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Chapter 14
Chemical Kinetics
Chemical Kinetics
14
14.1 Reaction Rates
14.2 Collision Theory of Chemical Reactions
14.3 Measuring Reaction Progress and Expressing Reaction Rate
Average Reaction Rate
Instantaneous Rate
Stoichiometry and Reaction Rate
14.4 Dependence of Reaction Rate on Reactant Concentration
The Rate Law
Experimental Determination of the Rate Law
14.5 Dependence of Reactant Concentration on Time
First-Order Reactions
Second-Order Reactions
14.6 Dependence of Reaction Rate on Temperature
The Arrhenius Equation
Chemical Kinetics
14
14.7 Reaction Mechanisms
Elementary Reactions
Rate-Determining Step
Experimental Support for Reaction Mechanisms
14.6 Catalysis
Heterogeneous Catalysis
Homogeneous Catalysis
Enzymes: Biological Catalysts
Reaction Rates
14.1
Chemical kinetics is the study of how fast reactions take place.
Some happen almost instantaneously, while others can take millions of years.
Increasing the rate of a reaction is important to many industrial processes.
Collision Theory of Chemical Reactions
Most reactions happen faster at higher temperature.
Chemical reactions generally occur as a result of collisions between reacting molecules.
According to collision theory of chemical kinetics, the reaction rate is directly proportional to the number of molecular collisions per second:
14.2
Collisions that result in a chemical reaction are called effective collisions.
The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.
Molecules must also be oriented in a way that favors reaction.
Cl + NOCl → Cl2 + NO
Collision Theory of Chemical Reactions
An effective collision results in reaction.
Correct orientation to facilitate reaction
Collisions that result in a chemical reaction are called effective collisions.
The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.
Molecules must also be oriented in a way that favors reaction.
Cl + NOCl → Cl2 + NO
An ineffective collision results in no reaction.
Incorrect orientation does not favor reaction
Collision Theory of Chemical Reactions
When molecules collide in an effective collision, they form an activated complex (also called the transition state).
Collision Theory of Chemical Reactions
Measuring Reaction Progress and Expressing Reaction Rate
Chemical kinetics is the study of how fast reactions take place.
14.3
A → B
[A] decreases
Measuring Reaction Progress and Expressing Reaction Rate
A → B
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction Rate
Measuring Reaction Progress and Expressing Reaction Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
First 50 seconds:
First 100 seconds:
Measuring Reaction Progress and Expressing Reaction Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction Rate
The instantaneous rate is the rate for a specific instant in time.
Measuring Reaction Progress and Expressing Reaction Rate
Time (s) [Br2](M) Rate (M/s)
0.0 0.0120 4.20 x 10–5
50.0 0.0101 3.52 x 10–5
250.0 0.00500 1.75 x 10–5
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
k is called the rate constant.
Measuring Reaction Progress and Expressing Reaction Rate
Time (s) [Br2](M) Rate (M/s)
0.0 0.0120 4.20 x 10–5
50.0 0.0101 3.52 x 10–5
250.0 0.00500 1.75 x 10–5
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
at t = 50.0 s
Measuring Reaction Progress and Expressing Reaction Rate
2H2O2(aq) → 2H2O(l) + O2(g)
Measuring Reaction Progress and Expressing Reaction Rate
aA + bB → cC + dD
Measuring Reaction Progress and Expressing Reaction Rate
Write the rate expressions for the following reaction:
CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)
Solution:
Use the equation below to write the rate expressions.
Worked Example 14.1
Strategy For reactions containing gaseous species, progress is generally monitored by measuring pressure. Pressures are converted to molar concentrations using the ideal gas equation, and rate expressions are written in terms of molar concentrations.
Write the rate expressions for each of the following reactions:
(a) I-(aq) + OCl-(aq) → Cl-(aq) + OI-(aq)
(b) 2O3(g) → 3O2(g)
(c) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Solution (a) All the coefficients in this equation are 1. Therefore,
rate = = = =
Δ[I-]
Δt
Δ[OCl-]
Δt
Δ[Cl-]
Δt
Δ[OI-]
Δt
Worked Example 14.1 (cont.)
Solution
(b) rate = =
(c) rate = = = =
Δ[O3]
Δt
Δ[O2]
Δt
1
2
1
3
Δ[NH3]
Δt
1
4
Δ[O2]
Δt
1
5
Δ[NO]
Δt
1
4
Δ[H2O]
Δt
1
6
Think About It Make sure that the change in concentration of each species is divided by the corresponding coefficient in the balanced equation. Also make sure that the rate expressions written in terms of reactant concentrations have a negative sign in order to make the resulting rate positive.
Worked Example 14.2
Strategy Determine the rate of reaction and, using the stoichiometry of the reaction, convert to rates of change for the specified individual species.
Consider the reaction
4NO2(g) + O2(g) → 2N2O5(g)
At a particular time during the reaction, nitrogen dioxide is being consumed at the rate 0.00130 M/s. (a) At what rate is molecular oxygen being consumed (b) At what rate is dinitrogen pentoxide being produced
Solution
rate = = =
We are given
= 0.00130 M/s
where the minus sign indicates that the concentration of NO2 is decreasing the time.
Δ[NO2]
Δt
1
4
Δ[O2]
Δt
Δ[N2O5]
Δt
1
2
Δ[NO2]
Δt
Worked Example 14.2 (cont.)
Solution The rate of reaction, therefore, is
rate = = ( 0.00130 M/s)
= 3.25×10-4 M/s
(a) 3.25×10-4 M/s =
= 3.25×10-4 M/s
Molecular oxygen is being consumed at a rate of 3.25×10-4 M/s.
(b) 3.25×10-4 M/s =
2(3.25×10-4 M/s) =
= 6.50×10-4 M/s
Dinitrogen pentoxide is being produced at a rate of 6.50×10-4 M/s.
Δ[NO2]
Δt
1
4
1
4
Δ[O2]
Δt
Δ[O2]
Δt
Δ[N2O5]
Δt
1
2
Δ[N2O5]
Δt
Δ[N2O5]
Δt
Think About It Remember that the negative sign in a rate expression indicates that a species is being consumed rather than produced. Rates are always expressed as positive quantities.
Dependence of Reaction Rate on Reactant Concentration
The rate law is an equation that relates the rate of reaction to the concentrations of reactants.
aA + bB → cC + dD
k: the rate constant
x: order with respect to A
y: order with respect to B
x + y: represents the overall reaction order.
14.4
rate = k[A]x[B]y
determined experimentally
F2(g) + 2ClO2(g) → 2FClO2(g)
The initial rate is the rate at the beginning of the reaction.
Dependence of Reaction Rate on Reactant Concentration
Experiment [F2](M) [ClO2](M) Initial Rate (M/s)
1 0.10 0.010 1.2 x 10–3
2 0.10 0.040 4.8 x 10–3
3 0.20 0.010 2.4 x 10–3
Initial Rate Data for the Reaction between F2 and ClO2
F2(g) + 2ClO2(g) → 2FClO2(g)
Dependence of Reaction Rate on Reactant Concentration
Initial Rate Data for the Reaction between F2 and ClO2
[F2] doubles
[ClO2] constant
Rate doubles
The reaction is first order in F2; x = 1
rate = k[F2][ClO2]y
rate = k[F2]x[ClO2]y
F2(g) + 2ClO2(g) → 2FClO2(g)
Dependence of Reaction Rate on Reactant Concentration
Initial Rate Data for the Reaction between F2 and ClO2
[F2] constant
[ClO2]
x 4
Rate
x 4
The reaction is first order in ClO2; y = 1
rate = k[F2][ClO2]y
rate = k[F2][ClO2]
aA + bB → cC + dD
Dependence of Reaction Rate on Reactant Concentration
Experiment [A](M) [B](M) Initial Rate (M/s)
1 0.10 0.015 2.1 x 10–4
2 0.20 0.015 4.2 x 10–4
3 0.10 0.030 8.4 x 10–4
Initial Rate Data for the Reaction between A and B
rate = k[A]x[B]y
Dependence of Reaction Rate on Reactant Concentration
[A]
x 2
[B]
constant
Rate
x 2
aA + bB → cC + dD
Initial Rate Data for the Reaction between A and B
The reaction is first order in A; x = 1
rate = k[A]x[B]y
rate = k[A][B]y
Dependence of Reaction Rate on Reactant Concentration
[A] constant
[B]
x 2
Rate
x 4
aA + bB → cC + dD
Initial Rate Data for the Reaction between A and B
The reaction is second order in B; y = 2
rate = k[A][B]y
rate = k[A][B]2
Dependence of Reaction Rate on Reactant Concentration
Three important things to remember about the rate law:
The exponents in a rate law must be determined from a table of experimental data.
Comparing changes in individual reactant concentrations with changes in rate shows how the rate depends on each reactant concentration.
Reaction order is always defined in terms of reactant concentrations, never product concentrations.
The reaction of peroxydisulfate ion (S2O82–) with iodide ion (I–) is:
S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq)
Determine the rate law and calculate the rate constant, including its units.
Dependence of Reaction Rate on Reactant Concentration
Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)
1 0.080 0.034 2.2 x 10–4
2 0.080 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
Initial Rate Data for the Reaction between S2O82– and I–
Solution:
Step 1: In experiments 1 and 2, [S2O82–] is constant. The [I–] is doubled, and rate doubles.
Dependence of Reaction Rate on Reactant Concentration
Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)
1 0.080 0.034 2.2 x 10–4
2 0.080 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
Initial Rate Data for the Reaction between S2O82– and I–
The reaction is first order in I–
Solution:
Step 2: In experiments 2 and 3, [S2O82–] is doubled, [I–] is constant, and rate doubles.
Dependence of Reaction Rate on Reactant Concentration
Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)
1 0.080 0.034 2.2 x 10–4
2 0.080 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
Initial Rate Data for the Reaction between S2O82– and I–
The reaction is first order in S2O82–
Solution:
Step 2: In experiments 2 and 3, [S2O82–] is doubled, [I–] is constant, and rate doubles.
Dependence of Reaction Rate on Reactant Concentration
Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)
1 0.080 0.034 2.2 x 10–4
2 0.080 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
Initial Rate Data for the Reaction between S2O82– and I–
The rate law is: rate = k [S2O82–] [I–]
Solution:
Step 3: Use the data from any experiment to calculate k.
Dependence of Reaction Rate on Reactant Concentration
Experiment [S2O82–](M) [I–](M) Initial Rate (M/s)
1 0.080 0.034 2.2 x 10–4
2 0.080 0.017 1.1 x 10–4
3 0.16 0.017 2.2 x 10–4
Initial Rate Data for the Reaction between S2O82– and I–
Worked Example 14.3
Strategy Compare two experiments at a time to determine how the rate depends on the concentration of each reactant.
The rate law is rate = k[NO]x[H2]y.
The gas-phase reaction of nitric oxide with hydrogen at 1280°C is
2NO(g) + 2H2(g) → N2(g) +2H2O(g)
From the following data collected at 1280°C, determine (a) the rate law, (b) the rate constant, including units, and (c) the rate of the reaction when [NO] = 4.8×10-3 M and [H2] = 6.2×10-3 M.
Experiment [NO] (M) [H2] (M) Initial rate (M/s)
1 5.0×10-3 2.0×10-3 1.3×10-5
2 1.0×10-2 2.0×10-3 5.0×10-5
3 1.0×10-2 4.0×10-3 1.0×10-4
Worked Example 14.3 (cont.)
Solution The rate of reaction, therefore, is
= ≈ 4 =
Canceling identical terms in the numerator and denominator gives
Therefore, x = 2. The reaction is second order in NO.
Dividing the rate from experiment 3 by the rate from experiment 2, we get
= = 2 =
Canceling identical terms in the numerator and denominator gives
Therefore, y = 1. The reaction is first order in H2. The overall rate law is
rate = k[NO]2[H2]
5.0×10-5 M/s
1.3×10-5 M/s
rate2
rate1
k(1.0×10-2 M)x(2.0×10-3 M)y
k(5.0×10-3 M)x(2.0×10-3 M)y
(1.0×10-2 M)x
(5.0×10-3 M)x
= 2x = 4
1.0×10-4 M/s
5.0×10-5 M/s
rate3
rate2
k(1.0×10-2 M)x(4.0×10-3 M)y
k(1.0×10-2 M)x(2.0×10-3 M)y
(4.0×10-3 M)y
(2.0×10-3 M)y
= 2y = 2
Worked Example 14.3 (cont.)
Solution We can use data from any of the experiments to calculate the value and units of k. Using the data from experiment 1 gives
k = =
(c) Using the rate constant determined in part (b) and the concentrations of NO and H2 given in the problem statement, we can determine the reaction rate as follows:
rate = (2.6×102 M-2 s-1)(4.8×10-3 M)2(6.2×10-3 M)
= 3.7×10-5 M s-1
1.3×10-5 M/s
(5.0×10-3 M)2(2.0×10-3 M)
rate
[NO]2[H2]
= 2.6×102 M-2 s-1
Think About It The exponent for the concentration of H2 in the rate law is 1, whereas the coefficient for H2 in the balanced equation is 2. It is a common error to try to write a rate law using the stoichiometric coefficients as the exponents. Remember that, in general, the exponents in the rate law are not related to the coefficients in the balanced equation. Rate laws must be determined by examining a table of experimental data.
Dependence of Reactant Concentration on Time
The rate law can be used to determine the rate of a reaction using the rate constant and the reactant concentrations:
A rate law can also be used to determine the concentration of a reactant at a specific time during a reaction.
14.5
rate = k[A]x[B]y
rate
rate constant
rate law
Dependence of Reactant Concentration on Time
A first-order reaction is a reaction whose rate depends on the concentration of one of the reactants raised to the first power.
C2H6 → 2 ·CH3 rate = k[C2H6]
2N2O5(g) → 2NO2(g) + O2(g) rate = k[N2O5]
Dependence of Reactant Concentration on Time
In a first-order reaction of the type
A → products
The rate can be expressed as the rate of change in reactant concentration,
as well as in the form of the rate law:
rate = k[A]
Setting the two expressions equal to each other yields:
Dependence of Reactant Concentration on Time
Using calculus, it is possible to show that:
ln is the natural logarithm
[A]0 and [A]t refer to the concentration of A at times 0 and t
The equation above is sometimes called the integrated rate law for a first order reaction.
Dependence of Reactant Concentration on Time
The rate constant for the reaction 2A → B is 7.5 x 10–3 s–1 at 110°C. The reaction is first order in A. How long (in seconds) will it take for [A] to decrease from 1.25 M to 0.71 M
Solution:
Step 1: Use the equation below to calculate time in seconds.
t = 75 seconds
Worked Example 14.4
Strategy Use ln ([A]t/[A]0) = –kt to find [H2O2]t where t = 3 h, and then solve for t to determine how much time must pass for [H2O2]t to equal 0.10 M. [H2O2]0 = 0.75 M; time t for part (a) is (3 h)(60 min/h)(60 s/min) = 10,800 s.
The decomposition of hydrogen peroxide is first order in H2O2.
2H2O2(aq) → 2H2O(l) + O2(g)
The rate constant for this reaction at 20°C is 1.8×10-5 s-1. If the start concentration of H2O2 is 0.75 M, determine (a) the concentration of H2O2 remaining after 3 h and (b) how long it will take for the H2O2 concentration to drop to 0.10 M.
Solution
(a) ln
(b) ln
[H2O2]t
[H2O2]0
= –kt
[H2O2]t
0.75 M
= –(1.8×10-5 s-1)(10,800 s) = –0.1944
Worked Example 14.4 (cont.)
Solution Take the inverse natural logarithm of both sides of the equation to get
[H2O2]t = (0.823)(0.75 M) = 0.62 M
The concentration of H2O2 after 3 h is 0.62 M.
(b) ln
The time required for the peroxide concentration to drop to 0.10 M is 1.1×105 s or about 31 h.
[H2O2]t
0.75 M
= e–0.1944 = 0.823
0.10 M
0.75 M
= –2.015 = –(1.8×10-5 s-1)t
2.015
1.8×10-5 s-1
= t = 1.12×105 s
Think About It Don’t forget the minus sign. If you calculate a concentration at time t that is greater than the concentration at time 0 (or if you get a negative time required for the concentration to drop to a specified level), check your solution for this common error.
Dependence of Reactant Concentration on Time
Rearrangement of the first-order integrated rate law gives:
Rearrangement in this way has the form
of the linear equation y = mx + b.
ln[A]t = –kt + ln[A]0
ln[A]t = –kt + ln[A]0
Slope = –k
Intercept = ln[A]0
Dependence of Reactant Concentration on Time
The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time.
CH3—N=N—CH3(g) → N2(g) + C2H6(g)
The data obtained at 300°C are listed in the following table:
Time (s) Pazomethane (mmHg)
0 284
100 220
150 193
200 170
250 150
300 132
Dependence of Reactant Concentration on Time
Plotting the data gives a straight line, indicating the reaction is first order.
ln[A]t = –kt + ln[A]0
Slope = –2.55 x 10–3 s–1
Intercept = 5.65
Dependence of Reactant Concentration on Time
Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows:
C2H5I(g) → C2H4(g) + HI(g)
Determine the rate of the reaction, after verifying that the reaction is first order.
Time (s) [C2H5I] (M)
0 0.36
15 0.30
30 0.35
48 0.19
75 0.13
Dependence of Reactant Concentration on Time
Solution:
Plot ln[C2H5I] vs time. If a straight line results, the reaction is first order. The slope is equal to k.
Slope = –1.3 x 10–2 s–1; k = 1.3 x 10–2 s–1
Time (s) [C2H5I] (M) ln[C2H5I]
0 0.36 -1.02
15 0.30 -1.20
30 0.35 -1.39
48 0.19 -1.66
75 0.13 -2.04
Worked Example 14.5
Strategy We can use ln ([A]t/[A]0) = –kt only for first-order reactions, so we must first determine if the decomposition of azomethane is first order. We do this by plotting ln P against time. If the reaction is first order, we can use ln ([A]t/[A]0) = –kt and the data at any two of the times in the table to determine the rate constant.
The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time:
CH3 N=N CH3(g) → N2(g) + C2H6(g)
The data obtained at 20°C are listed in the following table:
Time (s) Pazomethane (mmHg)
0 284
100 220
150 193
200 170
250 150
300 132
Worked Example 14.5 (cont.)
Solution The table expressed in ln P is
Plotting these data gives a straight line, indicating that the reaction is indeed first order. Thus, we can use ln ([A]t/[A]0) = –kt in terms of pressure.
ln = –kt
Pt and P0 can be pressures at any two times during the experiment. P0 need not be the pressure at 0 s–it need only be at the earlier of the two times.
Pt
P0
Time (s) ln P
0 5.649
100 5.394
150 5.263
200 5.136
250 5.011
300 4.883
Worked Example 14.5 (cont.)
Solution Using data from times 100 s and 250 s of the original table (Pazomethane versus t), we get
ln = –kt
ln 0.682 = –k(150 s)
k = 2.55×10-3 s-1
150 mmHg
220 mmHg
Think About It We could equally well have determined the rate constant by calculating the slope of the plot of ln P versus t. Using the two points labeled on the plot, we get
slope =
= 2.55×10-3 s-1
Remember that slope = –k, so k = 2.55×10-3 s-1.
5.011 – 5.394
250 – 100
Dependence of Reactant Concentration on Time
The half-life (t1/2) is the
time required for the
reactant concentration
to drop to half its
original value.
Worked Example 14.6
Strategy Use t = 0.693/k to calculate t in seconds, and then convert to minutes.
The decomposition of ethane (C2H6) to methyl radicals (CH3) is a first-order reaction with a rate constant of 5.36×10-4 s-1 at 700°C.
C2H6 → CH3
Calculate the half-life of the reaction in minutes.
Solution t = = = 1293 s
1293 s × = 21.5 min
The half-life of ethane decomposition at 700°C is 21.5 min.
0.693
k
0.693
5.36×10-4 s-1
1 min
60 s
Think About It Half-lives and rate constants can be expressed using any units of time and reciprocal time, respectively. Track units carefully when you convert from one unit of time to another.
Dependence of Reactant Concentration on Time
The half-life (t1/2) is the time required for the reactant concentration to drop to half its original value.
t = t1/2 when [A]t = [A]0.
rearranges
simplifies
Dependence of Reactant Concentration on Time
Calculate the half-life of the decomposition of azomethane, k = 2.55×10–3 s-1.
Solution:
Step 1: Use the equation below to calculate half-life:
Dependence of Reactant Concentration on Time
A second-order reaction is a reaction whose rate depends on the concentration of one reactant raised to the second power or on the product of the concentrations of two different reactants (first order in each).
Second-order integrated rate law:
Second-order half-life:
Worked Example 14.7
Strategy Use 1/[A]t = kt + 1/[A]0 to determine [I]t at t = 2.0 min; use t = 1/k[A]0 to determine t when [I]0 = 0.60 M and when [I]0 = 0.42 M.
Iodine atoms combine to form molecular iodine in the gas phase:
I(g) + I(g) → I2(g)
This reaction is second order and has a rate constant of 7.0×109 M-1 s-1 at 23°C. (a) If the initial concentration of I is 0.086 M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction when the initial concentration of I is 0.60 M and when the initial concentration of I is 0.42 M.
Solution t = (2.0 min)(60 s/min) = 120 s
(a) = kt +
= (7.0×109 M-1 s-1)(120 s) +
1
[A]t
1
[A]0
1
0.086 M
Worked Example 14.7 (cont.)
Solution
= 8.4×1011 M-1
[A]t = = 1.2×10-12 M
The concentration of atomic iodine after 2 min is 1.2×10-12 M.
(b) When [I]0 = 0.60 M,
t = = = 2.4×10-10 s
When [I]0 = 0.42 M,
t = = = 3.4×10-10 s
1
8.4×1011 M-1
1
k[A]0
1
(7.0×109 M-1 s-1)(0.60 M)
1
k[A]0
1
(7.0×109 M-1 s-1)(0.42 M)
Think About It (a) Iodine, like the other halogens, exists as diatomic molecules at room temperature. It makes sense, therefore, that atomic iodine would react quickly, and essentially completely, to form I2 at room temperature. The very low remaining concentration of I after 2 min makes sense. (b) As expected, the half-life of this second-order reaction is not constant. (A constant half-life is a characteristic of first-order reactions.)
Dependence of Reactant Concentration on Time
The rate of a zero-order reaction is a constant.
Third-order and higher are rare.
The dependence of the rate constant on temperature can be expressed by the Arrhenius equation.
A represents the collision frequency and is called the frequency factor.
Ea is the activation energy (in kJ/mol).
R is the gas constant (8.314 J/mol K).
T is the absolute temperature.
e is the base of the natural logarithm.
Dependence of Reaction Rate on Temperature
14.6
Taking the natural log of both sides, the Arrhenius equation may be written as:
Rearrangement gives the linear form of the Arrhenius equation:
Dependence of Reaction Rate on Temperature
Dependence of Reactant Rate on Temperature
Rate constants for the reaction
CO(g) + NO2(g) → CO2(g) + NO(g)
were measured at four different temperatures. The data are shown in the table. Determine the activation energy for the reaction.
k (M–1·s–1) T (K)
0.0521 288
0.101 298
0.184 308
0.332 318
Dependence of Reactant Rate on Temperature
Solution:
Plot ln k versus 1/T and determine the slope of the line;
slope = –Ea/R.
slope = –5.6 x 103 K = –Ea/R
Ea = (5.6 x 103 K )(8.314 J/mol·K) = 46 kJ/mol
k (M–1·s–1) T (K) ln k 1/T (K–1)
0.0521 288 0.00347 -2.95
0.101 298 0.00336 -2.29
0.184 308 0.00325 -1.69
0.332 318 0.00314 -1.10
Worked Example 14.8
Strategy Plot ln k versus 1/T, and determine the slope of the resulting line. According to ln k = (–Ea/R)(1/T) + ln A, slope = –Ea/R. R = 8.314 J/mol K.
Rate constants for the reaction
CO(g) + NO2(g) → CO2(g) + NO(g)
were measured at four different temperatures. The data are shown in the table. Plot ln k versus 1/T, and determine the activation energy (in kJ/mol) for the reaction.
k (M-1 s-1) T (K)
0.0521 288
0.101 298
0.184 308
0.332 318
Worked Example 14.8 (cont.)
Solution Taking the natural log of each value of k and the inverse of each value of T gives
A plot of these data yields the following graph:
ln k 1/T (K-1)
–2.95 3.47×10-3
–2.29 3.36×10-3
–1.69 3.25×10-3
–1.10 3.14×10-3
Worked Example 14.8 (cont.)
Solution The slope is determined using the x and y coordinates of any two points on the line. Using the points that are labeled on the graph gives
slope = = –5.5×103 K
The value of the slope is –5.5×103 K. Because the slope = –Ea/R,
Ea = –(slope)(R)
= –(–5.5×103 K)(8.314 J/K mol)
= 4.6×104 J/mol or 46 kJ/mol
–1.4 – (–2.5)
3.2×10-3 K-1 – 3.4×10-3 K-1
Think About It Note that while k has units M-1 s-1, ln k has no units.
Dependence of Reactant Rate on Temperature
A two point form of the Arrhenius equation may be written:
If the rate constants at two different temperatures are known, it is possible to calculate the activation energy.
If the activation energy and the rate constant at one temperature are known, it is possible to determine the rate constant at any other temperature.
Worked Example 14.9
Strategy Rearrange and solve for Ea using the following
The rate constant for a particular first-order reaction is given for three different temperatures:
Using the data, calculate the activation energy of the reaction.
T (K) k (s-1)
400 2.9×10-3
450 6.1×10-2
500 7.0×10-1
Ea = R
k1
k2
ln
1
T2
1
T2
–
Worked Example 14.9 (cont.)
Solution
The activation energy of the reaction is 91 kJ/mol.
Think About It A good way to check your work is to use the value of Ea that you calculated (and Equation 14.11) to determine the rate constant at 500 K. Make sure it agrees with the value in the table.
Ea = 8.314 J/K mol
2.9×10-3
6.1×10-2
ln
1
450 K
1
400 K
–
= 91,173 J/mol = 91 kJ/mol
Worked Example 14.10
Strategy Rearrange and solve for k2 using the following
Ea = 8.3×104 J/mol, T1 = 423 K, T2 = 573 K, R = 8.314 J/K mol, and k1 = 2.1×10-2 s-1.
A certain first-order reaction has an activation energy of 83 kJ/mol. If the rate constant for this reaction is 2.1×10-2 s-1 at 150°C, what is the rate constant at 300°C
k2 =
k1
1
T2
1
T2
–
Ea
R
e
Solution
The rate constant of 300°C is 10 s-1.
k2 =
2.1×10-2 s-1
1
573 K
1
423 K
–
8.3×104 J/mol
8.314 J/K mol
e
= 1.0×101 s-1
Think About It Make sure that the rate constant you calculate at a higher temperature is in fact higher than the original rate constant. According to the Arrhenius equation, the rate constant always increases with increasing temperature. If you get a smaller k at a higher temperature, check your solution for mathematical errors.
Reaction Mechanisms
A balanced chemical equation does not indicate how a reaction actually takes place.
The sequence of steps that sum to give the overall reaction is called the reaction mechanism.
Step 1: A + B → C
Step 2: C + B → D
Overall reaction: A + 2B → D
14.7
Chemical species that appear in the reaction mechanism, but not in the overall chemical equation are called intermediates.
Reaction Mechanism
Step 1:
Step 2:
NO + NO
N2O2 + O2
N2O2
2NO2
Overall reaction:
2NO + O2
2NO2
Each step in a reaction mechanism represents an elementary reaction, one that occurs in a single collision of the reactant molecules.
The molecularity of an elementary reaction is essentially the number of reactant molecules involved in the collision.
unimolecular (one reactant molecule)
A → products rate = k[A] first order
bimolecular (two reactant molecules)
A + B → products rate = k[A][B] second order
A + A → products rate = k[A]2 second order
termolecular (three reactant molecules)
Reaction Mechanism
In a reaction mechanism consisting of more than elementary step, the rate law for the overall process is given by the rate-determining step.
The rate determining step is the slowest step in the sequence.
A proposed mechanism must satisfy two requirements:
The sum of the elementary reaction must be the overall balanced equation for the reaction.
The rate determining step must have the same rate law as that determined from the experimental data.
Reaction Mechanism
The decomposition of hydrogen peroxide can be facilitated by iodide ions:
Reaction Mechanism
Step 1: (slow)
Step 2:
H2O2 + I–
H2O2 + IO–
H2O + IO–
H2O + O2 + I–
Overall reaction:
2H2O2
2H2O + O2
Rate = k1[H2O2][I–]
Worked Example 14.11
Strategy Add the two equations, canceling identical terms on opposite sides of the arrow, to obtain the overall reaction. The canceled terms will be the intermediates if they were first generated and then consumed. Write rate laws for each elementary step; the one that matches the experimental rate law will be the rate-determining step.
The gas-phase decomposition of nitrous oxide (N2O) is believed to occur in two steps:
Step 1: N2O → N2 + O
Step 2: N2O + O → N2 + O2
Experimentally the rate law is found to be rate = k[N2O]. (a) Write the equation for the overall reaction. (b) Identify the intermediate(s). (c) Identify the rate-determining step.
k2
k1
Worked Example 14.11 (cont.)
Think About It A species that gets canceled when steps are added may be an intermediate or a catalyst. In this case, the canceled species is an intermediate because it was first generated and then consumed. A species that is first consumed and then generated, but doesn’t appear in the overall equation, is a catalyst.
Solution Step 1: N2O → N2 + O rate = k[N2O]
Step 2: N2O + O → N2 + O2 rate = k[N2O][O]
(a) 2N2O → 2N2 + O2
(b) O (atomic oxygen) is the intermediate.
(c) Step 1 is the rate-determining step because its rate law is the same as the experimental law: rate = k[N2O].
k2
k1
Worked Example 14.12
Strategy To establish the plausibility of a mechanism, we must compare the rate law of the rate-determining step to the experimentally determined rate law. In this case, the rate-determining step has an intermediate (N2O2) as one of its reactants, giving us a rate law of rate = k2[N2O2][O2]. Because we cannot compare this directly to the experimental rate law, we must solve for the intermediate concentration in terms of reactant concentrations.
Consider the gas-phase reaction of nitric oxide and oxygen that was described at the beginning of Section 14.5.
2NO(g) + O2(g) → 2NO2(g)
Show that the following mechanism is plausible. The experimentally determined rate law is rate = k[N2O]2[O2].
Step 1: NO(g) + NO(g) N2O2(g) (fast)
Step 2: N2O2(g) + O2(g) → 2NO2(g) (slow)
k2
k1
k-1
Worked Example 14.12 (cont.)
Solution The first step is a rapidly established equilibrium. Both the forward and reverse of step 1 are elementary processes, which enables us to write their rate laws from the balanced equation.
rateforward = k1[NO]2 and ratereverse = k-1[N2O2]
Because at equilibrium the forward and reverse processes are occurring at the same rate, we can set their rates equal to each other and solve for the intermediate concentration.
k1[NO]2 = k-1[N2O2]
[N2O2] =
Substituting the solution into the original rate law (rate = k[N2O2][O2]) gives
rate = k2 [O2] = k[NO]2[O2] where k =
k1[NO]2
k-1
k1[NO]2
k-1
k2k1
k-1
Think About It Not all reactions have a single rate-determining step. Analyzing the kinetics of reactions with two or more comparably slow steps is beyond the scope of this book.
Catalysis
14.8
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
A catalyst speeds up a reaction by providing a set of elementary steps with more favorable kinetics than those that exist in its absence.
A catalyst usually speeds up a reaction by lowering the activation energy.
Step 1: (slow)
Step 2:
H2O2 + I–
H2O2 + IO–
H2O + IO–
H2O + O2 + I–
Overall reaction:
2H2O2
2H2O + O2
Catalysis
In the presence of a catalyst the rate constant is kc, called the catalytic rate constant.
ratecatalyzed > rateuncatalyzed
uncatalyzed
catalyzed
Catalysis
In heterogeneous catalysis, the reactants and the catalysts are in different phases.
Catalysis
In homogeneous catalysis, the reactants and catalyst are dispersed in a single phase, usually liquid.
Advantages:
Reactions can be carried out under atmospheric conditions
Can be designed to function selectively
Are generally cheaper
Catalysis
Enzymes are biological catalysts.
Catalysis
The mathematical treatment of enzyme kinetics is complex, but can be simplified:
ES
E + P
E + S
ES
uncatalyzed
catalyzed
Catalysis
Generally, the rate of an enzyme catalyzed reaction is given by the equation:
Key Points
14
Average Reaction Rates
Instantaneous Rate
Stoichiometry and Reaction Rate
The Rate Law
Experimental Determination of the Rate Law
First-Order Reactions
Second-Order Reactions
Collision Theory
Elementary Reactions
Rate-Determining Step
Experimental Support for Reaction Mechanisms
Heterogeneous Catalysis
Homogeneous Catalysis
Enzymes: Biological Catalysts

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