Kinetics
Chapter 15
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Kinetics – ch 15
1. The average rate of disappearance of ozone in the reaction
2 O3(g) → 3 O2(g)
is found to be 8.9 × 10–3 atm/min. What is the rate of appearance of
O2 during this interval?
Kinetics – ch 15
2. The rate law for the following reaction
2NO(g) + O2(g)  2NO2(g)
was experimentally found to be in the form:
rate=k[NO]x[O2]y
It was also found that when the NO concentration was doubled, the
rate of the reaction increases by a factor of 4. In addition, when both
the O2 and the NO concentration were doubled, the rate increases by a
factor of 8. What is the reaction order of O2?
Kinetics – ch 15
3. Consider the following reaction: A + 3 B  2 C
a. The following data was collected for this reaction at
25C. What is the rate law?
b. What is the overall order for the reaction?
c. Calculate the rate constant?
d. What is the rate for experiment 4?
Experiment
[A]0
[B]0
Initial Rate (M/s)
1
0.0019
0.031
0.141
2
0.0019
0.0031
0.0141
3
0.019
0.031
14.10
4
0.65
0.24
?
Kinetics – ch 15
4. Given the following data for the reaction:
2A + B  2C
Determine the rate law, the rate constant and the missing
concentrations.
Experiment
Initial [A]
Initial [B]
Initial Rate (M/min)
1
0.15
0.72
0.00084
2
0.15
0.36
0.00021
3
0.60
0.36
0.00021
4
0.85
?
0.0046
Kinetics – ch 15
5. Consider the following reaction:
Cl2 (g) + CHCl3 (g)  HCl (g) + CCl4 (g)
The rate law for this reaction has been observed to be:
Rate = k[Cl2]1/2[CHCl3]
What are the units of the rate constant assuming time in seconds?
Kinetics – ch 15
6. What would you have to plot in order to get a straight line for the
following:
a. Zero order
b. First order
c. Second order
Kinetics – ch 15
7. Consider the decomposition of nitrogen dioxide:
2 NO2  2 NO + O2
Rate = k[NO2]2
a. What is the integrated rate law?
b. A plot of 1/[NO2] verses time yielded a straight line with a
slope of 1.8 x 10-3 Lmol-1s-1 at 500 K. If the initial concentration
is 1.2 M, how long will it take for the [NO2] to decrease by 38%?
Kinetics – ch 15
8. The thermal decomposition of phosphine (PH3) into phosphorus and
hydrogen is a first order reaction. The half-life for this reaction is
35 sec at 680C.
a. Calculate the time required for 95% of the phosphine to
decompose.
b. What percentage of phosphine reacted after 78 seconds?
c. What is the concentration of the hydrogen after 78 s if you
started with 0.86M PH3?
Kinetics – ch 15
9. What happens to the half-lives for zero, first and second order?
a. Get longer
b. Get shorter
c. Stays constant
What is the 3rd half-life for a second order reaction with k = 0.22 hr-1M-1
with an initial concentration of 12 M.
Kinetics – ch 15
10. For the reaction A  products, successive half-lives are observed
to be 20.0, 10.0, and 5.0 min for an experiment in which [A]0 = 0.1
M. What is the concentration of A at 38.0 min?
Kinetics – ch 15
11. Consider two reaction vessels, one containing A and the other
containing B, with equal initial concentrations. If both substances
decompose by first-order kinetics where ka = 4.5 x 10-4 s-1 and kb =
3.7 x 10-3 s-1; how much time must pass to reach a condition such
that [A] = 4.00[B]?
Kinetics – ch 15
12. The rate of the reaction
O + NO2  NO + O2
was studied at a certain temperature.
a. In one experiment, [NO2] was in
large excess at a concentration of 1.0 x
1013 molecules/cm3 when the following
data collected. What is the order with
respect to the oxygen atoms?
b. The reaction is known to be first
order with respect to NO2. Determine
the overall rate law and the rate
constant.
Time
(s)
[O]
(atoms/cm3)
0
5.0 x 109
1.0 x 10-2
1.9 x 109
2.0 x 10-2
6.8 x 108
3.0 x 10-2
2.5 x 108
Kinetics – ch 15
13. Given the following mechanism:
H2O2  H2O + O
O + CF2Cl2  ClO + CF2Cl
ClO + O3  Cl + 2 O2
Cl + CF2Cl  CF2Cl2
a. Write the overall equation for the reaction?
b. Identify the reaction intermediates.
c. Identify the catalyst.
Kinetics – ch 15
14. Write the overall reaction, identify any intermediates and derive a
rate law given the following reaction mechanism:
Cl2 ⇌ 2 Cl
Cl + CHCl3  HCl + CCl3
Cl + CCl3  CCl4
(fast equilibrium)
(slow)
(fast)
Kinetics – ch 15
15. The following mechanism is proposed for the reduction of NO3- by
MoCl62-:
MoCl6
2-
NO3- + MoCl5-
k1
k-1
k2
MoCl5- + ClOMoCl5- + NO2-
a. What is the intermediate?
b. Derive an expression for the rate law (rate = d[NO2-]/dt)
for the overall reaction using steady-state approximation.
Kinetics – ch 15
16. The forward activation energy for an elementary step is 42 kJ/mol
and the reverse activation energy is 32 kJ/mol. Calculate E for
the step.
Kinetics – ch 15
17. The rate constants of a certain second-order reaction are
3.20x10-2/Ms at 24C and 0.945 /Ms at 150C.
a. Calculate the activation energy for this reaction.
b. What is the rate constant at 100C?
ln
𝑘2
𝑘1
=
𝐸𝑎 1
𝑅 𝑇1
−
1
𝑇2
Kinetics – ch 15
18. The activation energy for the reaction
H2 (g) + I2 (g) → 2 HI (g)
is changed from 184 kJ/mol to 59.0 kJ/mol both at 600. K by the
introduction of a Pt catalyst. Calculate the ratio of the catalyzed
rate constant to the un-catalyzed rate constant. Assume A is
constant.
Kinetics – ch 15
You have completed ch. 15
Ch 15 – Answer Key
1. The average rate of disappearance of ozone in the reaction
2 O3(g) → 3 O2(g)
is found to be 8.9 × 10–3 atm/min. What is the rate of appearance of O2
3 𝑚𝑜𝑙 𝑂2
during this interval? 8.9 × 10–3 atm/min
= 1.33 x 10-2 atm/min
2 𝑚𝑜𝑙 𝑂3
Ch 15 – Answer Key
2. The rate law for the following reaction
2NO(g) + O2(g)  2NO2(g)
was experimentally found to be in the form:
rate=k[NO]x[O2]y
It was also found that when the NO concentration was doubled, the rate of the
reaction increases by a factor of 4. In addition, when both the O2 and the NO
concentration were doubled, the rate increases by a factor of 8. What is the
reaction order of O2? k is always constant and if O2 is also constant the rate
law becomes ⇒ rate α [NO]x ⇒ 4 α 2x ⇒ x = 2 now if O2 is not constant the
rate law is rate α [NO]2[O2]y ⇒ 8 α [2]2[2]y ⇒ y = 1
Ch 15 – Answer Key
3. Consider the following reaction: A + 3 B  2 C
a. The following data was collected for this reaction at 25C. What is the
rate law?
b. What is the overall order for the reaction?
c. Calculate the rate constant?
d. What is the rate for experiment 4?
a. Looking at expts 1 and 2 ⇒ the conc of
𝑟𝑎𝑡𝑒 1
=
𝑟𝑎𝑡𝑒 2
A is constant ⇒
0.141
0.031 x
=
0.0141
0.0031
𝑟𝑎𝑡𝑒 1
⇒ 𝑟𝑎𝑡𝑒 3=
𝐴
𝐴
1
3
𝐵
𝐵
1
𝑥
⇒
2
⇒ x = 1 ⇒ Looking at expts 1 and 3 ⇒ the conc. of B is constant
𝑦
0.141
⇒ 14.10=
0.0019 y
⇒
0.019
y = 2 ⇒ rate law is rate=[A]2[B]1
Ch 15 – Answer Key
3. continued
b. 3
c. since rate=[A]2[B]1 ⇒ using data from expt 1 ⇒
0.141M/s=k(0.0019M)2(0.031M) ⇒ k = 1.26x106 M-2s-1
d. since rate=[A]2[B]1 ⇒ rate=(1.26x106 M-2s-1)(0.65M) 2(0.24M)= 1.28x105M/s
Ch 15 – Answer Key
4. Given the following data for the reaction:
2A + B  2C
Determine the rate law, the rate constant and the missing concentrations.
Using expts 1 and 2
0.00084
0.72 x
=
0.00021
0.36
⇒ x=2
Using expts 2 and 3
0.00021
0.15 y
=
0.00021
0.60
⇒ y=0
So the rate law ⇒ rate = k[B]2
Using expt 1 ⇒ 0.00084M/min = k(0.72M)2 ⇒ k = 0.00162M-1min-1
For expt 4 ⇒ 0.0046M/min = (0.00162M-1min-1)[B]2 ⇒ [B] = 1.68M
Ch 15 – Answer Key
5. Consider the following reaction:
Cl2 (g) + CHCl3 (g)  HCl (g) + CCl4 (g)
The rate law for this reaction has been observed to be:
Rate = k[Cl2]1/2[CHCl3]
What are the units of the rate constant assuming time in seconds?
units for k = t-1M1-overall order
The overall order for this reaction is 1.5 ⇒ the units for k ⇒ s-1M-0.5
Ch 15 – Answer Key
6. What would you have to plot in order to get a straight line for the following:
a. Zero order ⇒ conc. vs time ⇒ k = -slope
b. First order ⇒ ln(conc.) vs time ⇒ k = -slope
c. Second order ⇒ 1/conc. vs time ⇒ k = slope
Ch 15 – Answer Key
7. Consider the decomposition of nitrogen dioxide:
2 NO2  2 NO + O2
Rate = k[NO2]2
a. What is the integrated rate law? 2nd order ⇒
1
[𝐴]
= kt +
1
𝐴0
b. A plot of 1/[NO2] verses time yielded a straight line with a slope of 1.8
x 10-3 Lmol-1s-1 at 500 K. If the initial concentration is 1.2 M, how long
will it take for the [NO2] to decrease by 38%? for 2nd order k = slope
[A] = 1.2M and if A has decreased by 38% there’s 62% left so
[A]0 = (1.2M)(0.62) = 0.744M
1
[0.744]
= 1.8x10−3t +
1
1.2
⇒ t = 284s
Ch 15 – Answer Key
8. The thermal decomposition of phosphine (PH3) into phosphorus and
hydrogen is a first order reaction. The half-life for this reaction is 35 sec
at 680C.
a. Calculate the time required for 95% of the phosphine to decompose.
for 1st order reactions ⇒ t1/2 =0.693/k ⇒ k = 0.693/35s ⇒ k = 0.0198s-1
if 95% has decomposed there’s 5% left
ln(5) = -(0.0198)(t) + ln(100) ⇒ t = 151s
NOTE ⇒ %s can only be used in 1st order rate laws
b. What percentage of phosphine reacted after 78 seconds?
ln[A] = -(0.0198)(78s) + ln(100) ⇒ [A] = 21.3 ⇒ 21.3% remains so 78.7%
reacted
Ch 15 – Answer Key
8. continued
c. What is the concentration of the hydrogen after 78 s if you started
with 0.86M PH3? the balanced reaction is
2 PH3  2 P + 3 H2
2 PH3

2P
3 H2
0.86
0
0
-(0.86)(0.787)
+(0.86)(0.787)
+(3/2)(0.86)(0.787)
0.18
0.677
1.02
from part b
78.7% reacts after 78 s
molar ratio from
coefficients
Ch 15 – Answer Key
9. What happens to the half-lives for zero, first and second order?
a. Get longer ⇒ 2nd order
b. Get shorter ⇒ zero order
c. Stays constant ⇒ 1st order
What is the 3rd half-life for a second order reaction with k = 0.22 hr-1M-1 with
1
an initial concentration of 12 M. the first half-life ⇒ t½ =
𝑘𝐴
t½ =
1
0.22 12
0
=0.379 hr ⇒ each successive half-life is 2x the preceded
so the 3rd half-life is 4x the 1st half-life or 1.52 hr
Ch 15 – Answer Key
10. For the reaction A  products, successive half-lives are observed to be
20.0, 10.0, and 5.0 min for an experiment in which [A]0 = 0.1 M. What is the
concentration of A at 38.0 min? Since the half-life times is getting shorter ⇒
zero order
Use the half-life to get k ⇒ t½ =
𝐴0
2𝑘
⇒ 20min =
0.1
2𝑘
⇒ k = 0.0025min-1M-1
Use the zero order integrated rate law ⇒ [A]=-kt+[A]0 ⇒
[A]=-(0.0025min-1M-1)(38min) + (0.1M)
[A]=0.005M
Ch 15 – Answer Key
11. Consider two reaction vessels, one containing A and the other containing
B, with equal initial concentrations. If both substances decompose by firstorder kinetics where ka = 4.5 x 10-4 s-1 and kb = 3.7 x 10-3 s-1; how much time
must pass to reach a condition such that [A] = 4.00[B]?
since they are both 1st order ⇒ ln[A]=-kat + ln[A]0 and ln[B] = -kbt + ln[B]0
⇒ since they have the same initial concentrations ⇒ ln[A] + kat = ln[B] + kbt
since [A]=4[B] ⇒ ln4[B] + kat = ln[B] + kbt ⇒ ln4 = kbt – kat ⇒
ln4 = t(3.7 x 10-3 s-1 - 4.5 x 10-4 s-1 ) ⇒ t = 427s
Ch 15 – Answer Key
12. The rate of the reaction
O + NO2  NO + O2
was studied at a certain temperature.
a. In one experiment, [NO2] was in large excess at a concentration of 1.0
x 1013 molecules/cm3 when the following data collected. What is the
order with respect to the oxygen atoms?
b. The reaction is known to be first order with respect to NO2. Determine
the overall rate law and the rate constant.
a. if O is 1st order you should get a straight line if you plot ln[O] vs. t
⇒ if the slopes are the same using the first 2 data points and the last
2 data points you know it’s a line without actually plotting it ⇒
Δ𝑦
ln 5.0𝑥109 −ln(1.9𝑥109)
slope = Δ𝑥 =
= -98s-1 vs. slope =
0𝑠−0.01𝑠
since the 2 slopes are very close it must be 1st order
ln 6.8𝑥108 −ln(2.5𝑥108)
0.02𝑠−0.03𝑠
= -100s-1
Ch 15 – Answer Key
b. since both O and NO2 are 1st order the rate law is ⇒ rate = k[O][NO2] ⇒ however
when the concentration of one substance is really high you can write a pseudo rate law
⇒ rate = k’[O] ⇒ where k’ is the -slope of the line ⇒ setting the two rate laws equal to
each other ⇒ k[O][NO2] = k’[O] or k[NO2] = k’ ⇒ k(1.0 x 1013 molecules/cm3) =
99s-1 ⇒ k = 9.9x10-12s-1molecules-1cm3
Ch 15 – Answer Key
13. Given the following mechanism:
H2O2  H2O + O
O + CF2Cl2  ClO + CF2Cl
ClO + O3  Cl + 2 O2
Cl + CF2Cl  CF2Cl2
a. Write the overall equation for the reaction? sum of the elementary steps ⇒ if
you add them up the O, CF2Cl2, ClO, CF2Cl and Cl cancel out leaving you with
H2O2 + O3  H2O + 2 O2
b. Identify the reaction intermediates. ⇒ O, ClO, CF2Cl and Cl ⇒ reaction
intermediates are temporarily made then consumed
c. Identify the catalyst. ⇒ CF2Cl2 ⇒ catalysts speed up the reaction by changing
the pathway and lowering the activation energy ⇒ catalysts are temporarily
consumed then regenerated
Ch 15 – Answer Key
14. Write the overall reaction, identify any intermediates and derive a rate law given
the following reaction mechanism:
Cl2  2 Cl
(fast equilibrium)
Cl + CHCl3  HCl + CCl3
(slow)
Cl + CCl3  CCl4
(fast)
the rate law can be derived from the slowest elementary step ⇒ rate = k2[Cl][CHCl3]
⇒ the coefficient=order ONLY for elementary steps ⇒ the rate law is not
practical if it contains intermediates such as in this case the Cl ⇒ in step 1 it is a
fast equilibrium ⇒ so the rate forward=rate reverse ⇒ k1[Cl2] = k-1[Cl]2 ⇒
k [Cl ]
[Cl] = 1 2 ½⇒ now substituting this into the original rate law ⇒
k−1
k [Cl ] ½
½
rate = k2 1 2
[CHCl3] or rate = k[Cl2] [CHCl3]
k−1
Ch 15 – Answer Key
15. The following mechanism is proposed for the reduction of NO3- by MoCl62-:
a. What is the intermediate? MoCl5b. Derive an expression for the rate law (rate = d[NO2-]/dt) for the overall reaction
using steady-state approximation. the rate law rate = d[NO2-]/dt is for the second step
⇒ rate = k2[NO3-][MoCl5-] ⇒ in the steady state approximation the assumption is that
the concentration of the intermediate stays constant or the rate the intermediate is
produced = the rate the intermediate is consumed ⇒
k1[MoCl62-]= k-1[MoCl5-][Cl-] + k2[NO3-][MoCl5-]
k1[MoCl62−]
solving for [MoCl5 ⇒ [MoCl5 =
k−1[MoCl5−][Cl−]+k2[NO3−]
k2[NO3−]k1[MoCl62−]
substituting back into the rate law ⇒ rate =
k−1[MoCl5−][Cl−]+k2[NO3−]
-]
-]
Ch 15 – Answer Key
16. The forward activation energy for an elementary step is 42 kJ/mol and the
reverse activation energy is 32 kJ/mol. Calculate E for the step.
ΔE°= forward Ea – reverse Ea
ΔE°= 42kJ/mol – 32kJ/mol = 10kJ/mol
Ch 15 – Answer Key
17. The rate constants of a certain second-order reaction are
3.20x10-2/Ms at 24C and 0.945 /Ms at 150C.
a. Calculate the activation energy for this reaction.
ln
𝑘2
𝑘1
=
𝐸𝑎 1
𝑅 𝑇1
−
1
𝑇2
0.945/𝑀𝑠
𝑘
⇒ Ea =
𝑅𝑙𝑛 𝑘2
1
1
1
−𝑇
𝑇
1
⇒ Ea =
(0.0083145𝑘𝐽/𝑚𝑜𝑙𝐾)𝑙𝑛 0.032/𝑀𝑠
2
1
1
−
297𝐾 423𝐾
Ea = 28.1kJ/mol
b. What is the rate constant at 100C? ln
ln
𝑘2
0.032/𝑀𝑠
=
28.1kJ/mol
1
0.0083145𝑘𝐽/𝑚𝑜𝑙𝐾 297𝐾
−
1
373𝐾
𝑘2
𝑘1
=
𝐸𝑎 1
𝑅 𝑇1
⇒ k2 = 0.32/Ms
−
1
𝑇2
⇒
⇒
Ch 15 – Answer Key
18. The activation energy for the reaction
H2 (g) + I2 (g) → 2 HI (g)
is changed from 184 kJ/mol to 59.0 kJ/mol both at 600. K by the
introduction of a Pt catalyst. Calculate the ratio of the catalyzed
rate constant to the un-catalyzed rate constant. Assume A is
Ae−Ea(cat)/RT
e−Ea(cat)/RT
𝑘𝑐𝑎𝑡
-Ea/RT
constant. k=Ae
⇒𝑘
= −Ea(uncat)/RT = −Ea(uncat)/RT
Ae
e
𝑢𝑛𝑐𝑎𝑡
Ea(cat)/RT = (59kJ/mol)/(0.0083145kJ/molK x 600K) = 11.827
Ea(uncat)/RT = (184kJ/mol)/(0.0083145kJ/molK x 600K) = 36.883
𝑘𝑐𝑎𝑡
e−11.827
= −36.883 =
𝑘𝑢𝑛𝑐𝑎𝑡
e
7.6x1010