Spring, 2006
Solutions to Chemical Equilibrium Supplementary Sheet
1.
Consider the reaction at 700°K CO(g) + H
2
O(g)
CO
2
(g) + H
2
(g)
for which K
c
=
5.10.
Calculate the equilibrium concentration of all species present if 6.00 moles of
CO(g) and 6.00 moles of H
2
O(g) are mixed in a 2.00 liter flask at 700°K.
This is a typical equilibrium problem in which a dependent variable is assigned to represent one of the desired
concentrations and the solved for from the equilibrium expression.
[CO]
[H
2
O]
[CO
2
]
[
H2
]
6.00 moles
6.00 moles
initial
 = 3.00 M
 = 3.00 M
0
0
2.00 L
2 L
change
 X
 X
+ X
+ X
equil
3.00  X
3.00  X
X
X
Let X = the [H
2
] produced
[CO
2
][H
2
]
(X)(X)
x
2
Kc = =

=
 =
5.10
[CO][H
2
O]
(3.00  X)(3.00  X)
(3.00  X)
2
Taking the square root of both sides of the equation gives
X
 =
√5.10 = 2.26
(3.00  X)(2.26) =
X
(3.00)(2.26)  2.26X = X
3.00  x
6.78 = (2.26 + 1)X
6.78 = 3.26 X
X = 2.08
[CO]
[H
2
O]
[CO
2
]
[H
2
]
equil
3.00  2.08 =
0.92
3.00  2.08 =
0.92
2.08
2.08
Check K
c
= (2.08)(2.08)/(0.92)(0.92) = 5.11 (Note: If X is rounded off differently, the value of K
c
may vary slightly.)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Equilibrium Supplementary Sheet Solutions Page 2
2
.
Consider the reaction N
2
O
4
2NO
2
(g) at a temperature where K
p
= 0.131 atm.
A flask
initially contains N
2
O
4
(g) at 1.000 atm. pressure.
Calculate the pressures of NO
2
(g) and
N
2
O
4
(g) at equilibrium.
This problem is similar to the last problem except that since K
p
is given the partial pressures of the gases can be used rather than
the concentrations.
K
p
= P
NO
2
2
/P
N
2
O
4
.
There are two routes to the solution of this problem depending on how the dependent
variable is defined.
Route 1:
Let x = the amount of N
2
O
4
that reacts.
P
N
2
O
4
P
NO
2
initial
1.00
0
change
 x
+2x
NOTE: the 2 results from the reaction soichiometry
equil
1.00  x
+2x
K
p
= (2x)
2
/(1  x) = 0.131
Since this is not a perfect square, the quadratic equation must be used to solve for x.
4x
2
= 0.131(1x)
4x
2
= 0.131  0.131 X
4x
2
+ 0.131X  0.131 = 0
x = {0.131±√[(0.131)
2
 (4)(4)(0.131)]}/(2)(4)
X = {0.131 + 1.454}/8 = 0.165
(Note: the other route is negative and can be ignored)
P
N
2
O
4
= 1.00  0.165 =
0.835
P
NO
2
= 2(0.165) =
0.330
Check K
p
= (0.33)2/(0.835) = 0.130
Route 2:
Let y = the amount of NO
2
produced.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Martin
 Equilibrium, Mole, Kc, Equilibrium Supplementary Sheet

Click to edit the document details