The problems are on the chemical equilibrium worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Chemical equilibrium is reached when forward and reverse reactions occur at the same rate, so concentrations stay constant. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on chemical equilibrium, from equilibrium words to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the expression, the substitution and the result with units where needed. Check that you used equilibrium (not initial) concentrations, that coefficients became powers, and that your predicted shift agrees with the sign of ΔH and the number of gas moles.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R2.3
Problem 1 Equilibrium words
1. A state in which the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations stay constant.2. False: both reactions continue at equal rates.
3. In an open system products or reactants can escape, so equilibrium cannot be established.
Problem 2 Writing Kc
1. Kc = [NO]² / ([N₂][O₂]); no units.2. Kc = [N₂O₄] / [NO₂]²; dm³ mol⁻¹.
3. Kc = [CO][H₂]³ / ([CH₄][H₂O]); mol² dm⁻⁶.
4. Kc = [HBr]² / ([H₂][Br₂]); no units.
Problem 3 Solids in the expression
1. Kc = [CO₂]; Kp = p(CO₂).2. Kc = [CO]² / [CO₂]; Kp = p(CO)² / p(CO₂).
3. The concentration of a pure solid is constant, so it is included in the value of K: solids are omitted.
Problem 4 Reading equilibrium concentrations
1. [NO] = 0.050, [Cl₂] = 0.040, [NOCl] = 0.200 mol/dm³.2. Kc = [NOCl]² / ([NO]²[Cl₂]).
3. Kc = 0.200² / (0.050² × 0.040) = 0.0400 / 1.0 × 10⁻⁴ = 400 dm³ mol⁻¹.
Problem 5 Size of K
1. C: the huge K means products dominate.2. A: the tiny K means almost no products form.
3. B contains comparable amounts of reactants and products.
Problem 6 Predicting the shift
1. Left (opposes the added Cl₂).2. Left: 1 mole of gas on the left against 2 on the right.
3. Right: the forward reaction is endothermic.
4. None.
5. Right.
Problem 7 Does K change?
1. No: the position shifts but the ratio at equilibrium is the same.2. No: the position shifts only.
3. Yes: temperature changes the value of K.
4. No: both rates increase equally.
Problem 8 Kp from partial pressures
1. Kp = p(N₂O₄) / p(NO₂)².2. Kp = 30 / 12² = 30 / 144 = 0.208 kPa⁻¹.
3. Total = 42 kPa; x(NO₂) = 12 / 42 = 0.286.
Problem 9 Decomposition of PCl₅
1. [PCl₅] = 0.060, [PCl₃] = [Cl₂] = 0.040 mol/dm³.2. Kc = [PCl₃][Cl₂] / [PCl₅].
3. Kc = 0.040² / 0.060 = 2.67 × 10⁻² mol dm⁻³.
Problem 10 Reaction quotient
1. Q = (1.5 × 1.5) / (0.50 × 0.50) = 9.0, which is greater than 6.0, so the mixture shifts left.2. Q = (0.30 × 0.30) / (0.20 × 0.20) = 2.25, which is less than 6.0, so the mixture shifts right.
Problem 11 Finding a concentration
1. Kc = [H₂][I₂] / [HI]².2. [H₂]² = 0.020 × 0.50² = 0.0050, so [H₂] = 0.0707 mol/dm³.
3. The reverse reaction has the reciprocal constant: Kc = 1 / 0.020 = 50 (no units).
Problem 12 Temperature and K
1. K decreases as temperature rises, so the forward reaction is exothermic.2. 8000 / 4.5 = 1.8 × 10³ times.
3. A higher temperature shifts the equilibrium in the endothermic (reverse) direction to absorb heat, so less product forms.
Problem 13 Methanol yield
1. Yield increases with pressure: there are 3 moles of gas on the left and 1 on the right, so higher pressure favours the side with fewer moles.2. Yield is lower at 600 K, so the forward reaction is exothermic (ΔH negative).
3. About 58 % at 500 K and 29 % at 600 K.
Problem 14 Iron and thiocyanate colours
1. Deeper red: extra Fe³⁺ shifts the equilibrium right.2. Paler: removing SCN⁻ shifts the equilibrium left.
3. Paler: all concentrations halve, so Q = 0.5 / (0.5 × 0.5) = 2 times K. Q is greater than K, so the system shifts to the left (the side with more ions).
Problem 15 Kp from mole amounts
1. Total = 1.20 mol: x(COCl₂) = 0.50, x(CO) = x(Cl₂) = 0.25.2. p(COCl₂) = 75.0 kPa; p(CO) = p(Cl₂) = 37.5 kPa.
3. Kp = p(CO) p(Cl₂) / p(COCl₂) = 37.5² / 75.0 = 18.8 kPa.
Problem 16 Kp from Kc
1. Δn = 2 − 1 = +1.2. RT = 8.314 × 500 = 4157 kPa dm³ mol⁻¹.
3. Kp = 0.0267 × 4157 = 111 kPa.
Problem 17 Esterification
1. Kc = [ester][water] / ([acid][ethanol]).2. Volume cancels: x² / ((1 − x)(2 − x)) = 4.0, so x² = 4(2 − 3x + x²), giving 3x² − 12x + 8 = 0.
3. x = (12 − √48) / 6 = (12 − 6.928) / 6 = 0.845 mol (the root 3.15 is impossible).
Problem 18 Water gas shift
1. Initial: 1.00, 1.00, 0, 0. Change: −x, −x, +x, +x. Equilibrium: 1 − x, 1 − x, x, x.2. Kc = x² / (1 − x)² = 6.0, so x / (1 − x) = √6 = 2.449.
3. x = 2.449 / 3.449 = 0.710 mol; [H₂] = 0.710 / 5.0 = 0.142 mol/dm³.
Problem 19 K at a new temperature
1. 3.3557 × 10⁻³ − 2.8571 × 10⁻³ = 4.986 × 10⁻⁴ K⁻¹.2. ln(K₂/K₁) = (57 200 / 8.314)(4.986 × 10⁻⁴) = 6880 × 4.986 × 10⁻⁴ = 3.43.
3. K₂ = 4.6 × 10⁻³ × e^3.43 = 4.6 × 10⁻³ × 30.9 = 0.142 mol dm⁻³; K increases because the forward reaction is endothermic.
Problem 20 Spot the errors
1. Solids are omitted: Kc = [CO₂].2. A catalyst does not change Kc; it only speeds up the approach to equilibrium.
3. It shifts towards the side with fewer moles of gas.
4. The coefficient becomes a power: Kc = [B] / [A]².
