1. Chapter 14 Chemical Kinetics
Lecture Presentation
Chapter 14 Chemical Kinetics
James F. Kirby
Quinnipiac University
Hamden, CT

2. 14.1 Factors Affecting Reaction Rates
Chemical kinetics is the study of rate or speed of chemical reactions.
Chemical kinetics also gives us information about the reaction mechanism, that is, what really goes on during a chemical reaction.

3. 14.2 Reaction Rates
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
D[A] = change in concentration of A over
time period Dt
A B
rate = −
D[A] Dt
Because [A] decreases with time, D[A] is negative.
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
Average rate is the rate over a period of time Dt.
Instantaneous rate is the rate at a particular time t.
Initial rate is the rate at time t ≈ 0.

4. 14.2 Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
D[A] Dt
D[B] Dt 1 2
rate = −
rate =
aA + bB cC + dD 1 a
D[A] Dt 1 b
D[B] Dt 1 c
D[C] Dt 1 d
D[D] Dt
rate = −
= − = =

5. Consider the reaction 4NO2 (g) + O2 (g) 2N2O5 (g).
EXERCISE #1
Consider the reaction 4NO2 (g) + O2 (g) N2O5 (g).
a) Write the rate expression for the reaction.
− 1 4
∆[NO 2 ] ∆t =
− ∆[O 2 ] ∆t
= 1 2
∆[N 2 O 5 ] ∆t
rate =
Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of M/s.
b) At what rate is N2O5 being formed?
− ∆[𝑂 2 ] ∆𝑡 =
0.024 M/s
− ∆[O 2 ] ∆t
= 1 2
∆[N 2 O 5 ] ∆t
= M/s
∆[N 2 O 5 ] ∆t =
2 (− ∆[O 2 ] ∆t )
= 2(0.024 M/s)
= M/s
c) At what rate is NO2 reacting?
− 1 4
∆[NO 2 ] ∆t =
− ∆[O 2 ] ∆t
= M/s
− ∆[NO 2 ] ∆t =
4 (− ∆[𝑂 2 ] ∆𝑡 )
= 4(0.024 M/s)
= M/s

6. 14.3 Rate Law
Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
rate = k [A]x[B]y
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall

7. 14.3 Determining the Form of the Rate Law
Method of Initial Rates
F2 (g) + 2ClO2 (g) FClO2 (g)
Rate law:
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Quadruple [ClO2] with [F2] constant
Rate doubles
Rate quadruples
x = 1
y = 1
rate = k [F2][ClO2]

8. 14.3 Determining the Form of the Rate Law
Conclusion about Rate Law
Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2][ClO2] 1

9. EXERCISE #2
Consider the reaction of nitric oxide with hydrogen at 1280oC:
The following data were collected at this temperature:
a) Determine the rate law.
rate = k[NO]x[H2]y
= x 10 −3 M 5.0 x 10 −3 M
k [H2] 3 k [H2] 2
= 4.0 x 10 −3 M 2.0 x 10 −3 M
k [NO] 2 k [NO] 1
= 2
= 2
= 5.0 x 10 −5 M/s 1.3 x 10 −5 M/s
rate 2 rate 1
rate 3 rate 2
= x 10 −5 M/s 5.0 x 10 −5 M/s
= 3.85
= 2
2x = 3.85
x = 1.94
≈ 2
2y = 2
x = 1
rate = k[NO]2[H2]

10. EXERCISE #2
b) Determine the rate constant.
rate = k[NO]2[H2]
rate NO 2[H2]
= 5.0 x 10 −5 M/s x 10 −3 M 2(2.0 x 10 −3 M)
k =
= 2.5 x 102 M−2 s−1
c) Determine the rate of the reaction when [NO] = 12.0 x 10−3 M and [H2] = 6.0 x 10 −3 M.
rate =
k[NO]2[H2] =
(2.5 x 102 M−2 s−1)
(12.0 x 10−3 M)2
(6.0 x 10−3 M)
= 2.2 x 10−4 M/s
d) Determine the order of the reaction in regards to NO and to H2, respectively, and the overall order of the reaction.
2nd order in NO
1st order in H2
3rd order overall

11. 14.4 First-Order Rate Law rate = − D[A] Dt A product
[A]t = concentration of A at time t
[A]o = initial concentration of A
k = rate constant
t = time
Rate law: rate = k[A]
Integrated rate law:
ln[A]t = ln[A]o – kt
Half-life t1/2
Time required for the concentration of a reactant to decrease to half of its initial concentration.
t ½ = ln(2) k
Half–life does not depend on the concentration of reactants.

12. 14.4 Second-Order Rate Law rate = − D[A] Dt A product
[A]t = concentration of A at time t
[A]o = initial concentration of A
k = rate constant
t = time
Rate law: rate = k[A]2
Integrated rate law:
1 A 𝑡 = 1 [A] o + kt
Half-life:
t ½ = 1 k [A] o
Half–life gets longer as the reaction progresses and the concentration of reactants decrease.

13. 14.4 Zero-Order Rate Law rate = − D[A] Dt A product
[A]t = concentration of A at time t
[A]o = initial concentration of A
k = rate constant
t = time
Rate law: rate = k[A]0 = k
Integrated rate law:
[A]t = [A]o − kt
Half-life:
t ½ = [A] o 2k
Half–life gets shorter as the reaction progresses and the concentration of reactants decrease.

14. EXERCISE #3
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for [A] to decrease from 0.88 M to 0.14 M ?
ln[A]t = ln[A]0 – kt
ln A 0 − ln A t k =
ln − ln⁡(0.14) 2.8 x 10 −2 𝑠 −1
kt = ln[A]0 – ln[A]t
t =
= 66 s
EXERCISE #4
What is the half-life of N2O5 if it decomposes as first order reaction with a rate constant of 5.7 x 10−4 s−1?
t1/2 = ln(2) k
= 𝑥 10 −4 𝑠 −1 =
1200 s =
20 minutes

15. Concentration-Time Equation
14.4 The Rate Laws
Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A]t = [A]0 − kt t½
ln(2) k = 1
rate = k [A]
ln[A]t = ln[A]0 − kt 1
[A]t =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2

16. 14.4 Determining the Form of the Rate Law
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2][ClO2] 1
Why?

17. NO2(g) + CO(g) NO(g) + CO2(g)
14.6 Reaction Mechanisms
Reaction Mechanisms
NO2(g) + CO(g) NO(g) + CO2(g)
NO2
NO3 NO
Elementary step: +
NO3 CO
NO2
CO2
Elementary step:
Overall reaction:
NO2 + CO NO + CO2
The sequence of elementary steps that leads to product formation is the reaction mechanism.
An intermediate is not reflected in the rate law or in the overall reaction equation. It is always formed in an early elementary step and consumed in a later elementary step NO3

18. 14.6 Reaction Mechanisms Writing plausible reaction mechanisms:
The sum of the elementary steps must give the overall balanced equation for the reaction.
The rate-determining step should predict the same rate law that is determined experimentally.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.
EXERCISE #5
The proposed mechanism for the reaction NO2 + CO NO + CO2 is
NO2 + NO NO3 + NO (slow)
NO3 + CO NO2 + CO2 (fast)
Write a plausible rate law for the reaction.
rate-determining step
rate = k[NO2]2

19. EXERCISE #6
The experimental rate law for the reaction between NO2 and O3 to produce N2O5 and O2 is rate = k[NO2][O3]. The proposed reaction mechanism is as follows:
Step 1: NO2 + O NO3 + O2
Step 2: NO3 + NO N2O5
a) Write the equation for the overall reaction.
2NO2 + O N2O5 + O2
b) Identify the intermediate/s.
NO3
c) Identify the rate-determining step.
The rate law reflects step 1; therefore step 1 is the rate-determining step.

20. 14.5 A Model for Chemical Kinetics
We have seen how concentrations of the reactants affect the rate of the reaction. Let us see how the rate constant can affect the rate of the reaction.
rate ∝ k
Collision Model
Collision --- Molecules must collide to react.
Activation energy (Ea) --- Collision must involve enough energy to produce a reaction.
Orientation --- Relative orientation of the reactant molecules must allow formation of any new bonds necessary to produce products.
Temperature --- The rate constant is temperature dependent.

21. 14.5 Activation Energy and Transition State
Activation Energy (Ea)
The minimum amount of energy required to initiate a chemical reaction.
A + B AB C + D +
Exothermic Reaction
Endothermic Reaction

22. 14.5 Temperature Dependence of the Rate Constant
Linear form of Arrhenius equation
Arrhenius equation
ln 𝑘=− E a R 1 T + ln 𝐴
Taking into account the collision model, the rate constant k can be represented as follows:
k=A e − E a / RT
Arrhenius equation
Ea = activation energy (J/mol)
R = gas constant (8.314 J/K∙mol)
T = absolute temperature (in K)
A = frequency factor

23. 14.5 Temperature Dependence of the Rate Constant
Arrhenius equation at two different temperatures
k1 = rate constant at temperature T1
ln k 1 k 2 = E a R T 1 − T 2 T 1 T 2
k2 = rate constant at temperature T2
Ea = activation energy (J/mol)
R = gas constant (8.314 J/K∙mol)
EXERCISE #7
T1, T2 = absolute temperatures (in K)
When the temperature of a certain reaction was increased from 25oC to 35oC, the rate of the reaction doubled. Calculate the activation energy of this reaction.
rate ∝ k
k1 = 1
k2 = 2
T1 = 25oC = 298 K
T2 = 35oC = 308 K
ln k1 k2 R T 1 − T 2 T 1 T =
298 K− 308 K (298 K)(308 K)
Ea =
ln 1 2
(8.314 J/K∙mol)
= J
= 53 kJ

24. 14.7 Catalysis
Catalyst
A substance that speeds up a reaction without being consumed itself.
Provides a new pathway for the reaction with a lower activation energy.
ln 𝑘=− E a R 1 T + ln 𝐴 Ea k
Ea (catalyzed) < Ea (uncatalyzed)
ratecatalyzed > rateuncatalyzed

25. END OF CHAPTER 14