The position of equilibrium
3 marks · NormalThe following equilibrium is established in a sealed container at constant temperature:
The forward reaction is exothermic.
The system is at equilibrium with the following amounts:
| Substance | Amount (mol) |
|---|---|
| D | 0.80 |
| E | 1.60 |
| F | 0.60 |
| G | 0.60 |
The pressure on the system is increased. Predict, with a calculation, which direction the equilibrium will shift. Show your working.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Increasing pressure shifts equilibrium towards the left (reactants) because there are 3 moles of gas (D + 2E) on the left and 2 moles (F + G) on the right. Equilibrium moves to the side with fewer moles of gas.
Common mistakes: Forgetting to sum moles; Confusing which side has more moles; Mixing up direction of equilibrium shift
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly counts the total number of moles of gas on each side: left = 1 (D) + 2 (E) = 3, right = 1 (F) + 1 (G) = 2.
M2 (a) · 1 mark · Identifies that increasing pressure favours the side with fewer moles of gas.
A1 (a) · 1 mark · Correctly concludes equilibrium shifts to the right (products) since there are fewer moles of gas on the right, OR describes shift with clear reference to calculation.
Worked solution
Moles on left (reactants): D + 2E = 1 + 2 = 3 mol.
Moles on right (products): F + G = 1 + 1 = 2 mol.
Increasing pressure favours the side with fewer moles of gas (the right), so equilibrium shifts to the right (towards F and G).