Unit 5, Lesson 06: Equilibrium Problems #1 Worksheet With Answers
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5Unit 5, Lesson 06: Equilibrium Problems #1, Answers
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1)
For the reaction:
CO
+
Cl
COCl
(g)
2 (g)
2 (g)
0.200 moles of CO, 0.100 moles of Cl
and 0.250 moles of COCl
are at equilibrium in a 5.00 L container at
2
2
a constant temperature.
a)
What is the value of K
?
eq
[CO] = n/V = 0.0400 M
[Cl
] = 0.0200 M
[COCl
] = 0.0500 M
2
2
Keq =
[COCl
]__
2
[CO] [Cl
]
2
Keq =
62.5
b)
Does this reaction favour the reactants or products at these conditions? K >1, so the products are favoured
c)
Which has higher entropy, the reactants or the products? There are two moles of gas on the reactant side of
the equation, and only one mole of gas on the product side, so entropy favours the reactants.
d)
Considering your answer to part “c”, is the forward reaction endothermic or exothermic? Why? The forward
reaction must be exothermic. The reaction is an equilibrium reaction, so it must be reversible. This means
that the driving forces must act in opposite directions. Since entropy favours the reactants, enthalpy must
favour the products so ∆H for the forward reaction must be negative.
e)
Calculate the value for K
for the reverse reaction. K
reverse = 1/ K
forward = 1/62.5 = 0.0160
eq
eq
eq
f)
What is the relationship between the K
forward and the K
reverse for a reaction?
eq
eq
K
reverse = 1/ K
forward
eq
eq
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↔ ↔
↔
2)
For the reaction: COCl
CO
+
Cl
2 (g)
(g)
2 (g)
-10
At 100˚C the reaction has an equilibrium constant of 2.2 x 10
. If 1.00 mole of phosgene (COCl
) is placed
2
in a 10.0 L flask, calculate the concentration of carbon monoxide at equilibrium.
COCl
CO
Cl
2 (g)
(g)
2 (g)
I
C = n/V = 0.10 M
0
0
– x
+ x
+ x
C
0.10 - x
x
x
E
Can we ignore the –x?
Keq
= [CO] [Cl
]
2
[COCl
]
2
0.10
is much greater than 500, so ignore -x
-10
-10
2
2.2 x 10
2.2 x 10
=
x
__
0.10
-6
x = 4.7 x 10
mol/L
-6
so [CO] at eq’m = 4.7 x 10
mol/L
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