Solution Set 8

CHEM 3411, Fall 2010
Solution Set 8
In this solution set, an underline is used to show the last significant digit of numbers. For instance in
x = 2.51693
the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are not
significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in
calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps.
Numbers without underlines (including final answers) are shown with the proper number of sig figs.
1 Exercise 21.7b pg 826
Given
At T = 400 K, the rate of decomposition of a gaseous compound initially at a pressure of p 0 = 12.6 kPa, was
v 1 = 9.71 Pa s −1 when 10.0 percent had reacted (extent of reaction ξ 1 = 0.100) and v 2 = 7.67 Pa s −1 when 20.0
percent had reacted (ξ 2 = 0.200).
In terms of given variables, this is written:
T = 400 K
p 0 = 12.6 kPa
v 1 = 9.71 Pa s −1
ξ 1 = 0.100
v 2 = 7.67 Pa s −1
ξ 2 = 0.200
Find
Determine the order of this reaction.
Strategy
In general, we know the rate of a reaction with a single reactant obeys an expression of the form
v = dp
dt = −kpn
where p is the pressure of the reactants, n is the order of the reaction, and k is the rate constant.
Additionally, we know the rate of the reaction at two different times (different reactant pressures) which gives us two
the equations.
1

CHEM 3411, Fall 2010
Solution Set 8
v 1 = −kp n 1
v 2 = −kp n 2
With some algebra we can solve for n from these two equations as we know v 1 , p 1 , v 2 , and p 2 .
v 1
p n 1
= −k = v 2
p n 2
p n 2
p n 1
= v 2
v 1
(
p2
p 1
) n
= v 2
v 1
( ) ( )
p2 v2
n ln = ln
p 1 v 1
n =
ln
(
v2
v 1
)
ln
(
p2
p 1
)
We can relates the pressures of the reactant, p 1 and p 2 , to the extent of reaction at each point in time, ξ 1 and ξ 2
respectively. The extent of reaction ξ i at the i-th point in time is related to moles of reactant remaining n i and the
initial moles of the reactant n 0 by
n i = n 0 (1 − ξ i )
From the ideal gas law we know that the moles of a gas and the pressure of the gas are proportional. This tells us
the pressure of gas reactant p i at each i-th point in time is given by
p i = p 0 (1 − ξ i )
Substituting the expressions for p 1 and p 2 , p 1 = p 0 (1 − ξ 1 ) and p 2 = p 0 (1 − ξ 2 ) respectively, allows us to solve for
the value of n.
( )
v
ln 2
v 1
n = (
ln
p2
ln
=
ln
)
p 1
( )
v 2
v 1
[
p0 (1−ξ 2 )
p 0 (1−ξ 1 )
( )
v
ln 2
v 1
= [
ln 1−ξ2
ln
= [
ln
= 2.0023
]
]
1−ξ 1
( )
7.67✘✘✘
Pa s
9.71✘✘✘
−1
Pa s −1
1−0.200
1−0.100
]
2

CHEM 3411, Fall 2010
Solution Set 8
This tells us that within the uncertainty of the measurements n = 2.00, which is important as n should generally be
an integer. With n = 2 the reaction kinetics are second order.
Solution
Second order (n = 2.00)
3

CHEM 3411, Fall 2010
Solution Set 8
2 Exercise 21.10b pg 826
Given
A second-order reaction of the type A + 2B → P was carried out in a solution that was initially 0.050 mol dm −3 in
A and 0.030 mol dm −3 in B. After t = 1.0 h the concentration of B had fallen to 0.010 mol dm −3
(In out textbook there is an error where the last sentence had A written in place of B.)
In terms of given variables, this is written:
[A] 0 = 0.050 mol dm −3
[B] 0 = 0.030 mol dm −3
[B] t=1.0 h = 0.010 mol dm −3
Find
• (a) Calculate the rate constant k r .
• (b) What is the half-life of the reactants?
Strategy
For part (a) we can use the integrated rate law equation for a second order reaction of the form A + 2B → P as given
in Table 21.3 (pg 295) of our textbook.
k r × t =
1
[B] 0 − 2[A] 0
ln [A] 0 ([B] 0 − x)
([A] 0 − x) [B] 0
where x is the extent of reaction (in concentration units) and thereby also the concentration of product P at time t.
We know that at t = 1.0 h (t = 3600 s) the concentration of B is [B] = 0.010 mol dm −3 , which implies 0.020 mol dm −3
of B have reacted at this point in time. The chemical equation for this reactions tells us that for each 2 mol of B
that react, 1 mol of product will be formed. Hence, x = [P] = 0.010 mol dm −3 at this point in time (t = 3600 s).
Using this value of x at time t = 3600 s we can solve for the rate constant k r .
k r = 1 t
1
ln [A] 0 ([B] 0 − 2x)
[B] 0 − 2[A] 0 ([A] 0 − x) [B] 0
= 1
3600 s × 1
0.030 mol dm −3 − 2 × 0.050 mol dm −3 × ln 0.050 mol ( dm−3 0.030 mol dm −3 − 2 × 0.010 mol dm −3)
(
0.050 mol dm −3 − 0.010 mol dm −3) 0.030 mol dm −3
= 0.003474 dm 3 mol −1 s−1
In determining the half-lives of each reactant we will again use the integrated rate law equation from our textbook.
For each reagent, we’ll determine the value of x (concentration of product [P]) at the point where half of the reactant
is used up. We can then solve for the time t when the reaction reaches this value of x.
For reactant A, the half-life concentration is [A] = 1 2 [A] 0 = 0.025 mol dm −3 and using the chemical equation for the
reaction we find x = 0.025 mol dm −3 when [A] reaches this half-life concentration.
4

CHEM 3411, Fall 2010
Solution Set 8
Solving for the time t when this value of x is reached
t 1/2 A = 1 1
ln [A] 0 ([B] 0 − 2x)
k r [B] 0 − 2[A] 0 ([A] 0 − x) [B] 0
1
=
0.003474 s dm 3 mol −1 × 1
0.030 mol dm −3 − 2 × 0.050 mol dm −3 ×
ln 0.050 mol dm−3 ( 0.030 mol dm −3 − 2 × 0.025 mol dm −3)
(
0.050 mol dm −3 − 0.025 mol dm −3) 0.030 mol dm −3
=D.N.E.
Here we find an answer does not exist (D.N.E.) due to a negative value in the natural logarithm. On further inspection
we find this lack of solution results from insufficient initial concentration of B for this amount of product P to be
formed. Therefore the reaction never proceeds far enough for half of the initial concentration of A to be consumed.
This explains why t 1/2 A does not exist.
Repeating this process for B we find x = 0.0075 mol dm −3 when the concentration of B reaches its half-life value
[B] = 1 2 [B] 0 = 0.015 mol dm −3 .
t 1/2 B = 1 1
ln [A] 0 ([B] 0 − 2x)
k r [B] 0 − 2[A] 0 ([A] 0 − x) [B] 0
1
=
0.003474 s dm 3 mol −1 × 1
0.030 mol dm −3 − 2 × 0.050 mol dm −3
× ln 0.050 mol dm−3 ( 0.030 mol dm −3 − 2 × 0.0075 mol dm −3)
(
0.050 mol dm −3 − 0.0075 mol dm −3) 0.030 mol dm −3
=2181 s
Solution
k r = 3.5 × 10 −3 dm 3 mol −1 s −1
t 1/2 A = D.N.E.
t 1/2 B = 2.2 × 10 3 s = 0.61 h
5

CHEM 3411, Fall 2010
Solution Set 8
3 Exercise 21.13b pg 827
Given
The equilibrium A ⇋ B+C at T = 25 ◦ C is subjected to a temperature jump that slightly increases the concentrations
of B and C. The measured relaxation time is τ = 3.0 µs. The equilibrium constant for the system is K = 2.0×10 −16 at
T = 25 ◦ C, and the equilibrium concentration of B and C at T = 25 ◦ C are both [B] eq
= [C] eq
= 2.0 × 10 −4 mol dm −3 .
In terms of given equations and variables, this is written:
A ⇋ B + C
T = 25 ◦ C
τ = 3.0 µs = 3.0 × 10 −6 s
K = 2.0 × 10 −16 mol dm −3
[B] eq
= [C] eq
= 2.0 × 10 −4 mol dm −3
Find
Calculate the rate constants for the forward and reverse steps, k r and k ′ r respectively.
Strategy
In solving this problem we’ll follow a strategy similar to that of Example 21.4 (pg 798) of our textbook. We’ll
begin by constructing the following equation for the rate of change of reactant A in terms of reactant and products
concentrations.
d [A]
dt
= −k r [A] + k ′ r [B] [C]
Next, we’ll relates the instantaneous concentration of reactant and products to their equilibrium concentrations and
the instantaneous deviation from equilibrium x.
[A] = [A] eq
+ x
[B] = [B] eq
− x
[C] = [C] eq
− x
Substituting these expressions into our reaction rate equation gives the following.
6

CHEM 3411, Fall 2010
Solution Set 8
d [A]
= −k r [A] + k r ′ [B] [C]
(
dt
)
d [A] eq
+ x
) (
= −k r
([A]
dt
eq
+ x + k r
′
dx
(
)
dt = −k r [A] eq
+ k r ′ [B] eq
[C] eq
+ x
) ( )
[B] eq
− x [C] eq
− x
[
−k r − k r
′
([B] eq
+ [C] eq
)]
− k ′ rx 2
The first right-hand term in parentheses can be set to zero by the condition that forward and backwards rates are
equal at equilibrium.
k r [A] eq
= k ′ r [B] eq
[C] eq
−k r [A] eq
+ k ′ r [B] eq
[C] eq
= 0
Additionally, the k ′ rx 2 term can be neglected by assuming deviations from equilibrium are small enough to make x 2
negligible relative to x. Therefore our expression for dx/dt simplifies to.
dx
[
)]
dt = x −k r − k r
([B] ′ eq
+ [C] eq
This is simply the rate-equation for a first-order reaction and therefore the solution of this differential equation takes
the form
where the relaxation time-constant τ is given by
x(t) = x 0 e −t/τ
1
)
τ = k r + k r
([B] ′ eq
+ [C] eq
As we know τ, [B] eq
[C] eq
we can solve for k r and k ′ r if we have one more equation with only these two terms as
unknowns. Such an equation is found from the relationship of equilibrium constant K and forward reverse rates, k r
and k ′ r respectively (Equation 21.27 pg 797)
K = k r
k ′ r
Solving for k ′ r = k r
K and substituting into our time-constant equation allows us to solve for k r.
1
τ = k r + k )
r
([B]
K eq
+ [C] eq
k r =
τ
[
1 + 1 K
1
)]
([B] eq
+ [C] eq
1
=
[
(
3.0 × 10 −6 1
s 1 +
2.0×10 −16 mol dm 2.0 × 10
−4
mol dm −3 + 2.0 × 10 −4 mol dm −3)]
−3
= 1.667 × 10 −7 s −1 7

CHEM 3411, Fall 2010
Solution Set 8
Lastly, we’ll solve for k ′ r.
k ′ r = k r
K
= 1.667 × 10−7 s −1
2.0 × 10 −16 mol dm −3
= 8.335 × 10 8 dm 3 mol −1 s −1
Solution
k r = 1.7 × 10 −7 s −1
k ′ r = 8.3 × 10 8 dm 3 mol −1 s −1 8

CHEM 3411, Fall 2010
Solution Set 8
4 Exercise 21.18b pg 827
Given
The mechanism of a composite reaction consists of a fast pre-equilibrium step with forward and reverse activation
energies of E a (a) = 27 kJ mol −1 and E ′ a(a) = 35 kJ mol −1 , respectively, followed by an elementary step of activation
energy E a (b) = 15 kJ mol −1 .
In terms of given variables, this is written:
E a (a) = 27 kJ mol −1
E ′ a(a) = 35 kJ mol −1
E a (b) = 15 kJ mol −1
Find
What is the activation energy of the composite reaction?
Strategy
Our textbook has discussed the kinetics of a such a reaction in Section 21.8 (pgs 809-811). The following expression
was developed for the activation energy of the composite reaction.
E a = E a (a) + E a (b) − E ′ a(a)
= 27 kJ mol −1 + 15 kJ mol −1 − 35 kJ mol −1
= 7 kJ mol −1
Solution
E a = 7 kJ mol −1 9

CHEM 3411, Fall 2010
Solution Set 8
5 Exercise 22.4b pg 872
Given
Consider the second-order reaction D 2 (g) + Br 2 (g) → 2DBr(g) at T = 450 K which is assumed to be elementary
bimolecular. Further, take the collision cross-section as 0.30 nm 2 , the reduced mass as 3.930m u , and the activation
energy as 200 kJ mol −1 .
In terms of given equations and variables, this is written:
D 2 (g) + Br 2 (g) → 2DBr(g)
T = 450 K
σ = 0.30 nm 2 = 0.30 × 10 −18 m 2
µ = 3.930m u = 6.52592 × 10 −27 kg
E a = 200 kJ mol −1
Find
Use the collision theory of gas-phase-reactions to calculate the theoretical value of the second-order rate constant for
this reaction.
Strategy
We know the rate of a gas-phase bimolecular elementary (excluding steric requirements) is described by the following
equation (Equation 22.12 pg 835)
k r = N A σ¯c rel e −E a/RT
Additionally, we know the average relative speed ¯c rel can be calculated from the Maxwell distribution. In Justification
22.3 of our textbook this was calculated and this gives us a rate expression into which we can insert the given
parameters of this reaction.
( ) 1/2 8kT
k r = N A σ
e −Ea/RT
πµ
= 6.02214 × 10 23 mol −1 × 0.30 × 10 −18 m 2 ×
( 8 × 1.38065 × 10 −23 J K −1 ) 1/2
× 450 K
3.1415 × 6.52592 × 10 −27 ×
kg
exp [ −200 × 10 3 J mol −1 / ( 8.31447 J K −1 mol −1 × 450 K )]
= 1.715 × 10 −15 m 3 mol −1 s −1
= 1.715 × 10 −12 dm 3 mol −1 s −1 10

CHEM 3411, Fall 2010
Solution Set 8
Solution
k r = 1.7 × 10 −12 dm 3 mol −1 s −1 11

CHEM 3411, Fall 2010
Solution Set 8
6 Exercise 22.6b pg 872
Given
Suppose that the typical diffusion coefficient for a reactant in aqueous solution at T = 25 ◦ C is D = 4.2×10 −9 m 2 s −1 .
Further assume the critical reaction distance is R ∗ = 0.50 nm for the second-order diffusion controlled reaction with
this reactant.
In terms of given variables, this is written:
T = 25 ◦ C
D = 4.2 × 10 −9 m 2 s −1
R ∗ = 0.50 nm = 5.0 × 10 −10
Find
The rate constant for this reaction.
Strategy
The rate constant of a diffusion-controlled reaction is given by Equation 22.18 (pg 840) from our textbook.
k d = 4πR ∗ DN A
= 4 × 3.1415 × 5.0 × 10 −10 m × 4.2 × 10 −9 m 2 s −1 × 6.02214 × 10 23 mol −1
= 1.58915 × 10 7 m 3 mol −1 s −1
= 1.58915 × 10 10 dm 3 mol −1 s −1
Solution
k d = 1.6 × 10 10 dm 3 mol −1 s −1 12