Reversible Reactions and Equilibria
6 marks · NormalReversible reactions may respond differently to changes in pressure. Consider the information below for the following questions.
| Condition | Percentage of X | Percentage of Y |
|---|---|---|
| At 1 atm | 75 | 25 |
| At 3 atm | 85 | 15 |
The equilibrium is: State, with a reason, under which condition the position of equilibrium lies further to the right.
Explain why a change in pressure affects the relative percentages of X and Y in this equilibrium.
Predict and explain in which direction (left or right) the equilibrium will shift if the pressure is decreased.
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) At 1 atm, because there is a higher percentage of Y (25%) compared to 15% at 3 atm; so equilibrium is further to the right under lower pressure. · must show Correct reference to percentages of Y, States equilibrium position
(b) Changing pressure affects amounts of gases in equilibria where the number of gas molecules differs on each side. Here, 2 moles of X make 1 mole of Y. Increasing pressure favours the side with fewer gas molecules, so the equilibrium shifts left, changing the percentages. · must show References to number of gas molecules, States that pressure favours side with fewer molecules
(c) Decreasing pressure shifts the equilibrium to the right (towards Y), because the side with more gas molecules (left side, 2 X) is favoured as the pressure falls. · must show Shift to right, Explanation: side with more molecules favoured
Common mistakes: Assuming pressure does not affect solids or liquids; Believing equilibrium means equal percentages
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Correctly identifies condition where equilibrium is further to right
B2 (a) · 1 mark · Gives correct reason: higher percentage of Y means further to right
B3 (b) · 1 mark · Explains pressure only affects equilibria with different numbers of gas molecules
B4 (b) · 1 mark · Explains correctly how pressure change favours side with fewer/more gas molecules
B5 (c) · 1 mark · Predicts shift to right (towards Y) when pressure decreases
B6 (c) · 1 mark · Explanation refers to side with more gas molecules favoured at lower pressure
Worked solution
a) 1 atm, because Y makes up a higher proportion of the equilibrium mixture (25% vs 15% at 3 atm), so the equilibrium position is further to the right (towards Y).
b) The reaction involves different numbers of gaseous molecules (2 on the left, 1 on the right). Changing the pressure affects the position: higher pressure favours the side with fewer gas molecules (left), so more X and less Y.
c) If the pressure is decreased, the equilibrium shifts to the right (towards Y) because the side with more gas molecules (2 X) is favoured when pressure falls.