Test Bank – Rates of Reaction – Comprehensive Exercises and Solutions (2025 Update)

CHAPTER 13
Rates of Reaction


■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.

13.1. Rate of formation of NO2F = [NO2F]/t. Rate of reaction of NO2 = −[NO2]/t. Divide each
rate by the coefficient of the corresponding substance in the equation:
[NO 2 F] [NO 2 ] [NO 2 F] [NO 2 ]
½ = −½ ; or =−
t t t t

[I- ] [0.00101 M  0.00169 M]
13.2. Rate = − =− = 1.13  10−4 = 1.1  10−4 M/s
t 8.00 s  2.00 s

13.3. The order with respect to CO is zero and with respect to NO2 is 2. The overall order is 2, the sum
of the exponents in the rate law.

13.4. By comparing Experiments 1 and 2, you see the rate is quadrupled when the [NO2] is doubled.
Thus, the reaction is second order in NO2, and the rate law is
Rate = k[NO2]2
The rate constant may be found by substituting experimental values into the rate-law expression.
Based on values from Experiment 1,
rate 7.1  105 mol/(L  s)
k= 2
= = 0.710 = 0.71 L/(mols)
[NO 2 ] (0.010 mol/L) 2

13.5. a. For [N2O5] after 6.00  102 s, use the first-order rate law, and solve for the concentration at
time t:
[N 2 O5 ]t
ln = −(4.80  10−4 /s)(6.00  102 s) = −0.2880
[1.65  102 M]

Take the antilog of both sides to get
[N 2 O5 ]t
= e−0.2880 = 0.74976
[1.65  102 M]

Hence,
[N2O5]t = 1.65  10−2 M  0.74976 = 0.01237 = 0.0124 mol/L




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, Chapter 13: Rates of Reaction 453

b. Substitute the values into the rate equation to get
10.0%
ln = −(4.80  10−4/s)  t
100.0%
Taking the log on the left side gives ln(0.100) = −2.30258. Hence,
−2.30258 = −(4.80  10−4 /s)  t
or,
2.30258
t= = 4797 = 4.80  103 s (80.0 min)
4.80  104 /s

13.6. Substitute k = 9.2/s into the equation relating k and t½.
0.693 0.693
t½ = = = 0.0753 = 0.075 s
k 9.2/s
By definition, the half-life is the amount of time it takes to decrease the amount of substance
present by one-half. Thus, it takes 0.0753 s for the concentration to decrease by 50% and another
0.0753 s for the concentration to decrease by 50% of the remaining 50% (to 25% left), for a total
of 0.1506, or 0.151 s.

13.7. Solve for Ea by substituting the given values into the two-temperature Arrhenius equation:
2.14  102 Ea  1 1 
ln =   
1.05  103 8.31 J/(mol  K)  759 K 836 K 

Ea
3.0146 = (1.21  10−4 /K)
8.31 J/(mol  K)

3.0146  8.31 J/mol
Ea = = 2.06  105 = 2.1  105 J/mol
1.21  104

Solve for the rate constant, k2, at 865 K by using the same equation and using
Ea = 2.06  105 J/mol:
k2 2.06  105 J/mol  1 1 
ln =   = 0.994
 

2.14  102 / M 2  s
1
8.31 J/(mol  K)  836 K 865 K 


Taking antilogarithms,
k2
= e0.994 = 2.7
2
2.14  10 / M  s  1
2

k2 = 2.7  (2.14  10−2)/(M½s) = 5.8  10−2 = 6  10−2 (M½s)

13.8. The net chemical equation is the overall sum of the two elementary reactions:
H2O2 + I−  H2O + IO−
H2O2 + IO−  H2O + O2 + I−
2H2O2  2H2O + O2




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, 454 Chapter 13: Rates of Reaction


The IO− is an intermediate because it is produced in Step 1 and later consumed in Step 2; I− is a
catalyst because it is temporarily used in Step 1 but then regenerated in Step 2. Neither appears in
the net equation.

13.9. The reaction is bimolecular because it is an elementary reaction that involves two molecules.

13.10. For NO2 + NO2  N2O4, the rate law is
Rate = k[NO2]2
(The rate must be proportional to the concentration of both reactant molecules.)

13.11. The first step is the slow, rate-determining step. Therefore, the rate law predicted by the
mechanism given is
Rate = k1 [H2O2][I−]

13.12. According to the rate-determining (slow) step, the rate law is
Rate = k2[NO3][NO]
Eliminate NO3 from the rate law by looking at the first step, which is fast and reaches
equilibrium. At equilibrium, the forward rate and the reverse rate are equal.
k1[NO][O2] = k−1[NO3]
Therefore, [NO3] = (k1/k−1)[NO][O2], so
k 2 k1
Rate = [NO]2[O2] = k[NO]2[O2]
k -1

where k2(k1/k−1) has been replaced by k, which represents the experimentally observed rate
constant.


■ ANSWERS TO CONCEPT CHECKS
13.1. a. Since the slope is steeper at point A, point A must be a faster instantaneous rate.
b. Since the curve is not a flat line, the rate of reaction must be constantly changing over time.
Therefore, the rate for the reaction cannot be constant at all points in time.

13.2. a. Keeping in mind that all reactant species must be present in some concentration for a
reaction to occur, the reaction with [Q] = 0 is the slowest because no reaction occurs. The
other two reactions are equal in rate because the reaction is zero order with respect to [Q];
as long as there is some amount of Q present, the reaction rate depends on the [R], which is
constant in this case.
b. Since [Q]o = 1, you can rewrite the rate law as follows: Rate = k[R]2.

13.3. a. A possible rate law is Rate of aging = (diet)w(exercise)x(sex)y(occupation)z. Your rate law
probably will be different; however, the general form should be the same.
b. You will need a sample of people who have all the factors the same except one. For
example, using the equation given in part a, you could determine the effect of diet if you
had a sample of people who were the same sex, exercised the same amount, and had the



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, Chapter 13: Rates of Reaction 455

same occupation. You would need to isolate each factor in this fashion to determine the
exponent on each factor.
c. The exponent on the smoking factor would be 2 since you see a fourfold rate increase:
[2]2 = 4.

13.4. The half-life of a first-order reaction is constant over the course of the reaction. The half-life of a
second-order reaction depends on the initial concentration and becomes longer as time elapses.
Thus, the reaction must be second order because the half-life increases from 20 s to 40 s after
time has elapsed.

13.5. a. Since the “hump” is larger, the A + B reaction has a higher activation energy.
b. Since the activation energy is lower, the E + F reaction would have the larger rate constant.
Keep in mind the inverse relationship between the activation energy, Ea, and the rate
constant, k.
c. Since, in both cases, energy per mole of the reactants is greater than that of the products,
both reactions are exothermic.

13.6. a. Her finding should increase the rate since the activation energy, Ea, is inversely related to
the rate constant, k; a decrease in Ea results in an increase in the value of k.
b. This is possible because the rate law does not have to reflect the overall stoichiometry of
the reaction.
c. No. Since the rate law is based on the slow step of the mechanism, it should be Rate =
k[Y]2.


■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
13.1. The four variables that can affect rate are (1) the concentrations of the reactants, although in some
cases a particular reactant's concentration does not affect the rate; (2) the presence and
concentration of a catalyst; (3) the temperature of the reaction; and (4) the surface area of any
solid reactant or solid catalyst.

13.2. The rate of reaction of HBr can be defined as the decrease in HBr concentration (or the increase
in Br2 product formed) over the time interval, t:
 [HBr]  [Br2 ]  [HBr]  [Br2 ]
Rate = −¼ =½ or − =2
t t t t

13.3. Two physical properties used to determine the rate are color, or absorption of electromagnetic
radiation, and pressure. If a reactant or product is colored, or absorbs a different type of
electromagnetic radiation than the other species, then measurement of the change in color (change
in absorption of electromagnetic radiation) may be used to determine the rate. If a gas reaction
involves a change in the number of gaseous molecules, measurement of the pressure change may
be used to determine the rate.




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