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How does a reversible reaction reach equilibrium, and how does it respond to changing conditions?

Dynamic equilibrium, Le Chatelier's principle and the effect of concentration, pressure and temperature, the role of a catalyst, the equilibrium constant Kc for homogeneous equilibria, and the compromise conditions used in industry.

An OCR H432 module 3 answer on chemical equilibrium: dynamic equilibrium, Le Chatelier's principle, the equilibrium constant Kc, and the compromise conditions used in industrial processes such as the Haber process.

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  1. What this topic is asking
  2. Dynamic equilibrium
  3. Le Chatelier's principle
  4. The equilibrium constant Kc
  5. Compromise conditions in industry
  6. Examples in context
  7. Try this

What this topic is asking

OCR specification point 3.2.3 wants you to explain dynamic equilibrium, apply Le Chatelier's principle to predict how concentration, pressure and temperature shift an equilibrium, explain why a catalyst does not change the position, write and calculate the equilibrium constant KcK_c for homogeneous equilibria, and explain the compromise conditions used in industry. This qualitative and introductory-quantitative treatment is extended in Module 5.

Dynamic equilibrium

Le Chatelier's principle

A catalyst speeds the forward and reverse reactions equally. It lets equilibrium be reached faster but does not change the position of equilibrium or the yield.

The equilibrium constant Kc

A large KcK_c means the equilibrium lies toward the products; a small KcK_c means it lies toward the reactants. KcK_c changes only with temperature.

Compromise conditions in industry

Examples in context

Example 1. The contact process. Sulfur trioxide is made by 2SO2+O2⇌2SO32\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 (ΔH\Delta H negative); a moderate temperature and a vanadium(V) oxide catalyst give a high yield efficiently, the same compromise logic as the Haber process.

Example 2. Blood pH buffering. The carbon dioxide and hydrogencarbonate equilibrium in blood shifts to oppose changes in acidity, an example of Le Chatelier's principle keeping a biological system stable.

Try this

Q1. State what is meant by a dynamic equilibrium. [2 marks]

  • Cue. In a closed system, the forward and reverse reactions occur at the same rate, so the concentrations of all species remain constant.

Q2. For 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g), predict the effect of increasing the pressure on the yield of SO3\text{SO}_3. [1 mark]

  • Cue. The yield increases, because the equilibrium shifts toward the side with fewer gas moles (2 moles of product versus 3 of reactant).

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksThe Haber process is N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), ΔH=−92 kJ mol−1\Delta H = -92\ \text{kJ mol}^{-1}. Explain, using Le Chatelier's principle, why a high pressure but only a moderate temperature is used.
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High pressure: there are 4 moles of gas on the left and 2 on the right, so increasing pressure shifts the equilibrium to the side with fewer gas moles, the products, increasing the yield of ammonia (1). Very high pressures are expensive and hazardous, so a compromise is used (1).

The forward reaction is exothermic, so a low temperature would increase the yield, but it would make the rate too slow (1). A moderate temperature (around 450 ∘C450\ ^{\circ}\text{C}) is a compromise that gives an acceptable yield at an acceptable rate, helped by an iron catalyst (1).

Markers reward the mole-count argument for pressure, the exothermic argument for a low temperature, and the recognition that both are compromises between yield and rate or cost.

OCR 20214 marksFor the equilibrium H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g), the equilibrium amounts in a 1.00 dm31.00\ \text{dm}^3 vessel are 0.20 mol H20.20\ \text{mol}\ \text{H}_2, 0.20 mol I20.20\ \text{mol}\ \text{I}_2 and 1.60 mol HI1.60\ \text{mol}\ \text{HI}. Write the expression for KcK_c and calculate its value.
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Kc=[HI]2[H2][I2]K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} (1).

Concentrations equal amounts here because V=1.00 dm3V = 1.00\ \text{dm}^3: [HI]=1.60[\text{HI}] = 1.60, [H2]=[I2]=0.20 mol dm−3[\text{H}_2] = [\text{I}_2] = 0.20\ \text{mol dm}^{-3} (1).

Kc=(1.60)2(0.20)(0.20)=2.560.040=64K_c = \dfrac{(1.60)^2}{(0.20)(0.20)} = \dfrac{2.56}{0.040} = 64 (1). The expression has equal powers top and bottom, so KcK_c has no units (1).

Markers reward the correct expression, substitution of concentrations, the numerical value, and the statement that there are no units.

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