CHEMISTRY 150
QUIZ KINETICS 2
MAY 04, 2012
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Integrated Rate Equations: (First Order) ln[A]t = -kt + ln[A]o; (Second Order) 1/[A]t = kt + 1/[A]o;
(Zeroth Order) [A]t = -kt + [A]o; K = Ae-Ea/RT; Lnk1/k2 = Ea/R(T1-T2/T1T2) R = 8.314 J/mole-K =
0.08206 L-atm/mole-K8.314 j
1. The rate law for the reaction: 2NO(g) + Cl2(g) → 2NOCl (g) is given by
R = k[NO][Cl2]
Two
a. What is the overall order of the reaction?
b. A mechanism involving the following steps has been proposed for the reaction:
NO (g) + Cl2 (g) → NOCl2 (g)
NOCl2 (g) + NO (g) → 2 NOCl (g)
How do the rates of the individual steps relate to each other. In other words which one is
fast and which one is slow?
The rate law is defined by the slowest step. Since the rate law contains the reactants of
step #1, then the first step must be the slowest step.
2. The rate of a particular reaction increases four times when the temperature is increased from
25ºC to 35ºC. Calculate the activation energy for this reaction.
Ea mole-K 103j
(298.15-308.15)K
1
ln
=
kJ
298.15(308.15)K2
8.314 j
4
Ea = 110 kJ/mole
3. For the reaction of iodine atoms with hydrogen molecules in the gas phase, these rate constants
were obtained experimentally. 2 I (g) + H2 (g) → 2 HI (g)
Frequency factor at 1/T=0 is lnA
Rate constant X10-5
from graphing y intercept=6.5, giving
T (K)
(L2 mole-2S-1)
A=666
417.9
1.12
480.7
2.60
k = A -Ea/RT
520.1
3.96
3
22.4 kJ 10 j mole-K
633.2
9.38
=6.735
mole
Kj
8.314j 400K
666.8
11.50
-6.735
710.3
16.10
k = 666e
= 0.79
737.9
18.54
a. Calculate the activation energy and frequency factor for this reaction.
1.12
Ea
103J
(417.9-480.7)K
ln
=
Ea = 22.4 kJ/mole
417.9(480.7)K2
8.314 J kJ
2.60
close
b. Estimate the rate constant of the reaction at 400.0 K
-5
22.4 kJ 103 j
ln 1.12 x10 = mole
K2
kJ
mole-K (417.9- 400.0)K
8.314 j 417.9(400.0)K2
ln 1.12x10-5 – ln K = 0.28848
K = e -11.1688518 = 8.4x10-6