CHEMISTRY 1AA3
PROBLEM SET 5 – WEEK OF FEBRUARY 11, 2002
1.
What are the overall orders of the reactions to which the following rate laws apply?
(a)
(b)
(c)
rate = k[Cl2][H2]
rate = k[NO2]2[O2]
rate = k[H2]½[C2H4]
SOLUTION:
The overall order of a reaction comes from the sum of the exponents in the rate law,
i.e., for
rate = k[A]x [B]y
(a)
(b)
(c)
1+1=
2+1=
0.5 + 1 =
2
3
1.5
2.
Consider the reaction:
overall order = x + y
N2(g) + 3 H2 (g) → 2 NH3 (g)
(a)
Write expressions for the rate of the reaction in terms of the changes of
concentrations of N2, H2, and NH3.
(b)
Suppose that at a particular moment the rate of reaction with respect to H2 is
0.074 M-1 s-1. What is the rate at which N2 is being consumed?
SOLUTION:
(a)
rate = − ∆[N2] = − 1 ∆[H2] =
∆t
3 ∆t
(b)
∆[H2] = 0.074 M−1 s−1
∆t
+1 ∆[NH3]
2 ∆t
− ∆[N2] = − 1 ∆[H2]
∆t
3 ∆t
∆[N2] =
∆t
1 (0.074 M−1 s−1)
3
= 0.025 M−1 s−1
1
3.
The rate law for the reaction :
NH4+ (aq) + NO2− (aq) → N2 (g) + 2 H2O (l)
is given by the rate law: rate = k [NH4+][ NO2−]. At 25 °C, the rate constant for the
reaction is 3.0 × 10-4 M−1 s−1. Calculate the rate of the reaction at 25 °C if [NH4+] =
0.26 M and [ NO2−] = 0.080 M.
SOLUTION:
rate = k [NH4+][ NO2−] and k = 3.0 × 10−4 M−1 s−1
∴
4.
rate
rate
=
=
(3.0 × 10−4 M−1 s−1)(0.26 M)(0.080 M)
6.24 × 10−6 M/s
The molecule cyclopropane undergoes a first order isomerization to propene at 300 °C
with k = 0.540 h-1.
Cyclopropane ' propene
(a)
How many hours will have elapsed when half of the cyclopropane has reacted if
the initial concentration of cyclopropane is 0.0500 M?
(b)
If the initial concentration of cyclopropane were quadrupled to 0.200 M, how
long would it take for half the cyclopropane to react?
(c)
How long would it take for the [cyclopropane] to change from an initial
concentration of 0.200 M to 0.0200 M? What is the [propene] at this time?
SOLUTION:
(a)
Using the relationship:
t1/2 = ln 2
k
and k = 0.540 h-1
t1/2 = 0.693/0.540 h-1 = 1.28 h
(b)
For any first order reaction it does not matter how much material is present; the
half-life (t1/2) is always the same (at the same temperature).
That is, t1/2 = 1.28 h.
(c)
We need to use the formula which relates A0, At and time:
ln (A0/At) = kt
Rearranging this equation and substituting we get:
t = ln(0.200/0.0200) = 4.26 h
0.540 h-1
Since the reaction is 1:1, the amount of propene formed is equal to the amount of
cyclopropane used up, and since both species are in the same volume, the molarities will
also be equivalent. Thus, the cyclopropane concentration goes from 0.200 M to 0.0200
M, and so decreases by 0.180 M. Therefore, [propene] = 0.180 M.
2
5.
The decomposition of SO2Cl2 is first order in SO2Cl2,
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
and it has a half-life of 4.1 h. If the initial concentration of SO2Cl2 is 1.25 × 10−3 M,
how long does it take for the concentration to drop to 0.31 × 10−3 M?
SOLUTION:
Since the reaction is first order, use the half-life equation to find k:
t½ = 4.1 h = 0.693/k
which gives k = 0.17 h−1
Then we can use the first-order integrated rate law to solve for concentration at
time t:
ln [SO2Cl2] = −kt
[SO2Cl2]0
ln [0.31 × 10-3] = −(0.17 h−1) t which gives
[1.25 × 10-3]
6.
t = 8.2 h
The combination of CF3 "free radicals", 2 CF3• → C2F6, occurs with rate constants of
5.9 × 109 M−1•s−1 at 25 °C and 7.1 × 109 M−1• s−1 at 60 °C.
(a)
Use the Arrhenius form for the rate constant as a function of temperature to
obtain the activation energy.
(b)
At a certain temperature the product can be condensed to a liquid, and the rate
constant at this temperature is 2.30 × 109 M−1s−1. Determine the temperature, in
degrees Celsius, at which this rate constant applies.
SOLUTION:
(a)
Using k = A e(-Ea/RT) in natural log form: ln k = -Ea + lnA and substituting
RT
for the two temperatures, 298 K and 333 K (recall the Arrhenius relationship
uses Kelvin not Celsius temperatures).
ln (5.9 × 109) = ln A -Ea . 1
R 298
--------(1)
ln (7.1 × 109) = ln A -Ea . 1
R 333
Then (1) - (2) gives
--------(2)
ln 5.9 × 109
7.1 × 109
= -Ea . ( 1 - 1 )
R 298 333
3
or
ln (0.8310)
= -Ea
× (333 - 298)
-1
8.314 J mol
K-1 298 × 333
or
- 0.1851
= -Ea ×
35
8.314 298 × 333
Isolate for Ea,
Ea
= 0.1851 × 8.314 × 298 × 333
35
= 4363 J mol-1 or 4.36 kJ mol-1
(b)
If the rate constant at the cooler temperature is 2.30 × 109 M−1s−1, then we may
use the data either from the 25°C or 60 oC experiment together with the
calculated Ea to obtain the value of the temperature.
ln (5.9 × 109) = ln A - 4.36 kJ . 1
R
298
9
and
ln (2.30 × 10 ) = ln A - 4.36 kJ . 1
R
T2
Take equation (2) -(1)
--------(1)
Thus
ln(5.9 x 109) - ln (2.30 × 109)
or
= - 4.36 × 103 J mol-1 . ( 1 - 1 ) . 1
8.314 J mol-1 K-1
298
T2 K
0.9420 = 4.36 × 103 (3.356 × 10−3 - 1 )
8.314
T2
5.152 × 10−3 =
7.
--------(2)
1 or
T2
T2 = 194 K, and
194 – 273 = −79 °C
Biological reactions nearly always occur in the presence of enzymes, which are very
powerful catalysts. For example, the enzyme catalase that acts on peroxides reduces the
activation energy of peroxide decomposition from 72 kJ/mol (in the uncatalyzed
reaction) to 28 kJ/mol (in the catalyzed reaction) at 298 K. What is the increase in k
when the reaction is catalyzed? (Assume the frequency factor A remains constant).
SOLUTION:
Recall we can have the Arrhenius equation in the form
ln k = ln A - Ea
RT
So if we write this equation for both sets of experimental conditions,
ln k1 = ln A − Ea1 (eq. 1)
and ln k2 = ln A − Ea2 (eq.2)
RT
RT
4
Subtract eq. 1 from eq. 2 to get
−
1 (Ea1 − Ea2)
RT
ln k2
k1
=
ln k2
k1
=
ln k2
k1
=
∴
The rate increases 5.2 × 107 (52 million!) times with the catalyst.
Ea1
RT
Ea2
RT
=
1
(72 − 28 kJ/mol)(1 × 103 J/kJ)
(8.314 J/K•mol)(298 K)
17.76
which gives
k2
k1
=
5.16 × 107
5