Revision Notes
1) Equilibrium
quantities
 A
reversible reaction reaches a position of dynamic equilibrium where both
the forward and backward reactions are taking place at the same rate and
concentrations of chemicals are constant
 At
equilibrium there will always be a mixture of reactants and products
 The
moles present at equilibrium can be worked out using the equation and the
moles present at the start
Example
1
The hydrolysis of ethyl ethanoate is a reversible reaction. The equation
for the equilibrium is shown below.
CH_{3}COOC_{2}H_{5}
+ H_{2}O ¾
CH_{3}COOH + C_{2}H_{5}OH
A student mixed together 8.0 mol ethyl ethanoate and 5.0 mol water. He
also added a small amount of hydrochloric acid to catalyse the reaction.
The student left the mixture until it had reached equilibrium at constant
temperature. He found that 2.0 mol of ethanoic acid had formed.
The information in the question is summarised in the table below.
Component

CH_{3}COOC_{2}H_{5}

H_{2}O

CH_{3}COOH

C_{2}H_{5}OH

Initial mol

8.0

5.0

0.0

0.0

Equilibrium mol



2.0


The equation says that for every mole of CH_{3}COOH made, one
mole of C_{2}H_{5}OH is made so the equilibrium moles of
ethanol is also 2.0
The equation also says that for every mole of CH_{3}COOH made,
one mole of CH_{3}COOC_{2}H_{5} is used up so the
equilibrium moles of ethyl ethanoate is initial moles  2.0 = 6.0 mol
The equation also says that for every mole of CH_{3}COOH made,
one mole of H_{2}O is used up so the equilibrium moles of water is
initial moles  2.0 = 3.0 mol
The completed table is as follows.
Component

CH_{3}COOC_{2}H_{5}

H_{2}O

CH_{3}COOH

C_{2}H_{5}OH

Initial mol

8.0

5.0

0.0

0.0

Equilibrium mol

6.0

3.0

2.0

2.0

Source: OCR Module 2816/01 June 2008 part question
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