The effect of concentration
5 marks · HardThe reaction below is at equilibrium in a closed vessel:
A student increases the concentration of P(g).
Predict the effect on the position of equilibrium. State your answer as 'left', 'right' or 'no change'.
Explain your answer in (a), referring to the effect of changing concentration.
If instead some M(g) is removed from the vessel, what effect would this have on the equilibrium position? State your answer and explain.
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.
Show answer
(a) left
(b) The increase in P shifts the equilibrium to the left, towards reactants. The system responds by using up excess P to form more M and N.
(c) Shifts left; Removing M(g) decreases a reactant, so equilibrium shifts to form more M(g), i.e., towards the reactants (left).
Common mistakes: No effect; Wrong direction; Unclear explanation
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Correct direction given ('left').
B2 (b) · 1 mark · Mentions equilibrium shifts to reduce added product (P).
B3 (b) · 1 mark · States products react to form reactants (M/N), linking shift to left.
B4 (c) · 1 mark · Correctly predicts shift (left) due to removal of M(g).
B5 (c) · 1 mark · Explains system replaces removed M by shifting towards reactants.
Worked solution
a) Increasing P(g) shifts equilibrium to the left.
b) Adding product increases its concentration. The system tries to remove some P(g) by making more reactants (M and N), so shifts left.
c) Removing M(g) means the concentration of a reactant is decreased. The system will respond by replacing some M(g), shifting to the left (towards reactants).