Changing the pressure
5 marks · HardThe equation for a reaction that reaches equilibrium in a sealed container is:
A student investigates how changing the pressure affects this equilibrium.
State the total number of moles of gaseous molecules on each side of the equation.
Predict, with a reason, what will happen to the position of equilibrium if the pressure in the container is increased.
If the pressure is decreased, which side of the equilibrium will be favoured? Explain your answer.
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) Left: 3 moles (2 from A and 1 from B); Right: 3 moles (from 3C)
(b) There are the same number of moles of gaseous molecules on both sides. Increasing the pressure will have no effect on the position of equilibrium.
(c) The decrease in pressure will not affect the position of equilibrium because both sides have the same number of gaseous molecules.
Common mistakes: Incorrect count of gas molecules.; Saying pressure has no effect because both sides have gases.; Incorrect explanation of Le Chatelier's principle.
Show mark scheme and worked solution
Marking points
A1 (a) · 1 mark · Correctly counts and states the total number of gaseous moles on each side (3 left, 3 right).
M2 (b) · 1 mark · Mentions or uses the fact that the number of gaseous molecules is the same on both sides in the explanation.
A2 (b) · 1 mark · Predicts correctly that increasing pressure has no effect; or that equilibrium does not shift.
M3 (c) · 1 mark · States or uses the same number of gaseous molecules argument for decrease in pressure.
A3 (c) · 1 mark · Correctly states that decreasing pressure has no effect or equilibrium does not shift.
Worked solution
(a) On the left of the equation: 2A(g) + B(g) = 2 + 1 = 3 moles; on the right: 3C(g) = 3 moles.
(b) Since there are 3 moles of gaseous molecules on both sides, increasing the pressure will have no effect on the position of equilibrium. The number of gas molecules is equal, so equilibrium does not shift.
(c) For the same reason, decreasing the pressure will also not affect the equilibrium position, as there is no change in the number of gas molecules on either side.