South Pasadena • Honors Chemistry
Name
7 • Kinetics
Period
7.2
PROBLEMS
1. Which has a greater initial rate of disappearance?
Explain briefly.
concentration
A or B
B has a greater rate of disappearance, because it
has a greater slope.
2. This is a graph of the concentration of X over time.
concentration (M)
3. Which chemical equation is consistent with this
graph? Explain briefly.
(A) A  B
(B) B  A
(C) A  2B
(D) B  2A
(E) 2A  B
(F) 2B  A
A
concentration
time
B
time
B  2A. B is the reactant and A is the product
because our reaction mixture starts with some B
but no A. The slope of B is smaller than the
slope of A, so B disappears more slowly than A
appears.
0.80
0.60
0.40
0.20
0
REACTION RATES
A
B
0
–
Date
1
2
3
4
time (s)
(a) At which time (t = 0 s, 1 s, 2 s, 3 s, or 4 s) is the
concentration of X the greatest (or are they the
same concentration)? Explain briefly.
The concentration of X is greatest at t = 4 s
because that is when the value of the graph
is largest.
At which time (t = 0 s, 1 s, 2 s, 3 s, or 4 s) is the
rate of appearance of X the greatest (or is X
changing at the same rate)? Explain briefly.
The rate of appearance of X is greatest at t
= 0 s because that is when the slope is the
greatest.
4. For the reaction C4H8 (g)  2 C2H4 (g), the
following data were collected:
Time (s)
[C4H8] (mol/L)
0
1.000
10
0.913
20
0.835
30
0.763
40
0.697
50
0.637
(a) What is the average rate of the reaction
between 0 and 10 s? between 40 and 50 s?
∆[C4H8]
∆t
0.913 M ‒ 1.000 M
=‒
= 0.0087 M/s
10 s ‒ 0 s
Rate of Reaction = ‒
∆[C4H8]
∆t
0.637 M ‒ 0.697 M
=‒
= 0.0914 M/s
50 s ‒ 40 s
Rate of Reaction = ‒
(b) What is the rate of formation of C2H4 between
20 and 30 s?
∆[C2H4]
∆[C4H8]
= ‒2
∆t
∆t
0.763 M ‒ 0.835 M
= (‒2)
30 s ‒ 20 s

 = 0.0144 M/s
5. For the reaction:
Cl2 (g) + 3 F2 (g)  2 ClF3 (g)
∆[Cl2]
∆t = ‒0.012 M/s
∆[F2]
(a) Find ∆t .
∆[ClF3]
∆t .
∆[ClF3]
∆[ClF3]
= ‒2
= (‒2)(‒0.012 M/s)
∆t
∆t
= 0.024 M/s
∆[Cl2]
Rate of Reaction = ‒
= ‒ (‒0.012 M/s)
∆t
= 0.012 M/s
6. Consider the following reaction and the initial rate
data:
A  products
[A] (M) Initial rate (M/s)
Trial
1
0.15
0.008
2
0.60
0.032
3
0.90
?
(a) Find the rate law for the reaction.
Between trials 1 and 2, as [A] increases by a
factor of 4, the rate increases by a factor of 4
= 41, so the reaction is 1st order with respect
to A.

Rate Law: rate = k [A]1
(b) Find the value and units for k.
k=
7. Consider the following reaction and the initial rate
data:
2 NOCl (g)  2 NO (g) + Cl2 (g)
[NOCl] (M) Initial rate (M/s)
Trial
1
0.010
6.64 × 103
2
0.020
2.66 × 104
3
0.030
5.98 ×104
(a) Find the rate law for the reaction.
2.66 × 104 M/s k (0.020 M)x
=
6.64 × 103 M/s k (0.010 M)x

4 = 2x

x=2
Rate Law: rate = k[NOCl]2
(b) Find the value and units for k.
(c) Find Rate of Reaction.
0.032 M/s k (0.60 M)x
=
0.008 M/s k (0.15 M)x
 x=1
Rate = k [A]1 = (0.0533 s‒1)(0.90 M)1
= 0.048 M/s
Between trials 1 and 2, as [NOCl] increases
by a factor of 2, the rate increases by a
factor of 4 = 22, so the reaction is 2nd order
with respect to NOCl.
∆[F2]
∆[Cl2]
=3
= 3(‒0.012 M/s)
∆t
∆t
= ‒0.036 M/s
(b) Find
(c) Find the initial rate for Trial 3.
rate 0.008 M/s
=
= 0.053 s‒1
[A]1 (0.15 M)1
4 = 4x
k=
rate
6.64 × 103 M/s
=
[NOCl]2
(0.010 M)2
= 6.64 × 107 M‒1·s‒1
8. Consider the following reaction and the initial rate
data:
2 NO2 (g) + F2 (g)  2 NO2F (g)
Initial rate (M/s)
Trial
[NO2]
[F2]
1
0.100
0.100
0.026
2
0.200
0.100
0.051
3
0.200
0.200
0.103
4
0.400
0.400
0.411
(a) Find the rate law for the reaction.
Between trials 1 and 2, as [F2] is held
constant and [NO2] increases by a factor of
2, the rate increases by a factor of 2 = 21, so
the reaction is 1st order with respect to NO2.
0.051 M/s k (0.200 M)x(0.100 M)y
=
0.026 M/s k (0.100 M)x(0.100 M)y

2 = 2x

x=1
10. Consider the following reaction and data:
Between trials 2 and 3, as [NO2] is held
2 NO (g) + Br2 (g)  2 NOBr (g)
constant and [F2] increases by a factor of 2,
the rate increases by a
factor of 2 = 21, so
Initial rate (M/s)
Trial
[NO]
[Br2]
the reaction is 1st order with respect to F2.
1
0.10
0.20
24
0.103 M/s k (0.200 M)1(0.200 M)y
2
0.25
0.20
150
=
0.051 M/s k (0.200 M)1(0.100 M)y
3
0.10
0.50
60

2 = 2y

y=1
4
0.35
0.50
735
1
1
(a) Find the rate law for the reaction.
Rate Law: rate = k[NO2] [F2]
(b) Find the value and units for k.
k=
rate
0.026 M/s
1
1=
[NO2] [F2] (0.100 M)1(0.100 M)1
= 2.6 M‒1·s‒1
9. Consider the following reaction and data:
2 ClO2 (aq) + 2 OH‒ (aq) 
ClO3‒ (aq) + ClO2‒ (aq) + H2O (ℓ)
Initial rate (M/s)
Trial
[ClO2]
[OH‒]
1
0.060
0.030
0.0248
2
0.020
0.030
0.00276
3
0.020
0.090
0.00828
(a) Find the rate law for the reaction.
Between trials 1 and 2, as [OH‒] is held
constant and [ClO2] decreases by a factor of
3, the rate decreases by a factor of 9 = 32 so
the reaction is 2nd order with respect to
ClO2.
0.0248 M/s k (0.060 M)x(0.030 M)y
=
0.00276 M/s k (0.020 M)x(0.030 M)y

9 = 3x

x=2
Between trials 2 and 3, as [ClO2] is held
constant and [OH‒] increases by a factor of
3, the rate increases
by a factor of 3 =
31 so the reaction is 1st order with respect to
OH‒.
0.00828 M/s k (0.020 M)2(0.090 M)y
=
0.00276 M/s k (0.020 M)2(0.030 M)y

3 = 3y

y=1
Rate Law: rate = k[ClO2]2[OH‒]1
(b) Find the value and units for k.
k=
rate
0.0248 M/s
2
‒ 1=
[ClO2] [OH ] (0.060 M)2(0.030 M)1
= 229.6 M‒2·s‒1
Between trials 1 and 2, as [Br2] is held
constant and [NO] increases by a factor of
2.5, the rate increases
by a factor of 6.25
2
nd
= (2.5) , so the reaction is 2 order with
respect to NO.
150 M/s k (0.25 M)x(0.20 M)y
=
24 M/s k (0.10 M)x(0.20 M)y

6.25 = (2.5)x

x=2
Between trials 1 and 3, as [NO] is held
constant and [Br2] increases by a factor of
2.5, the rate increases by a factor of 2.5 =
(2.5)1, so the reaction is 1st order with respect
to Br2.
60 M/s k (0.10 M)2(0.50 M)y
=
24 M/s k (0.10 M)2(0.20 M)y

2.5 = (2.5)y

y=1
Rate Law: rate = k[NO]2[Br2]1
(b) Find the value and units for k.
k=
rate
24 M/s
=
[NO]2[Br2]1 (0.10 M)2(0.20 M)1
= 12,000 M‒2·s‒1