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Prepared by Prof. Anju Sharma for students in General Chemistry 2
1
CH116  General Chemistry 2  Solutions to HW Problems  Chapter 13
Chapter 13: Assigned Problems: 24, 28, 30, 32, 38, 42, 54, 62, 64, 66, 70, 84, 86, 92, 94, 106, 108.
24. For the reaction: H
2
(g) + Br
2
(g)
2HBr(g), Kp = 3.5 x 10
4
at 1495 K. What is the value of Kp for the
following reactions at 1495 K?
For the original reaction:
H
2
(g) + Br
2
(g)
2HBr(g), Kp =
Br2
H2
2
HBr
P
P
P
= 3.5 x 10
4
at 1495 K.
(a) HBr(g)
2
1
H
2
(g) +
2
1
Br
2
(g)
Kp(
a)
=
1/2
Kp
1
=
4
10
x
5
.
3
1
= 5.3452248 x 10
–3
(b) 2HBr(g)
H
2
(g) + Br
2
(g)
Kp
(b)
=
Kp
1
=
4
10
x
5
.
3
1
= 2.857142857 x 10
–5
(c)
2
1
H
2
(g) +
2
1
Br
2
(g)
HBr(g)
Kp
(c)
=
Kp
=
4
10
x
3.5
= 187.0828693
In general, when the equation is multiplied by “n”; the value of the equilibrium constant for the new equation is
K
n
. Thus, for the reverse reaction (see b), the equilibrium constant is K
–1
or
K
1
.
28. At a particular temperature a 2.00 L flask at equilibrium contains 2.80 x 10
–4
mol N
2
, 2.50 x 10
–5
mol
O
2
, and 2.00 x 10
–2
mol N
2
O. Calculate K at this temperature for the reaction: 2N
2
(g) + O
2
(g)
2N
2
O(g)
The amounts are mentioned in moles and the volume of the flask is 2.00 L, the molar concentration of each
species must be calculated:
K =
]
[O
]
[N
O]
[N
2
2
2
2
2
=
L
2.00
O2
mol
5

10
x
2.5
2
L
2.00
N2
mol
4

10
x
2.80
2
L
2.00
N2O
mol
2

10
x
2.00
x
=
)
(0.0000125
x
(0.00014)
(0.01)
2
2
= 408163265.3
K = 4.08 x 10
8
30. The following equilibrium pressures were observed at a certain temperature for the reaction: N
2
(g) +
3H
2
(g)
2NH
3
(g). P
(NH3)
= 3.1 x 10
–2
atm; P
(N2)
= 8.5 x 10
–1
atm; P
(H2)
= 3.1 x 10
–3
atm. Calculate the value
for the equilibrium constant Kp at this temperature.
The values given are the equilibrium partial pressures of the gases in the system:
Kp =
3
H2
N2
2
NH3
)
(P
x
)
(P
)
(P
=
3
3

1

2
2
)
10
x
(3.1
x
)
10
x
(8.5
)
10
x
(3.1
= 37950.66414 = 3.8 x 10
4
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32. At 1100. K, Kp = 0.25 for the reaction: 2SO
2
(g) + O
2
(g)
2SO
3
(g). What is the value of K at this
temperature?
The relationship between Kp and K is: Kp = K x (RT)
n
; where
n = n(products) – n(reactants)
For this reaction,
n = 2 – (2+1) =
1 and thus Kp = K x (RT)
1
K =
n
RT
Kp
=
3)

(2
1100)
x
(0.08206
0.25
= 0.25 x 0.08206 x 1100 = 22.5665
38. In a study of the reaction: 3Fe(s) + 4H
2
O(g)
Fe
3
O
4
(s) + 4H
2
(g) at 1200 K it was observed that when
the equilibrium partial pressure of water vapor is 15.0 torr, that total pressure at equilibrium is 36.3 torr.
Calculate the value of Kp for this reaction at 1200 K.
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This note was uploaded on 02/25/2011 for the course CH 116 taught by Professor Sharma during the Spring '10 term at Stevens.
 Spring '10
 SHARMA
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