171
CHAPTER 17
EQUILIBRIUM: THE EXTENT
OF CHEMICAL REACTIONS
FOLLOWUP PROBLEMS
17.1
Plan:
First, balance the equations and then use the coefficients to write the reaction quotient as outlined in
Equation 17.4.
Solution:
a) Balanced equation:
4 NH
3
(
g
) + 5 O
2
(
g
)
'
4 NO(
g
) + 6 H
2
O(
g
)
Reaction quotient:
Q
c
=
[
] [
]
[
]
4
6
2
4
5
3
2
NO
H O
NH
O
b) Balanced equation:
3 NO(
g
)
'
N
2
O(
g
) + NO
2
(
g
)
Reaction quotient:
Q
c
=
[
][
]
[
]
2
2
3
N O
NO
NO
Check:
Are products in numerator and reactants in denominator? Are all reactants and products expressed as
concentrations by using brackets? Are all exponents equal to the coefficient of the reactant or product in the
balanced equation?
17.2
Plan:
Add the individual steps to find the overall equation. Write the reaction quotient for each step and the overall
equation. Multiply the reaction quotients for each step and cancel terms to obtain the overall reaction quotient.
Solution:
(1)
Br
2
(
g
)
'
2 Br(
g
)
(2)
Br(
g
) + H
2
(
g
)
'
HBr(
g
) + H(
g
)
(3)
H(
g
) + Br(
g
)
'
HBr(
g
)
Br
2
(
g
) + 2 Br
(
g
)
+ H
2
(
g
) + H
(
g
)
'
2 Br
(
g
)
+ 2 HBr(
g
) + H
(
g
)
Canceling the reactants leaves the overall equation as Br
2
(
g
) + H
2
(
g
)
'
2 HBr(
g
).
Write the reaction quotients for each step:
Q
c1
=
[
]
[
]
2
2
Br
Br
Q
c2
=
[
][
]
[
][
]
2
HBr
H
Br
H
Q
c3
=
[
]
[
][
]
HBr
H
Br
and for the overall equation:
Q
c
=
[
]
[
][
]
2
2
2
HBr
H
Br
Multiplying the individual
Q
c
’s and canceling terms gives
Q
c1
Q
c2
Q
c3
=
[
]
[
]
2
2
Br
Br
[
][
]
[
][
]
2
HBr
H
Br
H
[
]
[
][
]
HBr
H
Br
=
[
]
[
][
]
2
2
2
HBr
H
Br
The product of the multiplication of the three individual reaction quotients equals the overall reaction quotient.
17.3
Plan:
When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor.
Solution:
a) All coefficients have been multiplied by the factor 1/2 so the equilibrium constant should be raised to the 1/2
power (which is the square root).
For reaction a)
K
c
= (7.6 x 10
8
)
½
= 2.7568 x 10
4
=
2.8x10
4
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172
b) The reaction has been reversed so the new
K
c
is the inverse of the original equilibrium constant.
The reaction
has also been multiplied by a factor of 2/3, so the inverse of the original
K
c
must be raised to the 2/3 power.
For reaction b)
K
c
=
2
3
8
1
7.6 x 10
= 1.20076 x 10
6
=
1.2 x 10
6
Check:
The order of magnitude of each answer can be estimated by examining the exponents: for part a) (10
8
)
1/2
=
10
4
and for part b) rounding 7.6 x 10
8
to 10
9
gives 1/10
9
= 10
9
and (10
9
)
2/3
= 10
6
. These estimates are within the
same order of magnitude as the answers under solution.
17.4
Plan:
K
p
and
K
c
for a reaction are related through the ideal gas equation as shown in
K
P
=
K
c
(RT)
∆
n
. Find
∆
n
gas
,
the change in the number of moles of gas between reactants and products (calculated as products minus reactants).
Then, use the given
K
c
to solve for
K
p
.
Solution:
The total number of product moles of gas is 1 and the total number of reactant moles of gas is 2.
∆
n
= 1  2 = 1.
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 Spring '11
 Equilibrium, Reaction, Kc

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