Unit VII(7)-I: Rates of Chemical Reactions (Chemical Kinetics) Chapter 12
Assigned Problems Answers
Exercises
Reaction Rates
19.
The coefficients in the balanced reaction relate the rate of disappearance of
reactants to the rate of production of products. From the balanced reaction, the rate
of production of P4 will be 1/4 the rate of disappearance of PH3, and the rate of
production of H2 will be 6/4 the rate of disappearance of PH3. By convention, all
rates are given as positive values.
= 2.4 × 10 3 mol/Ls
Rate =
Δ[P4 ]
1 Δ[PH 3 ]

= 2.4 × 10 3 /4 = 6.0 × 104 mol/Ls
Δt
4
Δt
21.
Δ[H 2 ]
6 Δ[PH 3 ]

= 6(2.4 × 10 3 )/4 = 3.6 × 10 3 mol/Ls
Δt
4 Δt
 Δ[H2O2 ]  (0.5 0 0M  1.0 0 0M)
a. average rate =
= 2.31

Δt
(2.1 6 1 04 s  0)
× 105 mol/Ls
From the coefficients in the balanced equation:
Δ[O2 ]
1 Δ[H2O2 ]
= 1.16 × 105 mol/Ls

Δt
2
Δt
b.
 Δ[H2O2 ]
 (0.250 0.500) M
= 1.16 × 105 mol/Ls

4
4
Δt
(4.3210  2.1610 ) s
Δ[O 2 ]
= 1/2 (1.16 × 105 ) = 5.80 × 106 mol/Ls
Δt
23.
Notice that as time goes on in a reaction, the average rate decreases.
a. The units for rate are always mol/Ls.
b. Rate = k; k must have units of mol/Ls.
c. Rate = k[A],
mol
 mol 
 k

Ls
 L 
k must have units of s 1
e. L2/mol2s
.
d. Rate = k[A]2,
mol
 mol 
 k

Ls
 L 
2
k must have units of L/mols.
1/ 2
24.
Rate =
k[Cl]1/2[CHCl
mol
 mol 
 k
3],

Ls
 L 
 mol 

 , k must have units of
 L 
L1/2/mol1/2s.
Rate Laws from Experimental Data: Initial Rates Method
25.
a. In the first two experiments, [NO] is held constant and [Cl2] is doubled. The rate
also doubled. Thus, the reaction is first order with respect to Cl2. Or
mathematically: Rate = k[NO]x[Cl2]y
0.36
k(0.10) x (0.20) y (0.20) y


, 2.0 = 2.0y, y = 1
0.18
k(0.10) x (0.10) y (0.10) y
We can get the dependence on NO from the second and third experiments. Here,
as the NO concentration doubles (Cl2 concentration is constant), the rate
increases by a factor of four. Thus, the reaction is second order with respect to
NO. Or mathematically:
1.45
k(0.20) x (0.20)
(0.20) x


, 4.0 = 2.0x, x = 2; So, Rate = k[NO]2[Cl2]
0.36
k(0.10) x (0.20)
(0.10) x
Try to examine experiments where only one concentration changes at a time.
The more variables that change, the harder it is to determine the orders. Also,
these types of problems can usually be solved by inspection. In general, we will
solve using a mathematical approach, but keep in mind you probably can solve
for the orders by simple inspection of the data.
b.
The rate constant k can be determined from the experiments. From experiment 1:
2
0.18mol
 0.10mol   0.10mol 
 k
 
 , k = 180 L2/mol2min
L min
L
L

 

From the other experiments:
k = 180 L2/mol2min (2nd exp.); k = 180 L2/mol2min (3rd exp.)
27.
The average rate constant is kmean = 1.8 × 102 L2/mol2min.
a. Rate = k[NOCl]n; Using experiments two and three:
2.66  104
(2.0  1016 )n
, 4.01 = 2.0n, n = 2; Rate = k[NOCl]2

k
6.64  103
(1.0  1016 )n
2
b.
 3.0  1 016 mo lecu les
5.9 8 1 04 mo lecu les

 , k = 6.6 × 1029 cm3/moleculess

k
3
3


cm s
cm


The other three experiments give (6.7, 6.6 and 6.6) × 1029 cm3/moleculess,
respectively.
The mean value for k is 6.6 × 1029 cm3/moleculess.
6.6 1 029 cm3
1L
6.0 2 2
1 023mo lecu les 4.0 1 08 L



mo lecu les
s
mo l
mo ls
1 0 0cm
0 3
a. Rate = k[Hb]x[CO]y; Comparing the first two experiments, [CO] is unchanged,
[Hb] doubles, and the rate doubles. Therefore, x = 1 and the reaction is first
order in Hb. Comparing the second and third experiments, [Hb] is unchanged,
[CO] triples. and the rate triples. Therefore, y = 1 and the reaction is first order
in CO.
c.
29.
b. Rate = k[Hb][CO]
c. From the first experiment:
0.619 µmol/Ls = k (2.21 µmol/L)(1.00 µmol/L), k = 0.280 L/µmols
The second and third experiments give similar k values, so kmean = 0.280 L/µmols.
d. Rate = k[Hb][CO] =
0.2 8 0L
3.3 6 μmo l 2.4 0 μmo l


= 2.26 µmol/Ls
μmo l s
L
L
Integrated Rate Laws
31.
The first assumption to make is that the reaction is first order because first-order
reactions are most common. For a first-order reaction, a graph of ln [H2O2] vs time
will yield a straight line. If this plot is not linear, then the reaction is not first order
and we make another assumption. The data and plot for the first-order assumption
follows.
Time
(s)
0
120.
300.
600.
1200.
1800.
2400.
3000.
3600.
[H2O2]
(mol/L)
1.00
0.91
0.78
0.59
0.37
0.22
0.13
0.082
0.050
ln H2O2]
0.000
-0.094
-0.25
-0.53
-0.99
-1.51
-2.04
-2.50
-3.00
Note: We carried extra significant figures in some of the ln values in order to reduce
round-off error. For the plots, we will do this most of the time when the ln function
is involved.
The plot of ln [H2O2] vs. time is linear. Thus, the reaction is first order. The rate law
and inte-grated rate law are: Rate = k[H2O2] and ln [H2O2] = -kt + ln [H2O2]o.
We determine the rate constant k by determining the slope of the ln [H2O2] vs time plot
(slope = -k). Using two points on the curve gives:
slope = -k =
Δy 0  (3.00)
= -8.3 × 10 4 s 1 , k = 8.3 × 10 4 s 1

Δx 0  3600.
To determine [H2O2] at 4000. s, use the integrated rate law where at t = 0, [H2O2]o = 1.00 M.
(33.)
 [H 2 O 2 ] 
 = -kt
ln [H2O2] = -kt + ln [H2O2]o or ln
 [H 2O 2 ]o 
 [H O ] 
ln 2 2  = -8.3 × 104 s 1 × 4000. s, ln [H2O2] = -3.3, [H2O2] = e 3.3 = 0.037 M
 1.00 
Assume the reaction is first order and see if the plot of ln [NO2] vs. time is linear. If
this isn’t linear, try the second-order plot of 1/[NO2] vs. time because second-order
reactions are the next most common after first-order reactions. The data and plots
follow.
Time (s)
[NO2] (M)
0
1.20 × 103
3.00 × 103
4.50 × 103
9.00 × 103
1.80 × 104
0.500
0.444
0.381
0.340
0.250
0.174
ln [NO2]
-0.693
-0.812
-0.965
-1.079
-1.386
-1.749
1/[NO2] ( M 1 )
2.00
2.25
2.62
2.94
4.00
5.75
The plot of 1/[NO2] vs. time is linear. The reaction is second order in NO2. The rate
law and integrated rate law are: Rate = k[NO2]2 and
1
1
 kt 
.
[ NO 2 ]
[ NO 2 ]o
The slope of the plot 1/[NO2] vs. t gives the value of k. Using a couple of points on
the plot:
slope = k =
Δy (5.75  2.00) M 1

= 2.08 × 104 L/mols
4
Δx (1.80  10  0) s
To determine [NO2] at 2.70 × 104 s, use the integrated rate law where 1/[NO2]o =
1/0.500 M
= 2.00 M 1 .
1
1
1
2.08 104 L
 kt 
,

× 2.70 × 104 s + 2.00 M 1
[ NO2 ]
[ NO2 ]o [ NO2 ]
mols
1
= 7.62, [NO2] = 0.131 M
[ NO 2 ]
35.
a. Because the [C2H5OH] vs. time plot was linear, the reaction is zero order in
C2H5OH. The
slope of the [C2H5OH] vs. time plot equals -k. Therefore, the rate law, the
integrated rate law and the rate constant value are: Rate = k[C2H5OH]0 = k;
[C2H5OH] = -kt + [C2H5OH]o; k = 4.00 × 105 mol/Ls.
b. The half-life expression for a zero-order reaction is: t1/2 = [A]o/2k.
t1/2 =
c.
Note: we could have used the integrated rate law to solve for t1/2 where
[C2H5OH] = (1.25 × 102 /2) mol/L.
[C2H5OH] = -kt + [C2H5OH]o , 0 mol/L = -(4.00 × 105 mol/Ls) t + 1.25 × 102 mol/L
t=
39.
41.
[C2 H5OH]o
1.25  102 mol/ L
= 156 s

2k
2  4.00  105 mol/ L  s
1.25  10 2 mol / L
= 313 s
4.00  10 5 mol / L  s
Because the 1/[A] vs. time plot is linear with a positive slope, the reaction is second
order with respect to A. The y-intercept in the plot will equal 1/[A]o. Extending the
plot, the y-intercept will be about 10, so 1/10 = 0.1 M = [A]o.
a. [A] = - kt + [A]o, [A] = -(5.0 × 102 mol/Ls) t + 1.0 × 103 mol/L
b. The half-life expression for a zero-order reaction is: t1/2 =
t1/2 =
[ A ]o
2k
1.0  10 3 mol / L
= 1.0 × 102 s
2  5.0  10 2 mol / L  s
c. [A] = -5.0 × 102 mol/Ls × 5.0 × 103 s + 1.0 × 103 mol/L = 7.5 × 104 mol/L
43.
Because 7.5 × 104 M A remains, 2.5 × 104 M A reacted, which means that 2.5 ×
104 M B has been produced.
a. When a reaction is 75.0% complete (25.0% of reactant remains), this represents
two half
lives (100% → 50%→ 25%). The first-order half-life expression is: t1/2 = (ln
2)/k. Because there is no concentration dependence for a first order half-life:
320. s = two half-lives, t1/2 = 320./2 = 160. s. This is both the first half-life, the
second half-life, etc.
ln 2
ln 2
ln 2
, k 

b. t1/2 =
= 4.33 × 103 s 1
k
t1/ 2
160. s
At 90.0% complete, 10.0% of the original amount of the reactant remains, so [A] =0.100[A]0.
 [A] 
0.100[A]0
ln 0.100
   kt, ln
= (4.33 × 103 s 1 )t, t =
= 532 s
ln
[
A
]
[
A
]
 4.33  103 s 1
o 
0

45.
Comparing experiments 1 and 2, as the concentration of AB is doubled, the initial
rate increases by a factor of 4. The reaction is second order in AB.
Rate = k[AB]2, 3.20 × 103 mol/ L  s = k1(0.200 M)2
k = 8.00 × 102 L/ mol s = kmean
For a second order reaction:
1
1

t1/2 =
= 12.5 s
2
k[AB]o 8.0 0 1 0 L / mol s  1.0 0 mol/ L
Reaction Mechanisms
50.
The observed rate law for this reaction is: Rate = k[NO]2[H2]. For a mechanism to
be plausible, the sum of all the steps must give the overall balanced equation (true
for all of the proposed mechanisms in this problem), and the rate law derived from
the mechanism must agree with the observed mechanism. In each mechanism (I III), the first elementary step is the rate-determining step (the slow step), so the
derived rate law for each mechanism will be the rate of the first step. The derived
rate laws follow:
Mechanism I: Rate = k[H2]2[NO]2
Mechanism II: Rate = k[H2][NO]
Mechanism III: Rate = k[H2][NO]2
51.
Only in Mechanism III does the derived rate law agree with the observed rate law.
Thus, only Mechanism III is a plausible mechanism for this reaction.
A mechanism consists of a series of elementary reactions where the rate law for
each step can be determined using the coefficients in the balanced equations. For a
plausible mechanism, the rate law derived from a mechanism must agree with the
rate law determined from experiment. To derive the rate law from the mechanism,
the rate of the reaction is assumed to equal the rate of the slowest step in the
mechanism.
Because step 1 is the rate-determining step, the rate law for this mechanism is: Rate
= [C4H9Br]. To get the overall reaction, we sum all the individual steps of the
mechanism.
Summing all steps gives:
C4H9Br → C4H9+ + Br
C4H9+ + H2O → C4H9OH2+
C4H9OH2+ + H2O → C4H9OH + H3O+
____________________________________
C4H9Br + 2 H2O → C4H9OH + Br + H3O+
Intermediates in a mechanism are species that are neither reactants nor products,
but that are formed and consumed during the reaction sequence.
The
intermediates for this mechanism are C4H9+ and C4H9OH2+.
Temperature Dependence of Rate Constants and the Collision Model
53.
In the following plot, R = reactants, P = products, Ea = activation energy and RC =
reaction coordinate which is the same as reaction progress. Note for this reaction
that ΔE is positive since the products are at a higher energy than the reactants.
E
Ea
P
E
R
RC
54.
When ΔE is positive, the products are at a higher energy relative to reactants and,
when ΔE is negative, the products are at a lower energy relative to reactants.
57.
The Arrhenius equation is: k = A exp (-Ea/RT) or in logarithmic form, ln k = -Ea/RT +
ln A.
Hence, a graph of ln k vs. 1/T should yield a straight line with a slope equal to -Ea/R
since the logarithmic form of the Arrhenius equation is in the form of a straight line
equation, y = mx + b. Note: We carried extra significant figures in the following ln k
values in order to reduce round off error.
T (K)
1/T ( K 1 )
k ( s 1 )
338
318
298
2.96 × 10 3
3.14 × 10 3
3.36 × 10 3
4.9 × 10 3
5.0 × 10 4
3.5 × 10 5
Slope =
 10.76  (5.85)
= -1.2 × 104 K = -Ea/R
(3.40  103  3.00  103 ) K 1
Ea = -slope × R = 1.2 × 104 K ×
ln k
-5.32
-7.60
-10.26
8.3145 J
, Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol
K mol
Catalysts
65.
a. NO is the catalyst. NO is present in the first step of the mechanism on the
reactant side, but it is not a reactant since it is regenerated in the second step.
b. NO2 is an intermediate. Intermediates also never appear in the overall balanced
equation. In a mechanism, intermediates always appear first on the product
side while catalysts always appear first on the reactant side.
c. k = A exp(Ea/RT);
k cat A exp [E a (cat) / RT ]
 E (un )  E a (cat) 

 exp a

k un
A exp[E (un ) / RT ]
RT




k cat
2100 J / mol
 = e0.85 = 2.3
 exp 
k un
8
.
3145
J
/
mol

K

298
K


The catalyzed reaction is 2.3 times faster than the uncatalyzed reaction at 25°C.