Kinetics ❮ 207❯ Answers and Explanations
- C—The exponent 2 means this is a second-
order rate law. Second-order rate laws give a
straight-line plot for 1/[A] versus t. - A—The value of k remains the same unless the
temperature is changed or a catalyst is added.
Only materials that appear in the rate law, in
this case NO 2 , will affect the rate. Adding NO 2
would increase the rate, and removing NO 2
would decrease the rate. CO has no effect on the
rate. - B—The half-life is 0.693/k = 0.693/0.049 s-^1 =
14 s. (It is possible to get this answer by round-
ing the values to 0.7/0.05 s-^1 .) The time given,
28 s, represents two half-lives. The first half-life
uses one-half of the beryllium, and the second
half-life uses one-half of the remaining mate-
rial, so only one-fourth of the original material
remains. - B—Slow reactions have high activation energies.
Low activation energies are typical of fast reac-
tions. A catalyst will increase the rate of a reac-
tion. Increasing the temperature will increase the
rate of a reaction. - A—Add the two equations together:
NO 2 (g) + CO(g) + NO(g) + O 3 (g) →
NO(g) + CO 2 (g) + NO 2 (g) + O 2 (g)Then cancel identical species that appear on
opposite sides:
CO(g) + O 3 (g) → CO 2 (g) + O 2 (g)
- C—The value will be decreased by one-half for
each half-life. Using the following table:
HALF-LIVES REMAINING
0 0.100
1 0.0500
2 0.0250
3 0.0125
4 0.00625Four half-lives = 4 (200 s) = 800 s
- C—This is the definition of the activation
energy.
8. B—The friction supplies the energy necessary to
start the reaction. This energy is the activation
energy. The free energy and the heat of reaction
for the reaction deal with the reactants and prod-
ucts. It is necessary to deal with the reactants and
transition state (activated complex), which are
separated by the activation energy.
9. C—Beginning with the generic rate law, Rate =
k[CO]m [Cl 2 ]n, it is necessary to determine the
values of m and n (the orders). Comparing
experiments 2 and 3, the rate doubles when the
concentration of CO is doubled. This direct
change means the reaction is first order with
respect to CO. Comparing experiments 1 and 3,
the rate doubles when the concentration of Cl 2
is doubled. Again, this direct change means the
reaction is first order. This gives Rate = k[CO]^1
[Cl 2 ]^1 = k[CO] [Cl 2 ].
10. C—The compound appears in the rate law, so a
change in its concentration will change the rate.
The reaction is first order in (CH 3 ) 3 CBr, so the
rate will change directly with the change in con-
centration of this reactant.
11. B—All substances involved, directly or indi-
rectly, in the rate-determining step will change
the rate when their concentrations are changed.
The ion is required in the balanced chemical
equation, so it cannot be a spectator ion, and
it must appear in the mechanism. Catalysts will
change the rate of a reaction. Since H+ does not
affect the rate, the reaction is zero order with
respect to this ion.
12. D—The rate law depends on the slow step of
the mechanism. The reactants in the slow step
are Cl and CHCl 3 (one of each). The rate law is
first order with respect to each of these. The Cl
is half of the original reactant molecule Cl 2. This
replaces the [Cl] in the rate law with [Cl 2 ]1/2. Do
not make the mistake of using the overall reac-
tion to predict the rate law. - A—Carbon monoxide, CO, does not appear in
the rate law. Since it does not appear in the rate
law, changing the concentration of CO will not
change the rate of the reaction.