bracketrightbigg
= 6
.
88395
006
1.0 points
A 0
.
835 mole sample of HI is placed in a
5.000 liter container and held at a constant
temperature until the gasphase reaction
2 HI
⇀
↽
H
2
+ I
2
comes to equilibrium.
At equilibrium, it is
found that the concentration of I
2
is 2
.
1
×
10
−
2
mol/L. What is the numerical value of the
equilibrium constant
K
c
at this temperature?
Correct answer: 0
.
028224.
Explanation:
[HI]
ini
=
0
.
835 mol
5 L
= 0
.
167 M
2 HI (g)
⇀
↽
H
2
(g) + I
2
(g)
ini, M
0
.
167
0
0
Δ, M

2
x
x
x
eq, M
0
.
167

2
x
x
x
[I
2
]
eq
=
x
= 2
.
1
×
10
−
2
M
[H
2
]
eq
=
x
= 2
.
1
×
10
−
2
M
[HI]
eq
= (0
.
167

2
x
) = 0
.
125 M
To (st22362) – Homework 3 – Fakhreddine – (52420)
3
K
c
=
[H
2
] [I
2
]
[HI]
2
=
(2
.
1
×
10
−
2
M)
2
(0
.
125 M)
2
= 0
.
028224
007
1.0 points
Consider the equilibrium
A(g)
⇀
↽
2 B(g) + 3 C(g)
at 25
◦
C. When A is loaded into a cylinder at
9
.
63 atm and the system is allowed to come to
equilibrium, the final pressure is found to be
16
.
17 atm. What is Δ
G
◦
r
for this reaction?
Correct answer:

12
.
5403 kJ
/
mol.
Explanation:
Considering the pressures (in atm),
A
⇀
↽
2 B
+
3 C
ini
9
.
63
0
0
Δ

x
+2
x
+3
x
fin
9
.
63

x
+2
x
+3
x
The total pressure is
P
total
= 9
.
63

x
+ 2
x
+ 3
x
16
.
17 atm = 9
.
63 + 4
x
x
= 1
.
635 atm
,
so
P
A
= 9
.
63

x
= 7
.
995 atm
P
B
= 2
x
= 3
.
27 atm
P
C
= 3
x
= 4
.
905 atm
The equilibrium expression is
K
=
P
2
B
·
P
3
C
P
A
=
(3
.
27)
2
(4
.
905)
3
7
.
995
= 157
.
832
and the energy is
Δ
G
◦
=

RT
ln
K
=

parenleftbigg
8
.
314
J
K
·
mol
parenrightbigg
(298 K)
×
ln (157
.
832)
=

12540
.
3 J
/
mol
=

12
.
5403 kJ
/
mol
.
008
1.0 points
At 25
◦
C,
K
c
= 1
.
58
×
10
−
8
for the reaction
NH
4
(NH
2
CO
2
)(s)
→
2 NH
3
(g) + CO
2
(g)
.
Calculate
K
at 25
◦
C for this reaction.
1.
3
.
87
×
10
−
7
2.
5
.
69
×
10
−
3
3.
9
.
45
×
10
−
5
4.
2
.
31
×
10
−
4
correct
5.
1
.
36
×
10
−
7
Explanation:
009
1.0 points
The equilibrium constant
K
p
for the reaction
I
2
(g) + Br
2
(g)
⇀
↽
2 IBr(g) + 11
.
7 kJ
is 280 at 150
◦
C. Suppose that a quantity of
IBr is placed in a closed reaction vessel and
the system is allowed to come to equilibrium
at 150
◦
C. When equilibrium is established,
the pressure of IBr is 0.200 atm. What is the
pressure of I
2
at equilibrium?
1.
0.168 atm
2.
0.067 atm
3.
0.012 atm
correct
4.
None of these
5.
0.096 atm
Explanation:
At equilibrium,
P
IBr
= 0
.
200 atm
K
p
= 280
I
2
(g) + Br
2
(g)
⇀
↽
2 IBr(g) + 11.7 kJ
0
0
y
x
x

2
x
x
x
y

2
x
To (st22362) – Homework 3 – Fakhreddine – (52420)
4
y

2
x
= 0
.
2
K
p
=
P
2
IBr
P
I
2
·
P
Br
2
= 280
0
.
2
2
x
2
= 280
x
=
0
.
2
√
280
= 0
.
0119523
P
I
2
= 0
.
0119523 atm
010
1.0 points
A 1.000 L vessel is filled with 2.000 moles of
N
2
, 1.000 mole of H
2
, and 2.000 moles of NH
3
.
When the reaction
N
2
(g) + 3 H
2
(g)
⇀
↽
2 NH
3
(g)
comes to equilibrium, it is observed that the
concentration of H
2
is 2
.
25 moles/L. What is
the numerical value of the equilibrium con
stant
K
c
?
Correct answer: 0
.
0494458.
Explanation:
[N
2
]
ini
=
2 mol
1 L
= 2 M
[NH
3
]
ini
=
2 mol
1 L
= 2 M
[H
2
]
ini
=
1 mol
1 L
= 1 M
Therefore equilibrium moves to the left.
N
2
(g) + 3 H
2
(g)
⇀
↽
2 NH
3
(g)
ini, M
2
1
2
Δ, M
+
x
+3
x

2
x
eq, M
2+
x
1+3
x
2

2
x
[H
2
]
eq
= 2
.
25 M , so
[H
2
]
eq
= 1
.
0 + 3
x
= 2
.
25 M
x
=
2
.
25 M

1 M
3
= 0
.
416667 M
[N
2
]
eq
= (2 +
x
) M = 2
.
41667 M
[NH
3
]
eq
= (2

2
x
) M = 1
.
16667 M
K
c
=
[NH
3
]
2
[N
2
] [H
2
]
3
=
(1
.
16667)
2
M
(2
.
41667 M) (2
.
25 M)
3
= 0
.
0494458
011
1.0 points
The vapor pressure of benzene at 298 K is
94.4 mm of Hg.
The standard molar Gibbs
free energy of formation of liquid benzene at
298 K is 124.5 kJ/mol.
What is the stan
You've reached the end of your free preview.
Want to read all 11 pages?
 Fall '10
 McCord
 Chemistry, Equilibrium, Reaction, Kc