Edexcel A
level Chemistry (2017)
Topic 11:
Equilibrium II:
Here is
the second learning objective:
11/II/2. To
be able to calculate a value, with units where appropriate, for the equilibrium
constant (Kc
and Kp) for homogeneous and heterogeneous reactions, from
experimental data
If you have been following this blog recently with the posts on calculation
of Kc and Kp then the following examples should help you
develop your understanding of this topic.
Example 1:
5 moles ethanol, 6 moles ethanoic acid, 6 moles ethyl ethanoate and 4 moles
of water were mixed together in a stoppered bottle at 15℃.
After equilibrium had been reached the bottle was found to contain only 4
moles of ethanoic acid.
Calculate Kc for this reaction
CH3CH2OH +
CH3COOH ⇌ CH3COOCH2CH3 + H2O
CH3CH2OH
|
CH3COOH
|
CH3COOCH2CH3
|
H2O
|
|
Initial number of moles
|
5
|
6
|
6
|
4
|
Change in number of moles
|
–2
|
–2
|
+2
|
+2
|
Equilibrium number of moles
|
3
|
4
|
8
|
6
|
∴ after substitution of the equilibrium values Kc
= 4
Now suppose that 1 mole of ethanol, 1 mole of ethanoic acid, 3 moles of
ethyl ethanoate and 3 moles of water are mixed together in stoppered flask at
15℃.
How many moles of each reagent exist in the bottle at equilibrium given
that Kc = 4.
Consider:
CH3CH2OH
|
CH3COOH
|
CH3COOCH2CH3
|
H2O
|
|
Initial number of moles
|
1
|
1
|
3
|
3
|
Change in number of moles
|
+𝝰
|
+𝝰
|
–𝝰
|
–𝝰
|
Equilibrium number of moles
|
1+𝝰
|
1+𝝰
|
3–𝝰
|
3–𝝰
|
Actual mole values at equilibrium
|
1.33
|
1.33
|
2.67
|
2.67
|
where 𝝰 is the degree of dissociation
of ethyl ethanoate and water. Since both these compounds are in the larger concentration the effect will be to push the equilibrium position to the left. The concentrations of water and ethyl ethanoate will decrease. Therefore these two species concentrations fall by the degree of dissociation 𝝰 where as the two reactants concentrations increase by the degree of dissociation, according to Le Chatelier’s principle.
∴ taking
roots
giving a value of 𝝰 = 0.333
Example 2:
At 200℃, Kc for the reaction:
PCl5(g) ⇌ PCl3(g) +
Cl2(g) ΔH =
+124kJ
is 8 × 10–3
mol.dm–3
A sample of pure PCl5(g) was introduced into an evacuated vessel
at 200℃. When the equilibrium had been
reached the concentration of PCl5(g) was 0.2 × 10–1
mol.dm–3
What are the concentrations of PCl3(g) and Cl2(g) at
equilibrium?
Equation: PCl5(g) ⇌ PCl3(g) +
Cl2(g) ΔH =
+124kJ
From this we see that [PCl3(g)]
= [Cl2(g)] because for
every mole of Cl2(g) formed a mole of PCl3(g) also forms.
Kc = 8 × 10–3 =
At equilibrium [PCl5] =
0.2 × 10–1 mol.dm–3
∴ [PCl3(g)] × [Cl2(g)] = [PCl5(g)] × 8 ×
10–3
= 0.2 × 10–1 × 8 × 10–3
∴ [PCl3(g)] = [Cl2(g)] = √0.2
× 10–1 × 8 × 10–3
=
1.265 mol.dm–3
Example 3: (More difficult)
This is taken from Chemistry in Context p331.
At 488K the equilibrium constant for the reaction:
COCl2 (g) ⇌ CO(g)
+ Cl2(g)
is 2 × 10–6 atm
Assuming that the total pressure at equilibrium is P and the degree of
dissociation of COCl2 (g) is 𝜶 , deduce a relationship
between Kp, 𝜶 and P.
(Degree of dissociation (𝜶) is the fraction
of COCl2(g) dissociated)
Let’s assume we have 1 mole of COCl2(g) initially in the
reaction vessel.
After simplifying
The equation can be simplified further and a quadratic solution avoided
since when 𝜶 is very small 1– 𝜶 or 1+ 𝜶 approximate to 1.
Therefore Kp ≈ P𝜶2
So at 1atm pressure P = 1atm
Then 𝜶 =
√Kp = √2 × 10–6 = 1.414 × 10–3
which as you can see is very small.
Example 4:
(From Selvaratnam and Frazer: Problem Solving in Chemistry)
Calculate the degree of dissociation when 0.500 mol of phosphorus
pentachloride is maintained at 573K and 2.00 × 105 Pa. The equilibrium
constant for this homogenous gas dissociation is 2.3 × 10–3 mol.dm–3 at 573K.
The previous problem provides a template with which to solve this example.
And
finally if you have become proficient in calculating Kp, Kc 𝛂 the
next question to ask is what does the value mean, what significance does it
hold?
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