Chapter 14 Homework Answers
14.47 The slope of the tangent to the curve at each time is the negative of the rate at each time:
Rate60 = 8.5 × 10–4 mol L–1 s–1
Rate120 = 4.0 × 10–4 mol L–1 s–1
14.49
From the coefficients in the balanced equation we see that, for every mole of B that reacts, 2 mol of A are
consumed, and three mol of C are produced. This means that A will be consumed twice as fast as B, and C
will be produced three times faster than B is consumed.
rate of disappearance of A = 2(–0.30) = –0.60 mol L–1 s–1
rate of appearance of C = 3(0.30) = 0.90 mol L–1 s–1
14.51 We rewrite the balanced chemical equation to make the problem easier to answer: N2O5 2NO2 + 1/2O2. Thus, the rates of formation of NO2 and O2 will be, respectively, twice and one
half the rate of disappearance of N2O5.
rate of formation of NO2 = 2(2.5 × 10–6) = 5.0 × 10–6 mol L–1 s–1
rate of formation of O2 = 1/2(2.5 × 10–6) = 1.3 × 10–6 mol L–1 s–1
14.53 (a)
(b)
−
1 d [PH3] d [P4] 1 d [H 2 ]
=
=
4 dt
dt
6 dt
Rate =
1
(0.34) M s −1 = 0.085 M s-1
4
14.55 rate = (1.3 × 1011 L mol–1 s–1)(1.0 × 10–7 mol/L)(1.0 × 10–7 mol/L)
rate = 1.3 × 10–3 mol L–1 s–1
14.59 Rate = k[Rn] = 0.0125 s-1 x 1.0 x 10-9 mol L-1 = 1.25 x 10-11 M s-1
14.63 Compare the data of the first and second experiments, in which the concentration of NO is held
constant and the concentration of O2 is increased by a factor of 4. Since this caused a rate
increase by a factor of 28.4/7.10 = 41, we conclude that the order of the reaction with respect to
O2 is one (case number six in Table 14.4). In the second and third experiments, an increase in
the concentration of NO by a factor of 3 (while holding the concentration of O2 constant) caused
a rate increase by a factor of 255.6/28.4 = 9. This is the eighth case in Table 14.4, and the order
is seen to be two.
rate = k[O2][NO]2
We can use any of the three sets of data to solve for k. Using the first data set gives:
7.10 mol L–1 s–1 = k[1.0 × 10–3 mol/L][1.0 × 10–3 mol/L]2
k = 7.10 × 109 L2 mol–2 s–1
14.67 Since it is the plot of 1/conc. that gives a straight line, the order of the reaction with respect to
CH3CHO is two. The rate constant is given by the slope directly:
k = 0.0771 M–1 s–1
14.75
1
1
−
= kt
[ A]t [ A]0
1
1
−
= 0.556 L mol−1 s −1 x t
0.025 M 0.25 M
t = 65 s
14.79 Since 1/2 of the Sr–90 decays every half–life, it will take 5 half–lives, or 5 × 28 yrs = 140 yrs, for
the Sr–90 to decay to 1/32 of its present amount.
14.85 Using equation 14.9 we may determine how long it has been since the tree died.
r 
ln  o  = 1.2 × 10−4 t
 rt 
(
)
where ro = 1.2 x 10−12
 1.2 × 10−12 
ln 
 = 1.21 × 10−4 t
 4.8 × 10−14 


Taking the natural log we determine:
(

1
t=
 1.21 × 10−4
14.89

 ln

)
 1.2 × 10−12

 4.8 × 10−14


 = 2.7 × 104 yr


The graph is prepared exactly as in example 14.15 of the text. The slope is found using linear regression, to
be:
–1.67 × 104 K. Thus –1.67 × 104 K = –Ea/R
Ea = –(–1.67 × 104 K)(8.314 J K–1 mol–1) = 1.39 × 105 J/mol = 139 kJ/mol
Using equation 13.12 we have:
ln
−E a  1
k2
1
=
−


k1
R  T2
T1 
 1.08 × 10−1 L mol−1 s−1 
−E a
1 
 1
ln 
−
 =
−2
−1 −1
−1 −1  503 K
478 K 

 1.91 × 10 L mol s  8.314 J mol K
1.732 =
1.04 × 10−4 K −1
8.314 J mol−1 K −1
× Ea
Ea = 1.38 × 105 J/mol = 138 kJ/mol
14.93
Substitute into the equation:
−E a  1
k
1
ln 2 =
−


k1
R  T2
T1 
k2

 −108 × 103 J mol−1  1
1 
ln 
=
−

−5 −1
−1 −1  318 K
308 K 

8.314 J mol K
 6.2 × 10 s 
ln(k2/6.2 × 10–5 s–1) = 1.33
k2 = 6.2 × 10–5 s–1 × exp(1.33) = 2.3 × 10–4 s–1
14.97 An intermediate is formed in one step of a mechanism and then used in another step.
Intermediates in this mechanism are Cl and CCl3
Sum the three steps to obtain the overall, balanced equation for the reaction.
Cl2(g) + CHCl3(g)
→
HCl(g) + CCl4(g)
The rate law is obtained from the slow step.
Rate = k[Cl][CHCl3]
However, Cl is an intermediate and does not occur in the overall reaction. To obtain a rate
equation in terms of reactants or products we need to replace [Cl].
From the first step,
K eq =
[ Cl]2
[ Cl2 ]
1/ 2
[ Cl] = {K eq [Cl2 ]}
{
1/ 2
} [CHCl3 ] = kK eq1/2 [Cl2 ]1/ 2 [CHCl3 ] = k ' [Cl2 ]1/ 2 [CHCl3 ]
Rate = k K eq [ Cl2 ]
14.99 k = 0.693/12.5 y = 5.54 × 10–2 y–1
ln
1
= 5.54 × 10−2 y −1 × t
0.1
(
t = 41.6 y
14.100
Reaction is exothermic
)
14.103 (a) The reaction rate is first order so as the concentration of reactants triples the rate will triple.
(b) The reaction rate is second order so as the concentration of reactants triples the rate will
increase by a factor of nine.
(c) The reaction rate is zeroth order so the rate is independent of concentration. There is no
change in the rate.
(d) The reaction rate is second order so as the concentration of reactants triple the rate
increases by a factor of nine
(e) The reaction rate is third order so as the concentration of reactants triple the rate increases
by a factor of twenty seven.
14.105 (a)
rate = k1[A]2
(b)
rate = k–1[A2]1
(c)
rate = k2[A2]1[E]1
(d)
2A + E B + C
(e)
The rates for the forward and reverse directions of step one are set equal to each other
in order to arrive at an expression for the intermediate [A2] in terms of the reactant [A]:
k1[A]2 = k–1[A2]
[A2] =
k1
[A]2
k −1
This is substituted into the rate law for question (c) above, giving a rate expression that
is written using only observable reactants: rate = k2 k1 [A]2[E]1
k −1
14.111 To solve this problem, plot the data provided as 1/T vs 1/t where T is the absolute temperature and 1/t is
proportional to the rate constant.
t (min)
T (K)
1/T
ln(1/t)
10
291
0.003436 –2.302585
9
293
0.003412 –2.197224
8
294
0.003401 –2.079441
7
295
0.003389 –1.945910
6
297
0.003367 –1.791759
13.6
288
0.003472 –2.608
The slope of the graph is equal to –Ea/R, therefore:
–7,704 = –Ea/R
7,704R = Ea
7,704 K(8.314 J/mol·K) = Ea
Ea = 64,050 J/mol
Ea = 64 kJ/mol
From the straight–line equation, we can determine the time needed to develop the film at 15 °C
is 14 min.
14.114 (a)
The first step, in which a free radical is produced, is the initiation step.
(b)
Both the second and third steps are propagating steps since HBr, the desired product,
and an additional free radical are produced.
(c)
The final step in which two bromine free radicals recombine to give a bromine molecule
is the termination step.
The presence of the additional reaction step serves to decrease the concentration of
HBr.