Unit 5, Lesson 06: Equilibrium Problems #1 Worksheet With Answers Page 4
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1
2
3
4
5↔
↔
↔ ↔
-7
7)
For the reaction:
2 CO
2 CO
+
O
K
= 6.40 x 10
at 2000˚C
2 (g)
(g)
2 (g)
eq
If 2.00 moles of carbon dioxide are placed in a 0.500 L container and allowed to come to equilibrium:
a)
What will be the concentration of O
at equilibrium?
2 ( g)
2 CO
2 CO
O
2 (g)
(g)
2 (g)
I
C = n/V = 4.0 M
0
0
C
– 2x
+ 2x
+ x
E
4.0 – 2x
2x
x
2
Can we ignore the –x?
Keq
=
[CO]
[O
]
2
2
[CO
]
2
4.0
is much greater than,
–7
–7
2
6.40 x 10
so we can ignore the –x
6.40 x 10
=
[2x]
[x]
2
[4.0 ]
–4
3
1.20 x 10
= 4x
(remember to square both terms)
16
x
= 0.0137
Substitute x into the last row of the ICE table to find the equilibrium concentrations:
at eq’m,
[O
] = 0.0137 M (3 sig digs)
2
b)
How many moles of O
will be in the container at equilibrium? ( n = CV = 0.00685 mol of O
)
2 (g)
2
c)
Does this reaction favour the reactants or products at 2000˚C? Explain.
This reaction favours the reactants
because the value of K
is much less than 1
eq
6
d)
What is the K
for the reverse reaction at 2000˚C?
1/Keq forward = 1.56 x 10
eq
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