Kinetics Review-
ANSWER KEY
Initial Rate Problem:
For the reaction A + B
→
products, the following initial rate data was
collected at 298 K:
Run
[A]
o
, M
[B]
o
, M
Initial Rate, M/s
1
0.500
0.100
6.0 x 10
-5
2
0.500
0.200
6.0 x 10
-5
3
0.250
0.100
1.5 X 10
-5
a. Find the rate equation, including k with the proper units.
E
a
for this reaction is 330J/mol.
First we should write our rate equations, in the form rate = k[A]
x
[B]
y
, for all three runs.
Run 1:
rate = 6.0 x 10
-5
M/s = k(0.500M)
x
(0.100M)
y
Run 2:
rate = 6.0 x 10
-5
M/s = k(0.500M)
x
(0.200M)
y
Run 3:
rate = 1.5 x 10
-5
M/s = k(0.250M)
x
(0.100M)
y
To solve for x, we divide the rate equations for Run 1 and Run 3:
5
rate1
rate3
5
k
6.0x10
M /s
4
1.5x10
−
−
==
=
xy
(0.500M) ( 0.100M)
k
(0.250M) (0.100M)
x
24
x2
,
⇒=
=
To solve for y, we do the same for Run 1 and Run 2
5
rate1
rate2
5
k
6.0x10
1
6.0x10
−
−
=
x
(0.500M)
y
(0.100M)
k
x
(0.500M)
y
y
y0
1
1,
2
(0.200M)
⎛⎞
=
⎜⎟
⎝⎠
=
⇒
(Remember, it doesn’t matter what order you solve for x and y, or for that matter which rate goes
in the numerator and denominator for your divisions)
Now that we have x and y, to solve for k we just use one of the rate equations.
For no particular
reason, I used Run 1:
6.0 X 10
-5
M/s = k(0.500M)
2
(0.100M)
0
Using a little algebra, k= 2.4 x 10
-4
M
-1
s
-1
.
Since I kept the units in the equation, the correct units
for k automatically come out in the answer.
Combining everything:
rate=(2.4 x 10
-4
M
-1
s
-1
)[A]
2
(no need to put [B]
0
since it always is 1)
b.
The experiment is repeated at 373 K.
What is the new value for Initial Rate 1?
Any time we are asked to relate rate to temperature, we use the Arrhenius equation.
Since the
only thing we change between part (a) and (b) is the temperature, only the rate constant changes,
and the ratio of the rate constants gives us the ratio of the rates.
a
2
12
1
E
k
11
ln
kR
T
T
−
=−
From the problem, T
2
=373K, and T
1
=298K (the temperature of the
original experiment).
E
a
=330 J, and for R we must use 8.314 J/Kmol since we are dealing with
energy in Joules.
Putting these numbers into the equation and undoing the logs (taking the exponent), we get: