4.
The initial rate for a reaction is equal to the slope of the tangent line at t≈0 in a plot of [A] versus
d[A]
dt
time. From calculus, initial rate =
d[A]
 k[A]n
Therefore, the differential rate law for a reaction is Rate =
. Assuming you have
dt

4.
some calculus in your background, derive the zero, first, and second-order integrated rate laws using
the differential rate law
Zero order:
 d[A]
 k,
dt
[ A ]t
[ A]
[ A ]t
t
[ A ]0
0

 d[A]    kdt
t
  kt , [A]t  [A]0   kt , [A]t = -kt + [A]0
[ A ]0
0
First order:
 d[A]
 k[A],
dt
[ A ]t
ln[ A]
[ A ]t
t
d[A]
 [A]    kdt
[ A ]0
0
  kt , ln[ A]t  ln[ A]0   kt , ln[A]t = -kt + ln[A]0
[ A ]0
Second order:
 d[A]
 k[A]2 ,
dt

1
[ A]
[ A ]t
[ A ]0
  kt, 
[ A ]t
t
d[A]
 [A]2    kdt
[ A ]0
0
1
1
1
1

  kt ,
 kt 
[A]t [A]0
[ A] t
[A]0
25. The reaction
2NO(g) + Cl2(g) → 2NOCl(g) was studied at -10˚C. The following results were obtained where
 [Cl2 ]
Rate =
t
[NO]0 (mol/L)
[Cl2]0 (mol/L)
Initial Rate
(mol/L • min)
0.10
0.10
0.18
0.10
0.20
0.36
0.20
0.20
1.45
a. What is the rate law?
b. What is the value of the rate constant?
25.
a. In the first two experiments, [NO] is held constant and [Cl 2] 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
x
y
y
0.18
k (0.10) (0.10)
(0.10)
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
0.18 mol
 0.10 mol   0.10 mol 
 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.)
The average rate constant is kmean = 1.8 × 102 L2/mol2min.
26. The reaction
2I-(aq) + S2O82-(aq) → I2(aq) + 2SO42-(aq)
was studied at 25˚C. The following results were obtained where
Rate = -[S2O82-]
t
[I-]0
(mol/L)
0.080
0.040
0.080
0.032
0.060
[S2O82-]0
(mol/L)
0.040
0.040
0.020
0.040
0.030
Initial Rate
(mol/L • s)
12.5 x 10-6
6.25 x 10-6
6.25 x10-6
5.00 x10-6
7.00 x 10-6
a. Determine the rate law
b. Calculate a value for the rate constant for each experiment and an average value for the rate constant.
1.25  10 6
k (0.080 ) x (0.040 ) y

, 2.00 = 2.0x, x = 1
26.
a. Rate = k[I]x[S2O82]y;
6.25  10 6
k (0.040 ) x (0.040 ) y
1.25  10 6
k (0.080 )(0.040 ) y

, 2.00 = 2.0y, y = 1; Rate = k[I][S2O82]
6.25  10 6
k (0.080 )(0.020 ) y
b. For the first experiment:
12.5  10 6 mol
 0.080 mol   0.040 mol 
3
 k

 , k = 3.9 × 10 L/mols
Ls
L
L



Each of the other experiments also gives k = 3.9 × 10 3 L/mols,
so kmean = 3.9 × 10 3 L/mols.
27. The decomposition of nitrosyl chloride was studied:
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
The following data were obtained where
Rate = -[NOCl]
t
[NOCl[0
Initial Rate
(molecules/cm3)
(molecules/cm3 • s)
3.0 x 1016
5.98 x 104
16
2.0 x 10
2.66 x 104
1.0 x 1016
6.64 x 103
16
4.0 x 10
1.06 x 105
a. What is the rate law?
b. Calculate the rate constant.
c. Calculate the rate constant when concentrations are given in moles per liter.
27.
a. Rate = k[NOCl]n; Using experiments two and three:
2.66  10 4
(2.0  1016 ) n

k
, 4.01 = 2.0n, n = 2; Rate = k[NOCl]2
6.64  10 3
(1.0  1016 ) n
2
b.
 3.0  1016 molecules 
5.98  10 4 molecules

 , 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.
c.
6.6  10 29 cm3
1L
6.022  10 23 molecules 4.0  10 8 L



molecules s
mol
mol s
1000 cm3
28. The following data were obtained for the gas-phase decomposition of dinitrogen pentoxide.
2N2O5(g) → 4NO2(g) + O2(g)
[N2O5]0
Initial Rate
(mol/L)
(mol/L • s)
0.0750
8.90 x 10-4
0.190
2.26 x 10-3
0.275
3.26 x 10-3
0.410
4.85 x 10-3
Defining the rate as - [N2O5]/ t, write the rate law and calculate the value of the rate constant.
28.
Rate = k[N2O5]x; The rate laws for the first two experiments are:
2.26 × 10 3 = k(0.190)x and 8.90 × 10 4 = k(0.0750)x
Dividing: 2.54 =
k=
(0.190 ) x
= (2.53)x, x = 1; Rate = k[N2O5]
(0.0750 ) x
8.90  10 4 mol / L  s
Rate

= 1.19 × 10 2 s 1 ; kmean = 1.19 × 10 2 s 1
[N 2 O 5 ]
0.0750 mol / L
29.
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.
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] =
30.
0.280 L 3.36 μmol 2.40 μmol
= 2.26 µmol/Ls


μmol s
L
L
a. Rate = k[ClO2]x[OH-]y; From the first two experiments:
2.30 × 10 1 = k(0.100)x(0.100)y and 5.75 × 10 2 = k(0.0500)x(0.100)y
Dividing the two rate laws: 4.00 =
(0.100 ) x
= 2.00x, x = 2
(0.0500 ) x
Comparing the second and third experiments:
2.30 × 10 1 = k(0.100)(0.100)y and 1.15 × 10 1 = k(0.100)(0.0500)y
Dividing:
2.00 =
(0.100 ) y
= 2.00y, y = 1
y
(0.0500 )
The rate law is: Rate = k[ClO2]2[OH]
2.30 × 10 1 mol/Ls = k(0.100 mol/L)2(0.100 mol/L), k = 2.30 × 102 L2/mol2s = kmean
2
b. Rate = k[ClO2]2[OH] =
2.30  10 2 L2
0.0844 mol
 0.175 mol 
= 0.594 mol/Ls
 
 
2
L
L
mol s


Integrated Rate Laws
31. The decomposition of hydrogen peroxide was studied, and the following data were obtained at a
particular temperature:
Time (s)
0
120 ± 1
300 ± 1
600 ± 1
1200 ± 1
1800 ± 1
[H2O2] (mol/L)
1.00
0.91
0.78
0.59
0.37
0.22
2400 ± 1
3000 ± 1
3600 ± 1
0.13
0.082
0.050
Assuming that
Rate = -∆[H2O2]
∆t
Determine the rate law, the integrated rate law, and the value of the rate constant. Calculate [H2O2] at 4000. s
after the start of the reaction.
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 firstorder assumption follows.
Time
(s)
[H2O2]
(mol/L)
ln H2O2]
0
120.
300.
600.
1200.
1800.
2400.
3000.
3600.
1.00
0.91
0.78
0.59
0.37
0.22
0.13
0.082
0.050
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.
 [H 2 O 2 ] 
 = -kt
ln [H2O2] = -kt + ln [H2O2]o or ln 
 [H 2 O 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 
32. A certain reaction has the following general form:
aA → bB
At a particular temperature and [A]0 = 2.00 x 10-2 M, concentration versus time data were collected for
this reaction, and a plot of ln[A] versus time resulted in a straight line with a slope value of -2.97 x 10-2
min-1.
a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.
b. Calculate the half-life for this reaction.
c. How much time is required for the concentration of A to decrease to 2.50 x 10-3 M?
32.
a. Because the ln[A] vs time plot was linear, the reaction is first order in A. The slope of the ln[A] vs.
time plot equals -k. Therefore, the rate law, the integrated rate law and the rate constant value are:
Rate = k[A]; ln[A] = -kt + ln[A]o; k = 2.97 × 10 2 min 1
b. The half-life expression for a first-order rate law is:
ln 0.5 0.6931
0.6931

t1/2 = 
, t1/2 =
= 23.3 min
k
k
2.97  10 2 min 1
c. 2.50 × 10 3 M is 1/8 of the original amount of A present, so the reaction is 87.5% complete. When a
first-order reaction is 87.5% complete (or 12.5% remains), the reaction has gone through 3 half-lives:
100%
→ 50.0% → 25%
t1/2
t1/2
→ 12.5%; t = 3 × t1/2 = 3 × 23.3 min = 69.9 min
t1/2
Or we can use the integrated rate law:
 2.50  10 3 M 
 [ A] 
2
1

   kt , ln 
ln 
 2.00  10  2 M  = -(2.97 × 10 min ) t
[
A
]
o 



ln( 0.125)
= 70.0 min
 2.97  10 2 min 1
33. The rate of the reaction
t=
NO2(g) + CO(g) → NO(g) + CO2(g)
Depends only on the concentration of nitrogen dioxide below 225˚C. At a temperature below 225˚C, the following
data were collected:
Time (s)
0
1.20 x 103
3.00 x 103
4.50 x 103
9.00 x 103
1.80 x 104
[NO2] (mol/L)
0.500
0.444
0.381
0.340
0.250
0.174
Determine the rate law, the integrated law, and the value of the rate constant. Calculate [NO2] at 2.70 x 104s after
the start of the reaction.
33.
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 firstorder reactions. The data and plots follow.
Time (s)
0
[NO2] (M)
0.500
ln [NO2]
-0.693
1/[NO2] ( M 1 )
2.00
1.20 × 103
3.00 × 103
4.50 × 103
9.00 × 103
1.80 × 104
0.444
0.381
0.340
0.250
0.174
-0.812
-0.965
-1.079
-1.386
-1.749
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 
[ NO2 ]
[ NO2 ]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
Δx (1.80  10 4  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 ]
mol s
1
= 7.62, [NO2] = 0.131 M
[ NO2 ]
34. A certain reaction has the following general form:
aA → bB
At a particular temperature and [A]0 = 2.80 x 10-3 M, concentration versus time data were collected for
this reaction, and a plot of 1/[A] versus time resulted in a straight line with a slope value of +3.60 x 10-2
L/mol · s.
a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.
b. Calculate the half-life for this reaction.
c. How much time is required for the concentration of A to decrease to 7.00 x 10-4 M?
34.
a. Because the 1/[A] vs. time plot was linear, the reaction is second order in A. The slope of the 1/[A] vs.
time plot equals the rate constant k. Therefore, the rate law, the integrated rate law and the rate constant
value are:
1
1
Rate = k[A]2;
; k = 3.60 × 10 2 L/mols
 kt 
[ A]
[A]o
1
b. The half-life expression for a second-order reaction is: t1/2 =
k [A]o
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 2 H 5 OH]o
1.25  10 2 mol / L

= 156 s
2k
2  4.00  10 5 mol / L  s
Note: we could have used the integrated rate law to solve for t1/2 where
[C2H5OH] = (1.25 × 10 2 /2) mol/L.
c. [C2H5OH] = kt + [C2H5OH]o , 0 mol/L = -(4.00 × 10 5 mol/Ls) t + 1.25 × 10 2 mol/L
t=
36.
1.25  10 2 mol / L
= 313 s
4.00  10 5 mol / L  s
From the data, the pressure of C2H5OH decreases at a constant rate of 13 torr for every 100. s. Since the
rate of disappearance of C2H5OH is not dependent on concentration, the reaction is zero order in C2H5OH.
k=
13 torr
1 atm
= 1.7 × 10 4 atm/s

100. s 760 torr
The rate law and integrated rate law are:
 1 atm 
 = -kt + 0.329 atm
Rate = k = 1.7 × 10 4 atm/s; PC2 H5OH = kt + 250. torr 
 760 torr 
At 900. s: PC2 H5OH = -1.7 × 10 4 atm/s × 900. s + 0.329 atm = 0.176 atm = 0.18 atm = 130 torr
For this reaction: t1/2 =
3.60  10
2
1
= 9.92 × 103 s
L / mol  s  2.80  10 3 mol / L
Note: We could have used the integrated rate law to solve for t1/2 where
[A] = (2.80 × 10 3 /2) mol/L.
c. Since the half-life for a second-order reaction depends on concentration, we will use the integrated rate
law to solve.
1
3.60  10 2 L
1
1
1

t
,
 kt 
4
mol  s
2.80  10 3 M
[ A]
[A]o 7.00  10 M
1.43 × 103 - 357 = 3.60 × 10 2 t, t = 2.98 × 104 s
42. The radioactive isotope 32P decays by first-order kinetics and has a half-life of 14.3 days. How long does
it take for 95% of a sample of 32P to decay?
42.
 [ A] 
ln 2 0.6931
   kt ; k 
= 4.85 × 10 2 d 1
ln 

t1/ 2 14.3 d
 [A]o 
If [A]o = 100.0, then after 95.0% completion, [A] = 5.0.
44.
44.
 5.0 
2
1
ln 
 = -4.85 × 10 d × t, t = 62 days
100
.
0


The rate law for the decomposition of phosphine (PH3)
Rate= - ∆[PH3] = k [PH3]
∆t
It takes 120. S for 1.00 M PH3 to decrease to 0.250 M. How much time is required for 2.00 M PH3 to
decrease to a concentration of 0.350 M.?
For a first-order reaction, the integrated rate law is: ln([A]/[A]o) = -kt. Solving for k:
 0.250 mol / L 
 = -k × 120. s, k = 0.0116 s 1
ln 
1
.
00
mol
/
L


 0.350 mol / L 
 = - 0.0116 s 1 × t, t = 150. s
ln 
 2.00 mol / L 