The questions are on the chemical equilibrium test. Take it under real conditions before reading the mark scheme.
Quick answer: Chemical equilibrium is reached when forward and reverse reactions occur at the same rate, so concentrations stay constant. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on chemical equilibrium, from writing Kc to direction of change, with the equivalent answers to accept.
Model answers and mark schemes follow. Award method marks even if a numerical slip occurs, and check that predicted shifts are justified by moles of gas or by the sign of ΔH.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R2.3
Question 1 Writing Kc (2 marks)
(a) Kc = [CO₂]² / ([CO]²[O₂])
(b) Units: mol² dm⁻⁶ / (mol² dm⁻⁶ × mol dm⁻³) = dm³ mol⁻¹.
Marking: 1 for the correct powers; 1 for the units.
Question 2 Calculating Kc (3 marks)
Kc = [C] / ([A][B]²) = 0.60 / (0.10 × 0.20²) = 0.60 / 0.0040 = 150 dm⁶ mol⁻².
Marking: 1 for the expression; 1 for the value; 1 for the units.
Question 3 Butane isomerisation (3 marks)
Let x mol of methylpropane form: butane = 1 − x. The volume cancels because the mole numbers on both sides are equal. x / (1 − x) = 2.5, so x = 2.5 / 3.5 = 0.714. Methylpropane = 0.714 mol and butane = 0.286 mol.
Marking: 1 for the ICE setup; 1 for the equation; 1 for both amounts.
Question 4 Nitrogen monoxide yield (3 marks)
(a) The yield increases: the forward reaction is endothermic, so a rise in temperature shifts the equilibrium to the right, and K increases.
(b) No effect: there are 2 moles of gas on each side, so pressure does not change the position.
Marking: 1 for the effect in (a); 1 for the explanation in (a); 1 for (b) with reason.
Question 5 Kc and temperature (3 marks)
(a) Exothermic, because Kc falls as temperature rises.
(b) A rise in temperature shifts the equilibrium in the endothermic direction, here the reverse, so less product forms.
(c) Kc keeps falling and approaches zero.
Marking: 1 mark per part.
Question 6 Calculating Kp (2 marks)
Kp = p(NO₂)² / p(N₂O₄) = 50² / 25 = 100 kPa.
Marking: 1 for the expression; 1 for the value with units.
Question 7 Role of a catalyst (2 marks)
A catalyst lowers the activation energy of both the forward and reverse reactions equally, so both rates increase by the same factor and K is unchanged. Equilibrium is reached sooner.
Marking: 1 for both reactions affected equally; 1 for K unchanged and faster approach.
Question 8 Direction of change (2 marks)
Q is less than K, so the mixture has too much reactant relative to equilibrium. It shifts to the right, forming more products until Q = K.
Marking: 1 for the direction; 1 for the comparison of Q with K.
