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bleiweiss (emb853) – H03: Equilibrium – Mccord – (90540)
1
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beFore answering.
001
10.0 points
An equilibrium in which processes occur con
tinuously, with NO NET change, is called
1.
homogeneous equilibrium.
2.
static equilibrium.
3.
heterogeneous equilibrium.
4.
dynamic equilibrium.
correct
Explanation:
±or a system at dynamic equilibrium, al
though the concentrations oF the components
do not change, the processes continue to oc
cur in the Foward and reverse directions at the
same rate.
002
10.0 points
The expression For
K
c
For the reaction at equi
librium is
4 NH
3
(g) + 5 O
2
(g)
±
²
4 NO(g) + 6 H
2
O(g)
1.
[NO]
4
[H
2
O]
6
[NH
3
]
4
[O
2
]
5
correct
2.
[NH
3
]
4
[O
2
]
5
3.
[NO]
4
[H
2
O]
6
4.
[NH
3
]
4
[O
2
]
5
[NO]
4
[H
2
O]
6
Explanation:
The equation must be written with the ap
propriate Formula and correctly balanced.
K
c
is the equilibrium constant For species in so
lution and equals the mathematical product
oF the concentrations oF the chemical prod
ucts, divided by the mathematical product oF
the concentrations oF the chemical reactants.
In this mathematical expression, each concen
tration is raised to the power oF its coe²cient
in the balanced equation. ±or
K
c
the molar
concentrations are used For the activities oF
the components.
003
10.0 points
At 600
◦
C, the equilibrium constant For the
reaction
2 HgO(s)
→
2 Hg(
³
) + O
2
(g)
is 2.8. Calculate the equilibrium constant For
the reaction
1
2
O
2
(g) + Hg(
³
)
→
HgO(s)
.
1.
0.36
2.
0.60
correct
3.
1.7
4.

1
.
7
5.
1.1
Explanation:
004
10.0 points
Calculate the equilibrium constant at 25
◦
C For
a reaction For which Δ
G
0
=

5
.
64 kcal
/
mol.
1.

13697
.
9
2.
13697
.
9
correct
3.
6848
.
96
4.
27395
.
9
5.
1
.
36979
×
10
5
Explanation:
T
= 25
◦
C + 273 = 298 K
Δ
G
0
=

5640 cal
/
mol
At equilibrium
Δ
G
0
=

RT
ln
K

5640 = (

1
.
987 cal
/
mol
·
K)
×
(298 K)(ln
K
)
K
= 13697
.
9
005
10.0 points
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View Full Documentbleiweiss (emb853) – H03: Equilibrium – Mccord – (90540)
2
Given the reversible reaction equation
2 CO(g) + O
2
(g)
±
²
2 CO
2
(g)
which is the relationship between
K
c
and
K
p
?
1.
K
p
=
K
c
(
RT
)

1
correct
2.
K
p
=
K
c
3.
K
p
=
K
c
4.
K
p
=
K
c
(
)
2
5.
K
p
=
K
c
(
)

2
Explanation:
The ideal gas law can be used to derive the
link between
K
c
and
K
p
:
P
=
n
V
=
M RT
M
=
P
K
c
=
[CO
2
]
2
[CO]
2
[O
2
]
=
P
2
CO
2
(
)
2
P
2
CO
(
)
2
·
P
O
2
=
P
2
CO
2
P
2
CO
·
P
O
2
=
K
p
K
p
=
K
c
(
)

1
006
10.0 points
The reaction
N
2
(g) + 3 H
2
(g)
±
²
2 NH
3
(g)
,
has an equilibrium constant of 4
.
0
×
10
8
at
25
◦
C. What will eventually happen if 44.0
moles of NH
3
, 0.452 moles of N
2
, and 0.108
moles of H
2
are put in a 10.0 liter container at
25
◦
C?
1.
Nothing; the system is at equilibrium.
2.
More N
2
and H
2
will be formed.
3.
More NH
3
will be formed.
correct
Explanation:
K
=4
.
0
×
10
8
[NH
3
]=
44
.
0 mol
10 L
[N
2
0
.
452 mol
10 L
[H
2
0
.
108 mol
10 L
Q
=
[NH
3
]
2
[N
2
] [H
2
]
3
=
(4
.
40 M)
2
(0
.
0452 M) (0
.
0108 M)
3
=3
.
4
×
10
8
Since
Q < K
equilibrium will shift to the
right, forming more NH
3
.
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 Summer '07
 Holcombe
 Equilibrium

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