5 Steps to a 5 AP Chemistry

10.The reaction (CH 3 ) 3 CBr(aq) + H 2 O(l) →
(CH 3 ) 3 COH(aq) +HBr(aq) follows the rate law:
Rate =k[(CH 3 ) 3 CBr]. What will be the effect of
decreasing the concentration of (CH 3 ) 3 CBr?

(A) The rate of the reaction will increase.

(B) More HBr will form.

(C) The rate of the reaction will decrease.

(D) The reaction will shift to the left.

(E) The equilibrium constant will increase.

11. When the concentration of H+(aq) is doubled for
    the reaction H 2 O 2 (aq) +2 Fe^2 +(aq) +2 H+(aq)
    →2 Fe^3 +(aq) +2 H 2 O(g), there is no change in
    the reaction rate. This indicates

(A) the H+is a spectator ion

(B) the rate-determining step does not involve H+

(C) the reaction mechanism does not involve H+

(D) the H+is a catalyst

(E) the rate law is first order with respect to H+

12.The mechanism below has been proposed for the

reaction of CHCl 3 with Cl 2.

Step l: fast

Step 2: Cl(g) +CHCl 3 (g)

→CCl 3 (g) +HCl(g) slow

Step 3: CCl 3 (g) +Cl(g) →CCl 4 (g) fast

Which of the following rate laws is consistent

with this mechanism?

(A) Rate =k[Cl 2 ]

(B) Rate =k[CHCl 3 ][Cl 2 ]

(C) Rate =k[CHCl 3 ]

(D) Rate =k[CHCl 3 ]/[Cl 2 ]

(E) Rate =k[CHCl 3 ][Cl 2 ]1/2

Cl g 2 () 2 Cl g()

Kinetics  207

 Answers and Explanations

1. C—The “2” exponent means this is a second-
   order rate law. Second-order rate laws give a
   straight-line plot for 1/[A] versus t.


2. A—The value of kremains 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.


3. B—The half-life is 0.693/k=0.693/0.049 s−^1 =14 s.
   The time given, 28 s, represents two half-lives. The
   first half-life uses one-half of the isotope, and the
   second half-life uses one-half of the remaining
   material, so only one-fourth of the original material
   remains.


4. B—Slow reactions have high activation energies.
   High activation energies are often attributed to
   strong bonds within the reactant molecules. All
   the other choices give faster rates.


5. 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)

6. 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 0.00625

Four half-lives =4(200 s) =800 s

7. C—This is the definition of the activation
   energy.


8. B—The friction supplies the energy needed to
   start the reaction. The energy needed to start the
   reaction is the activation energy.


9. E—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