The Foundations of Chemistry

(Marcin) #1
Initial Rate of
Initial Initial Formation of C
Experiment [A] [B] (Ms^1 )

1 1.0 10 ^2 M 1.0 10 ^2 M 1.5 10 ^6
2 1.0 10 ^2 M 2.0 10 ^2 M 3.0 10 ^6
3 2.0 10 ^2 M 1.0 10 ^2 M 6.0 10 ^6

Because we are describing the same reaction in each experiment, all the experiments
are governed by the same rate-law expression,

ratek[A]x[B]y

The initial concentration of A is the same in experiments 1 and 2, so any change in
the initial rates for these experiments would be due to different initial concentrations of
B. To evaluate y,we solve the ratio of the rate-law expressions of these two experiments
for y.We can divide the first rate-law expression by the corresponding terms in the second
rate-law expression.




r
r

a
a

t
t

e
e

1
2




k
k

[
[

A
A

]
]

1
2

x
x

[
[

B
B

]
]

1
2

y
y

The value of kalways cancels from such a ratio because it is constant at a particular temper-
ature. The initial concentrations of A are equal, so they too cancel. Thus, the expression
simplifies to




r
r

a
a

t
t

e
e

1
2




[
[

B
B

]
]

1
2




y

The only unknown in this equation is y.We substitute data from experiments 1 and 2 into
the equation, which gives us




y

0.5(0.5)y so y 1

Thus far, we know that the rate-law expression is

ratek[A]x[B]^1

Next we evaluate x.In experiments 1 and 3, the initial concentration of B is the same, so
any change in the initial rates for these experiments would be due to the different initial
concentrations of A. We solve the ratio of the rate-law expressions of these two experi-
ments for x.We divide the third rate-law expression by the corresponding terms in the
first rate-law expression.




r
r

a
a

t
t

e
e

3
1




k
k

[
[

A
A

]
]

3
1

x
x

[
[

B
B

]
]

3
1

1
 1

The value kcancels, and so do the concentrations of B because they are equal. Thus, the
expression simplifies to




r
r

a
a

t
t

e
e

3
1




[
[

A
A

]
]

3
1

x
x

[
[

A
A

]
]

3
1




x

1.0 10 ^2 M

2.0 10 ^2 M

1.5 10 ^6 Ms^1

3.0 10 ^6 Ms^1

It does not matter which way we take
the ratio. We would get the same value
for yif we divide the second rate-law
expression by the first—try it!


660 CHAPTER 16: Chemical Kinetics

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