Chapter 5
Chemical Equilibrium
51
1.
What are the units of K
p
and K
c
for each of the following:
a) 2H
2
S(g)
U
2H
2
(g) + S
2
(g)
×
==
=
22
2
2
2
HS
p
2
P
P
atm
atm
K
a
tm
Pa
t
m
×
=
c
2
[H ] [S ]
M
M
M
[H S]
M
b) 4NH
3
(g) + 3O
2
(g)
U
2N
2
(g) + 6H
2
O(g)
3
×
=
×
32
26
NH
O
p
43
4
NH
O
PP
atm
atm
a
P
P
atm atm
×
=
×
2
6
c
4
3
[N ] [H O]
M
M
M
[NH ] [O ]
M M
3. Determine K
c
values for the following at 298 K:
a) N
2
O
4
(g)
U
2NO
2
(g)
K
p
= 0.15 atm
Δ
×
n
g
cp
K=K (RT
) &
Δ
n
g
= 2  1 = 1, so
⋅
××
⋅
c
Kmo
l
1
K =0.15 atm
= 0.0061 M
0.0821 L atm
298 K
b) NO
2
(g) + NO(g)
U
N
2
O
3
(g)
K
p
= 0.86 atm
1
Δ
n
g
= 1, so
⋅
⋅
1
1
c
0.0821 L atm
K =0.86 atm
298 K = 21 M
mol K
5.
Determine the value of K
p
for each of the following:
Use Eq. 5.2, K
c
= K
p
(RT)
Δ
n
g
a)
CO(g) + H
2
O(g)
U
CO
2
(g) + H
2
(g)
K
c
= 23.2 at 600 K
Δ
n
g
= 0 for the reaction, so K
p
= K
c
= 23.2
b) 2 H
2
S(g)
U
2 H
2
(g) + S
2
(g)
K
c
= 2.3 x 10
4
M at 1405 K
Δ
n
g
= 1, so
(
)
⋅
⋅
×
1
4
2
mol
L atm
p
Lm
o
l
K
K =(2.3 10
) 0.0821
1450 K =2.7 10 atm
7.
a)
Determine K for: HI(g)
U
1
/
2
H
2
(g) +
1
/
2
I
2
(s) given H
2
(g) + I
2
(s)
U
2 HI(g)
K= 8.6
We are asked to find the value of K for a reaction given the value for the reaction that is
1
/
2
of the reverse of the original
reaction.
Consequently,
K
2
= (K
1
)
0.5
= 8.6
0.5
= 0.34
b) Determine K
c
for:
2 SO
2
(g) + O
2
(g)
U
2 SO
3
(g) given SO
2
(g) +
1
/
2
O
2
(g)
U
SO
3
(g)
K
c
= 3.61 M
1/2
This reaction is simply twice the original reaction:
()
===
2
2
21
K
K
3.61
13.0 M
9.
Given the following:
2 NO(g)
U
N
2
(g) + O
2
(g)
K
1
= 2.4 x 10
30
NO(g) +
1
/
2
Br
2
(g)
U
NOBr(g)
K
2
= 1.4
Determine K for
1
/
2
N
2
(g) +
1
/
2
O
2
(g) +
1
/
2
Br
2
(g)
U
NOBr(g)
1
/
2
N
2
(g) +
1
/
2
O
2
(g)
U
NO(g)
K
1
1/2
= (2.4 x 10
30
)
1/2
= 6.5 x 10
16
NO(g) +
1
/
2
Br
2
(g)
U
NOBr(g)
K
2
= 1.4
NO(g) cancels, giving net equation.
By rule 3, K = (K
1
1/2
) x K
2
= 9.0 x 10
16
11. Lead fluoride dissolves in strong acid by the following reaction:
PbF
2
(s) + 2H
3
O
1+
(aq)
U
Pb
2+
(aq) + 2HF(aq) +2H
2
O(l)
a)
What is the equilibrium constant expression for the reaction?
+
=
1+ 2
23
[Pb ][HF]
K
[PbF ][H O ]
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52
b)
Use the following to determine the value of the equilibrium constant of the above reaction:
PbF
2
(s)
U
Pb
2+
(aq) + 2F
1
(aq)
K
1
= 3.7
x
10
8
HF(aq) + H
2
O(l)
U
H
3
O
1+
+ F
1
(aq)
K
2
= 7.2
x
10
4
Reverse the second reaction and multiply it by two
2H
3
O
1+
(aq)
+
2F
1
(aq)
U
2HF(aq)
+
2H
2
O(l)
K = K
2
2
Addition of the above equation and the first equation yields the reaction of interest.
Therefore, the equilibrium constant is K =
×
=
×
8
2
1
2
4
2
2
K3
.
7
1
0
=
7.1 10
K(
7
.
2
1
0
)
13.
Equal numbers of moles of Cl
2
and NO are placed in a vessel at some temperature where they reach the following
equilibrium:
2 NO(g) + Cl
2
(g)
U
2 ClNO(g).
Indicate whether each of the following statements about the resulting
equilibrium mixture is true, false, or depends upon the value of the equilibrium constant.
2 NO(g)
+
Cl
2
(g)
U
2 ClNO(g)
Initial
a
a
0
Δ
 2x
 x
+ 2x
Equilibrium
a  2x
a  x
2x
a) [NO] > [ClNO]
(a  2x) > (2x) is true only if x is small, so it depends upon K
b) [Cl
2
] < [NO]
(ax) < (a2x) cannot be true.
It is always false.
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 Spring '10
 Wilson
 Chemistry, Equilibrium, pH, Kc

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