3CHO] .10 M .20 M .30 M .40 M

Rate (mol/L s)

.085 .34 .76 1.4

Using the given data determine the reaction order

Example 1

Rate2 = [A2]m 4 = 2m m = 2

Rate1 [A1]m

 

second order rate = k[CH3CHO]2

 

once the order of the rxn is known, rate constant can be calculated, let’s calculate the rate constant, k

Example 1

rate = k[CH3CHO]2 rate = .085 mol/L s

conc = .10 mol/L

 

k = rate = .085 mol/L s = 8.5 L/mol s

[CH3CHO]2 (.10 mol/L)2

 

now we can calc. the rate at any concentration, let’s try .55 M

Example 1

rate = k[CH3CHO]2

rate = 8.5 L/mol s [.55]2 = 2.6 mol/ L s

Rate when [CH3CHO] = .55 M

is 2.6 mol/L s

Order of rxn with more than 1 reactant

aA + bB prod

rate exp: rate = k [A]m [B]n

m = order of rxn with respect to A

n = order of rxn with respect to B

Overall order of the rxn is the sum, m + n

Key

• When more than 1 reactant is involved the order can be determined by holding the concentration of 1 reactant constant while varying the other reactant. From the measured rates you can calculate the order of the rxn with respect to the varying reactant

Example 2

(CH3)3CBr + OH- (CH3)3COH + Br-

Exp. 1

Exp. 2

Exp. 3

Exp. 4

Exp. 5

[(CH3)3CBr] .5 1.0 1.5 1.0 1.0

[OH-] .05 .05 .05 .10 .20

Rate (mol/L s) .005 .01 .015 .01 .01

Find the order of the reaction with respect to both reactants, write the rate expression, and find the overall order of the reaction

Reactant concentration and time Rate expression rate = k[A]

Shows how the rate of decomposition of A changes with concentration

More important to know the relation between concentration and time

Using calculus: Integrated rate equations relating react conc. to time

For the following rate law, what is the overall order of the reaction?

Rate = k [A]2 [B]

1. 1

2. 2

3. 3

4. 4

Order Rate Expression

Conc-TimeRelation

Half-life Linear Plot

0 Rate = k [A]0 – [A] = kt [A]0

2k[A] vs. t

1 Rate = k[A] ln [A]0 = kt [A]

0.693 k

ln [A] vs. t

2 Rate = k[A]2 1 – 1 = kt[A] [A]0

1 k[A]0

1 vs. t[A]

First Order

First Order: A Products

rate = k[A]

ln [A]o/[A] = kt t½ = .693/k

[A]o = original conc. of A

[A] = Conc. of A at time, t

k = first order rate constant

ln = natural logarithm

Second Order

Second order: A Products rate = k[A]2

1 – 1 = kt t½ = 1

[A] [A]0 k[A]0

Zero Order

Zero order: A Products rate = k

[A]0 – [A] = kt t½ = [A]0

2k

Example 3

The following data was obtained for the gas-phase decomp. of HI

Is this reaction zero, first, or second order in HI?

Hint: Graph each Conc. Vs. time corresponding to correct [A], ln [A], or 1/[A]

Time (h) 0 2 4 6

[HI] 1.00 0.50 0.33 0.25

[HI] vs. time not linear so it is not zero order

[HI] vs. t

0

0.5

1

1.5

0 2 4 6 8

Time (h)

[HI]

ln [HI] vs. time not linear so it is not first order

ln [HI] vs. t

-1.5

-1

-0.5

0

0 2 4 6 8

Time (h)

ln [

HI]

1/[HI] vs. time is linear so it is second order

1/[HI] vs. t

0

2

4

6

0 2 4 6 8

Time (h)

1/[

HI]

Activation Energy Activation Energy: Ea (kJ)For every rxn there is a certain minimum energy that

molecules must possess for collisions to be effective.

1. Positive quantity (Ea>0)

2. Depends only upon the nature of reaction

3. Fast rxn = small Ea

4. Is independent of temp and concentration

For the following reaction:

A + B C

If I double the concentration of A and hold B constant, the rate doubles. What is the order of the reaction with respect to A?

1. 0

2. 1

3. 2

Activation Energy

Reaction Rate and Temp

Reaction Rate and Temp

1. As temp increases rate increases, Kinetic Energy increases, and successful collisions increase

2. General rule for every 10°C inc. in temp, rate doubles

The Arrhenius Equation

The Arrhenius Equation

f = e-Ea/RT

f = fraction of molecules having an En. equal to or greater than Ea

R = gas constant A = constant T = temp in K

 

ln k = ln A –Ea/RT plot of ln k Vs. 1/T linear

slope = -Ea/R

Two-point equation relating k & T

ln k2 = Ea [1/T1 – 1/T2]

k1 R

Example 4

For a certain rxn the rate constant doubles when the temp increases from 15 to 25°C.

a) Calc. The activation energy, Ea

b) Calc. the rate constant at 100°C, taking k at 25°C to be 1.2 x 10-2 L/mol s

If I increase the temperature of a reaction from 110 K to 120 K, what happens to the rate of the reaction?

1. Stay the same

2. Doubles

3. Triples

Reaction Mechanism Reaction Mechanism

Description of a path, or a sequence of steps, by which a reaction occurs at the molecular level.

Simplest case- only a single step, collision between two reactant molecules

 

Reaction Mechanism

“Mechanism” for the reaction of CO with NO2 at high temp (above 600 K) 

CO(g) + NO2(g) NO(g) + CO2(g)

“Mechanism” for the reaction of CO with NO2 at low temp

NO2(g) + NO2(g) NO3(g) + NO(g) slow

CO(g) + NO3(g) CO2(g) + NO2(g) fast

CO(g) + NO2(g) NO(g) + CO2(g) overall

The overall reaction, obtained by summing the individual steps is identical but the rate expressions are different.

 

High temp: rate = k[CO][NO2]

Low temp: rate= k[NO2]2

Reaction MechanismElementary Steps:

Individual steps that constitute a reaction mechanism

Unimolecular A B + C rate = k[A]

Bimolecular A + A B + Crate = k[A][A] = [A]2

Termolecular A + B + C D + Erate = k[A][B][C]

The rate of an elementary step is equal to a rate constant, k, multiplied by the concentration of each reactant molecule. You can treat all reactants as if they were first order. If a reactant is second order it will appear twice in the general equation.

Reaction Mechanism

Slow Steps- A step that is much slower than any other in a reaction mechanism.

Rate-determining step - The rate of the overall reaction can be taken to be that of the slow step

 

Step 1: A Bfast

Step 2: B Cslow

Step 3: C Dfast

A D

The rate A D (overall reaction) is approx. equal to the rate of B C the slow step

Deducing a Rate Expression from a Proposed Mechanism 1. Find the slowest step and equate the rate of the overall reaction to the rate of that step.2. Find the rate expression for the slowest step. NO2(g) + NO2(g) NO3(g) + NO(g) slow

CO(g) + NO2(g) CO2(g) + NO2(g) fast

CO(g) + NO2(g) CO2(g) + NO(g)

Rate of overall reaction = rate of 1st step

= k[NO2] [NO2] = k[NO2]2

Elimination of Intermediates Intermediate

1. A species produced in one step of the mechanism and consumed in a later step.

2. Concentration too small to determine experimentally

3. Must be eliminated from rate expression

4. The final rate expression must include only those species that appear in the balanced equation for the overall reaction

Example 5

Find the rate expression for the following reaction mechanism

Step1: NO(g) + Cl2(g) NOCl2(g) fast

 

Step2: NOCl2(g) + NO(g) 2NOCl(g) slow

 

  2 NO(g) + Cl2(g) 2NOCl(g)

Example 6

The decomposition of ozone, O3, to diatomic oxygen, O2, is

believed to occur by a two-step mechanism:

Step 1: O3(g) O2(g) + O(g) fast

Step 2: O3(g) + O(g) 2 O2(g) slow

2 O3(g) 3 O2(g)

• Find the rate expression for this reaction

Which is an intermediate for the following multi step mechanism?

2 A + 2 B C + 2 D

Step 1 2 B E

Step 2 E + A D + F

Step 3 F + A C + D

1. E

2. F

3. E & F

4. A

Catalysts

Catalysis A catalyst increases the rate of a reaction without being consumed

by it. Changes the reaction mechanism to one with a lower activation energy.

1.      Heterogeneous catalysisa) Catalyst is in a different phase from the reaction

mixture. Most common, solid catalyst with gas or liquid mixture.

b) Solid catalyst is easily poisoned, foreign material deposited on the surface during reaction reduce or

destroy its effectiveness.

2.      Homogeneous Catalysis a) Same phase as the reactants

Which is the catalyst for the following multi step mechanism?

2 A + 2 B C + 2 D

Step 1 2 B + G EStep 2 E + A D + FStep 3 F + A C + D + G

1. E2. G3. E & F4. A

Summary Problem and Homework Problems

Chapter 12: Kinetics

Hydrogen peroxide decomposes to water and oxygen according to the following reaction

H2O2(aq) H2O + ½ O2(g)

It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate (KMnO4) at certain

intervals.

a)      Initial rate determinations at 40C for the decomposition give the following data:

[H2O2] Rate (mol/L min)

0.10 1.93 x 10-4

0.20 3.86 x 10-4

0.30 5.79 x 10-4

1. Order of rxn?

2. Rate expression?

3. Calc. k @ 40°C

4. Calc. half-life @ 40°C

b) Hydrogen peroxide is sold commercially as a 30.0% solution. If the solution is kept at 40C, how long will it take for the solution to become 10.0% H2O2?

c) It has been determined that at 50C, the rate constant for the reaction is

4.32 x 10-3/min. Calculate the activation energy for the decomposition of H2O2

d) Manufacturers recommend that solutions of hydrogen peroxide be kept in a refrigerator at 4C. How long will it take for a 30.0% solution to decompose to 10.0% if the solution is kept at 4C?

e) The rate constant for the uncatalyzed reaction at 25C is 5.21 x 10-4/min. The rate constant for the catalyzed reaction at 25C is 2.95 x 108/min.

1) What is the half-life of the uncatalyzed reaction at 25C?

2) What is the half-life of the catalyzed reaction?

 

1) Express the rate of reaction

2HI(g) H2(g) + I2(g)

a)  in terms of [H2]

b) in terms of [HI]

3) Dinitrogen pentaoxide decomposes according to the following equation:

2N2O5(g) 4NO2(g) + O2(g)

a) write an expression for reaction rate in terms of [ N2O5], [NO2], and [O2]

4) What is the order with respect to each reactant and the overall order of the reactions described by the following rate expressions?

a)      rate = k1[A]3

b)     rate = k2[A][B]

c)      rate = k3[A][B]2

d) rate = k4[B]

e) rate = k

5) Complete the following table for the reaction, which is first order in both reactants

A(g) + B(g) products

[A] [B] K (L/mol s) Rate (mol/L s)

.2 .3 1.5

.029 .78 .025

.45 .520 .033

7) The decomposition of ammonia on tungsten at 1100C is zero-order, with a rate constant of 2.5 x 10-4 mol/L min

a) write the rate expression

b) calculate the rate when the concentration of ammonia is 0.080M

c) At what concentration of ammonia is the rate equal to the rate constant?

8) In solution at constant H+ concentration, I- reacts with H2O2 to

produce I2

H+(aq) + I-

(aq) + ½ H2O2(aq) ½ I2(aq) + H2O

The reaction rate can be followed by monitoring iodine production. The following data apply:

[I-] [H2O2] Rate (mol/L s)

.02 .02 3.3 x 10-5

.04 .02 6.6 x 10-5

.06 .02 9.9 x 10-5

.04 .04 1.3 x 10-4

a) Order of I-

b) Order of H2O2

c) Calc. k

d) Rate? When [I-] = .01 M [H2O2] = .03 M

9) In the first-order decomposition of acetone at 500C it is found that the concentration is 0.0300 M after 200 min and 0.0200M after 400 min.

H3C-CO-CH3(g) products

 Calculate

a)      The rate constantb)      The half-lifec)      The initial concentration

10) The decomposition of hydrogen iodide is second-order. Its half-life is 85 seconds when the initial conc. is 0.15M

HI(g) ½ H2(g) + ½ I2(g)

a)  What is k for the reaction?

b) How long will it take to go from 0.300M to 0.100M?

11) Write the rate expression for each of the following elementary steps:

a)      K+ + HCl KCl + H+

b)     NO3 + CO NO2 + CO2

c)      2NO2 2NO + O2

12) For the reaction

2H2(g) + 2NO(g) N2(g) + 2H2O(g)

the experimental rate expression is rate = k[NO]2[H2]

The following mechanism is proposed:

2NO N2O2 fast

N2O2 + H2 H2O + N2O slow

N2O + H2 N2 + H2O fast

Show that the mechanism is consistent with the rate expression.