IB Chemistry R2.3 Exam Practice
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R2.3 Exam Practice

Test your knowledge on Equilibrium

Section B: Data Analysis (Paper 1B Style)

Calculator and Data Booklet permitted. Show all working clearly.

Question 1: Le Chatelier's Principle Predict

5 marks

The Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = −92 kJ mol⁻¹
A student collects data on NH₃ yield at different temperatures and pressures:

Temperature / °C200 atm yield / %400 atm yield / %
3006780
4003655
5001832

(a) Explain, using Le Chatelier's principle, why increasing pressure increases the yield of NH₃. [2]

(b) Explain why increasing temperature decreases the yield of NH₃. [2]

(c) Suggest why industry uses 400–500 °C despite the lower yield. [1]

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(a) There are 4 moles of gas on the left and 2 moles on the right [1]; increasing pressure shifts equilibrium towards the side with fewer moles of gas (right), increasing yield [1]

(b) The forward reaction is exothermic (ΔH negative) [1]; increasing temperature favours the endothermic (reverse) reaction, shifting equilibrium left and reducing NH₃ yield [1]

(c) Higher temperature increases the rate of reaction / allows equilibrium to be reached faster, making the process economically viable [1]

Examiner tip: Distinguish between the position of equilibrium (yield) and the value of K. Temperature changes both; pressure/concentration changes only the position, not K.

Question 2: Equilibrium Constant Calculation Calculate

4 marks

For the reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
At equilibrium at 450 °C: [H₂] = 0.040 mol dm⁻³, [I₂] = 0.040 mol dm⁻³, [HI] = 0.32 mol dm⁻³.

(a) Write the expression for Kc. [1]

(b) Calculate the value of Kc. [1]

(c) State and explain whether the equilibrium favours products or reactants. [1]

(d) Predict how Kc would change if the temperature were increased, given that the forward reaction is exothermic. [1]

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(a) Kc = [HI]² / ([H₂][I₂]) [1]

(b) Kc = (0.32)² / (0.040 × 0.040) = 0.1024 / 0.0016 = 64 [1]

(c) Kc >> 1, so equilibrium favours products [1]

(d) Kc would decrease because increasing temperature favours the endothermic (reverse) direction [1]

Section C: Structured Questions (Paper 2 Style)

Show all working. State answers with appropriate significant figures and units.

Question 3: Dynamic Equilibrium Define

5 marks

(a) Define dynamic equilibrium. [2]

(b) State two conditions necessary for a system to reach equilibrium. [1]

(c) Explain why a catalyst does not change the position of equilibrium or the value of K. [2]

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(a) The rates of the forward and reverse reactions are equal [1]; the concentrations of reactants and products remain constant (but are not necessarily equal) [1]

(b) Closed system AND reversible reaction [1]

(c) A catalyst speeds up the forward and reverse reactions equally [1]; since both rates increase by the same factor, the ratio of products to reactants (and hence K) is unchanged [1]

Examiner tip: "Concentrations remain constant" does NOT mean "concentrations are equal." Many students lose marks for this error.

Question 4: Kp and Partial Pressures HL Calculate

5 marks

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At 250 °C and total pressure of 2.00 atm, the mole fractions at equilibrium are: PCl₅ = 0.40, PCl₃ = 0.30, Cl₂ = 0.30.

(a) Calculate the partial pressure of each gas. [2]

(b) Write the expression for Kp and calculate its value. [2]

(c) State the units of Kp for this reaction. [1]

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(a) p(PCl₅) = 0.40 × 2.00 = 0.80 atm; p(PCl₃) = 0.30 × 2.00 = 0.60 atm [1]; p(Cl₂) = 0.30 × 2.00 = 0.60 atm [1]

(b) Kp = p(PCl₃) × p(Cl₂) / p(PCl₅) = (0.60 × 0.60) / 0.80 [1] = 0.45 [1]

(c) atm (Δn = 2 − 1 = +1, so units = atm¹) [1]

Examiner tip: Partial pressure = mole fraction × total pressure. Always check units: Kp has units of pressure^Δn, where Δn = moles of gas products − moles of gas reactants.

Question 5: Gibbs Free Energy and Equilibrium HL Deduce

4 marks

The relationship between Gibbs free energy and the equilibrium constant is: ΔG° = −RT ln K.

(a) Deduce what a large positive value of K implies about ΔG°. [1]

(b) A reaction has K = 0.0050 at 298 K. Calculate ΔG° in kJ mol⁻¹. (R = 8.31 J K⁻¹ mol⁻¹) [2]

(c) State whether this reaction is spontaneous under standard conditions. Justify your answer. [1]

Show Mark Scheme

(a) ΔG° is large and negative (since ln K is positive when K > 1, and the negative sign makes ΔG° negative) [1]

(b) ΔG° = −(8.31)(298) ln(0.0050) = −(2476.4)(−5.298) [1] = +13 120 J mol⁻¹ = +13.1 kJ mol⁻¹ [1]

(c) Not spontaneous under standard conditions because ΔG° is positive [1]

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