1. Chapter 13
Chemical Kinetics

2. 13.2 The Rate Of A Chemical Reaction
The speed of a chemical reaction is called its reaction rate.
The rate of a reaction is a measure of how fast the reaction makes products (or uses reactants).
Rate is how much a quantity changes in a given period of time.
The rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time
or product concentration increases

3. at t = 0
[A] = 8
[B] = 8
[C] = 0
at t = 0
[X] = 8
[Y] = 8
[Z] = 0
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1

5. H2 (g) + I2 (g)  2 HI (g)
Reactants decrease while products increase over time.
The average rate of reaction can be calculated by
As time goes on, the rate of a reaction generally slows down.
because the concentration of the reactants decreases.
At some time the reaction stops, either because the reactants run out or because the system has reached equilibrium.

6. H2 (g) + I2 (g)  2 HI (g)
The instantaneous rate is the change in concentration at any one particular time.
slope at one point of a curve
Determined by taking the slope of a line tangent to the curve at that particular point.
first derivative of the function

7. Using [H2], the instantaneous rate at 50 s is
Using [HI], the instantaneous rate at 50 s is

9. H2O2(aq) + 3 I–(aq) + 2 H+(aq)  I3–(aq) + 2 H2O(l)
EXAMPLE 13.1 Expressing Reaction Rates
Consider the following balanced chemical equation:
H2O2(aq) + 3 I–(aq) + 2 H+(aq)  I3–(aq) + 2 H2O(l)
In the first 10.0 seconds of the reaction, the concentration of I– dropped from M to M.
(a) Calculate the average rate of this reaction in this time interval.
(b) Predict the rate of change in the concentration of H+ (that is, [H+]/t) during this time interval.
(a) Use Equation 13.5 to calculate the average rate of the reaction.
SOLUTION
(b) Use Equation 13.5 again for the relationship between the rate of the reaction and [H+]/t. After solving for [H+]/t, substitute the computed rate from part (a) and compute [H+]/t.
FOR PRACTICE 13.1
For the above reaction, predict the rate of change in concentration of H2O2 ([H2O2]/t) and I3– ([I3–]/t) during this time interval.
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10. Measuring Reaction Rates
Spectrometer
Gas Chromatograph

11. 13.3 The Rate Law : The Effect of Concentration on Reaction Rate
Generally, the larger the concentration of reactant molecules, the faster the reaction
increases the frequency of reactant molecule contact
The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants.
The rate law must be determined experimentally!!
The rate of a reaction is directly proportional to the concentration of each reactant raised to a power.
For the reaction aA + bB  products the rate law would have the form given below
n and m are called the orders for each reactant
k is called the rate constant
n + m = overall reaction order

12. Reaction Orders
(n = 0) Zero Order = rate is independent on the concentration of A
(n = 1) First Order = rate is directly proportional to the concentration of A
(n = 2) Second Order = rate is proportional to the SQUARE of concentration of A

13. The exponent on each reactant in the rate law is called the order with respect to that reactant.
The sum of the exponents on the reactants is called the order of the reaction .
The rate law for the reaction
2 NO(g) + O2(g) ® 2 NO2(g)
is Rate = k[NO]2[O2]
The reaction is
second order with respect to [NO],
first order with respect to [O2],
and third order overall

14. Determining Orders of a Reaction
Rate = k[A]n
If a reaction is Zero Order, the rate of the reaction is always the same
doubling [A] will have no effect on the reaction rate
If a reaction is First Order, the rate is directly proportional to the reactant concentration
doubling [A] will double the rate of the reaction
If a reaction is Second Order, the rate is directly proportional to the square of the reactant concentration
doubling [A] will quadruple the rate of the reaction

15. NO2(g) + CO(g)  NO(g) + CO2(g)
EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction
Consider the reaction between nitrogen dioxide and carbon monoxide:
NO2(g) + CO(g)  NO(g) + CO2(g)
The initial rate of the reaction was measured at several different concentrations of the reactants with the following results:
From the data, determine:
(a) the rate law for the reaction
(b) the rate constant (k) for the reaction
continued…
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16. Write the overall rate expression. Rate = k[NO2]2[CO]0 = k[NO2]2
SOLUTION
(a) Begin by examining how the rate changes for each change in concentration. Between the first two experiments, the concentration of NO2 doubled, the concentration of CO stayed constant, and the rate quadrupled, suggesting that the reaction is second order in NO2.
Between the second and third experiments, the concentration of NO2 stayed constant, the concentration of CO doubled, and the rate remained constant (the small change in the least significant figure is simply experimental error), suggesting that the reaction is zero order in CO.
Between the third and fourth experiments, the concentration of NO2 again doubled, the concentration of CO halved, yet the rate quadrupled again, confirming that the reaction is second order in NO2 and zero order in CO.
Write the overall rate expression.
Rate = k[NO2]2[CO]0 = k[NO2]2
(b) To determine the rate constant for the reaction, solve the rate law for k and substitute the concentration and the initial rate from any one of the four measurements. In this case, we use the first measurement.
© 2011 Pearson Education, Inc.

17. Practice – Determine the rate law and rate constant for the reaction NH4+ + NO2−  N2 + 2 H2O given the data below
Rate = k[NH4+]n[NO2]m

18. 13.4 The Integrated Rate Law : Dependence of Concentration on TIME
Half-Life
The half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactant to fall to ½ its initial value.
The half-life of the reaction depends on the order of the reaction.

20. EXAMPLE 13.4 The First-Order Integrated Rate Law: Determining the Concentration of a Reactant at a Given Time
In Example 13.3, you determined that the decomposition of SO2Cl2 (under the given reaction conditions) is first order and has a rate constant of  10–4 s–1. If the reaction is carried out at the same temperature, and the initial concentration of SO2Cl2 is M, what will the SO2Cl2 concentration be after 865 s?
SORT You are given the rate constant of a first-order reaction and the initial concentration of the reactant and asked to find the concentration at 865 seconds.
GIVEN:
FIND:
STRATEGIZE Refer to the first-order integrated rate law to determine the information you are asked to find from the given information.
EQUATION
SOLUTION
SOLVE Substitute the rate constant, the initial concentration, and the time into the integrated rate law.
Solve the integrated rate law for the concentration of [SO2Cl2]t.
© 2011 Pearson Education, Inc.

21. EXAMPLE 13.6 Half-Life FOR PRACTICE 13.6
Molecular iodine dissociates at 625 K with a first-order rate constant of s–1. What is the half-life of this reaction?
Since the reaction is first order, the half-life is given by Equation Substitute the value of k into the expression and calculate t1/2.
SOLUTION
FOR PRACTICE 13.6
A first-order reaction has a half-life of 26.4 seconds. How long will it take for the concentration of the reactant in the reaction to fall to one-eighth of its initial value?
© 2011 Pearson Education, Inc.

22. 13.5 The Effect of Temperature on Reaction Rate
Rate = k [A]n and k is dependent on temperature.
When temperature is high, rate is higher.
k is also dependent on Activation Energy (Ea).
When activation energy is high, rate is lower.
Activation energy is an energy barrier (hump) that must be overcome for the reactants to be transformed into products.

25. Activation Energy and Transition State (Activated Complex)
The activation energy is the amount of energy needed to convert reactants into the activated complex
aka transition state
The activated complex is a chemical species with partially broken and partially formed bonds
always very high in energy because of its partial bonds

26. Reaction: For the reaction to occur, the H3C─N bond must break, and
a new H3C─C bond form

28. The Collision Model How do reactants transform into products?
A collision happens with enough energy, and the right orientation, to break bonds and form new bonds (changing reactants to products).

29. H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
13.6 Reaction Mechanisms
Most reactions occur not in a single step, but through several steps.
Chemical reactions are represented by chemical equations that represents the overall reaction, not the individual steps that took place for the reaction to occur.
Overall reaction:
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
Mechanism:
1. H2(g) + ICl(g)  HCl(g) + HI(g)
2. HI(g) + ICl(g)  HCl(g) + I2(g)
The reactions in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps

30. H2(g) + 2 ICl(g)  2 HCl(g) + I2(g) 1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
Notice that the HI is a product in Step 1, but then a reactant in Step 2.
Because HI is made but then consumed, HI does not show up in the overall reaction.
Materials that are products in an early mechanism step, but then a reactant in a later step, are called reaction intermediates.

31. Rate Laws for Elementary Steps
Elementary steps are characterized by their molecularity (the number of reactant particles involved in the step).
A  products Unimolecular (First order reaction)
A + A  products Bimolecular (Second order reaction)
A + B  products Bimolecular (Second order reaction)
Each step in the mechanism is like its own little reaction – with its own activation energy and own rate law.
The rate law for an overall reaction must be determined experimentally.
But the rate law of an elementary step can be deduced from the equation of the step.
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
1) H2(g) + ICl(g)  HCl(g) + HI(g) Rate = k1[H2][ICl]
2) HI(g) + ICl(g)  HCl(g) + I2(g) Rate = k2[HI][ICl]

33. Rate-Determination Step and Overall Reaction Rate Laws
In most mechanisms, one step occurs slower than the other steps.
The result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction.
We call the slowest step in the mechanism the rate determining step.
the slowest step has the largest activation energy
The rate law of the rate determining step (without intermediates) = the rate law of the overall reaction.
The reaction is as fast as the slowest step. (Think relay race, the team is only as fast as the slowest runner)

34. NO2(g) + CO(g)  NO(g) + CO2(g) Rateobs = k[NO2]2
1. NO2(g) + NO2(g)  NO3(g) + NO(g) Rate = k1[NO2]2 Slow
2. NO3(g) + CO(g)  NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast
The first step is slower than the second step because its activation energy is larger.
The first step in this mechanism is the rate determining step.
The rate law of the first step is the same as the rate law of the overall reaction.

35. intermediates of the following mechanism
Practice – Mechanism
Determine the overall reaction, the rate determining step, the rate law, and identify all
intermediates of the following mechanism
1. A + B2 ® AB + B Slow
2. A + B ® AB Fast

36. Overall reaction 2 A + B2 ® 2 AB B is an intermediate
Practice – Mechanism
Determine the overall reaction, the rate determining step, the rate law, and identify all intermediates of the following mechanism
1. A + B2 ® AB + B Slow
2. A + B ® AB Fast
Overall reaction 2 A + B2 ® 2 AB
B is an intermediate
The first step is the rate determining step
Rate = k[A][B2], same as RDS

37. 13.7 Catalysis mechanism without catalyst
Catalysts are substances that affect the rate of a reaction without being consumed.
Catalysts work by providing an alternative mechanism for the reaction.
with a lower activation energy
Catalysts are consumed in an early mechanism step, then made in a later step.
Catalyst : Cl
Intermediate: ClO
Reactant: O3, O
Product: O2
mechanism without catalyst
O3(g) + O(g)  2 O2(g) V. Slow
mechanism with catalyst
Cl(g) + O3(g)  O2(g) + ClO(g) Fast
ClO(g) + O(g)  O2(g) + Cl(g) Slow

39. 1. HQ2R2 + R− Û Q2R2− + HR Fast 2. Q2R2− ® Q2R + R− Slow
Practice – Mechanism
Determine the overall reaction, the rate determining step, the rate law, and identify all catalysts and intermediates of the following mechanism
1. HQ2R2 + R− Û Q2R2− + HR Fast
2. Q2R2− ® Q2R + R− Slow
Overall reaction HQ2R2 ® Q2R + HR
R− is a catalyst
Q2R2− is an intermediate
The second step is the rate determining step
Rate = k[Q2R2−] = k[HQ2R2][R−][HR]-1

40. Types of Catalysts