Chem 162 2012 final exam solutions 7 6 chem 162 2012

Chem 162-2012 final exam + solutions 7
6 Chem 162-2012 Final Exam + Answers
Chapter 14 – Chemical Kinetics
Reaction Mechanisms, Activation Energy & Catalysts Calculations
The rate of a certain reaction triples when the temperature is raised from 25 °C to 37 °C.
Calculate the activation energy of this reaction.
A
. 70.3
mol
B. 52.5
kJ
mol
C. 31.2
kJ
mol
D. 86.5
kJ
mol
kJ

Chem 162-2012 final exam + solutions 8
7 Chem 162-2012 Final Exam + Answers
Chapter 14 – Chemical Kinetics
Rates, Rate Constants, Reaction Orders, Half-Lives Calculations
2HgCl
2
(aq) + C
2
O
4
2-
→ 2Cl
-
+ 2CO
2
(g) + Hg
2
Cl
2
(s)
Experiment Initial
[HgCl
2
] M
Initial
[C
2
O
4
2-
] M
Rate
M/min
1 0.0836 M 0.202 0.52x10
-4
2 0.0836 M 0.404 2.08x10
-4
3 0.0418 0.404 1.04x10
-4
4 0.0627 0.606 ?
Find the initial rate for experiment 4.
A. 8.75x10
-3
M/min
B. 1.20x10
-4
M/min
C. 6.00x10
-4
M/min
D
. 3.51x10
-4
M/min
E. 1.25x10
-4
M/min
Use the “Initial Rate Method”.
Rate = k[HgCl
2
]
m
[C
2
O
4
2-
]
n
Exp. 2: 2.08x10
-4
= k[0.0836]
m
[0.404]
n
Exp. 1: 0.52x10
-4
= k[0.0836]
m
[0.202]
n
4 = 2
n
n = 2
Exp. 2: 2.08x10
-4
= k[0.0836]
m
[0.404]
n
Exp. 3: 1.04x10
-4
= k[0.0418]
m
[0.404]
n
2 = 2
m
m = 1
Rate = k[HgCl
2
]
1
[C
2
O
4
2-
]
2
Find k using data from experiment 1, 2 or 3.
Experiment 1:
0.52x10
-4
= k[0.0836][0.202]
2
(0.52x10
-4
) = (k x [0.0836] x ([0.202]
2
))
k = 0.01524
Rate = 0.01524 x [HgCl
2
]1[C
2
O
4
2-
]
2
Now find the rate of experiment 4:
Experiment 4
Rate = 0.01524 x [0.0627]
1
[0.606]
2
Rate = 3.51 x 10
-4

Chem 162-2012 final exam + solutions 9
8 Chem 162-2012 Final Exam + Answers
Chapter 14 – Chemical Kinetics
Reaction Mechanisms, Activation Energy & Catalysts Concepts
Which mechanism for the overall reaction A → C is consistent with the reaction profile
below?
A → B is a rapid, reversible step (which can be called a “fast” step), and B → C is a
relatively slow step.
A
B
B
C