Ch 1 · Chemical equilibria: reversible reactions, dynamic equilibrium · AS & A Level Chemistry 9701
Topic 1 of 2 in Equilibria — AS & A Level Chemistry 9701 — 48 lessons.
Introduction
Many important reactions, including the industrial synthesis of ammonia and sulfuric acid, do not go to completion but settle into a state of dynamic equilibrium. Cambridge 9701 requires you to explain this state qualitatively using Le Chatelier's principle, and quantitatively using the equilibrium constants K_c and K_p, and to apply both to the industrial conditions chosen for the Haber and Contact processes.
Key ideas — Reversible reactions and dynamic equilibrium
A reversible reaction is one that can proceed in both the forward and reverse directions, shown using leftharpoons. If a reversible reaction is carried out in a closed system, so that no reactants or products can enter or leave, it reaches dynamic equilibrium: the forward and reverse reactions continue to occur, but at equal rates, so the concentrations of reactants and products remain constant, though not necessarily equal to one another.
Key ideas — Le Chatelier's principle
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature or pressure, the position of equilibrium shifts so as to minimise the effect of that change. Increasing the concentration of a reactant shifts equilibrium towards the products, to remove some of the added reactant; increasing pressure shifts equilibrium towards the side with fewer gas molecules, to reduce the pressure; increasing temperature shifts equilibrium in the endothermic direction, to absorb the added heat. A catalyst speeds up the forward and reverse reactions equally, so it causes no shift in the position of equilibrium; it only allows equilibrium to be reached faster.
🧑🏫 Teacher's explanation
Imagine you are playing a game of tug-of-war where both sides are pulling with equal strength. This is 'equilibrium'—the net force is zero, even though both sides are working hard. If one side suddenly pulls harder (like adding more reactant), the game shifts. The other side must pull harder to compensate and restore the balance. This 'compensation' is exactly what Le Chatelier's principle describes.
📖 Definitions
🔍 Worked example — Easy
Consider the reaction N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g). If the concentration of N_2 is increased at constant temperature and pressure, what is the qualitative effect on the yield of NH_3?
🔍 Worked example — Standard
For the reaction N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g), Δ H = -92 kJ mol^-1. Predict the effect of increasing the temperature on the equilibrium constant K_c and the yield of NH_3.
🔍 Worked example — Exam-style
In the reaction PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g), the equilibrium constant K_p is 0.040 at 500 K. If the total pressure is doubled at constant temperature, what happens to the value of K_p?
⚠️ Common errors
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🎯 Exam question
Key ideas — Equilibrium constants Kc and Kp
For a homogeneous equilibrium aA + bB leftharpoons cC + dD, the equilibrium constant in terms of concentration, K_c, is:
🧑🏫 Teacher's explanation
Imagine a busy shop where people are constantly entering and leaving through the front and back doors. If the number of people inside stays exactly the same every minute, the shop is in 'equilibrium'. Even though individuals are moving (the reaction is 'dynamic'), the overall crowd size doesn't change. In chemistry, this happens in a closed container where the forward and reverse reactions occur at the same rate.
📖 Definitions
🔍 Worked example — Standard
For the reaction PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g), the K_p value is 0.040 at 500 K. Calculate the partial pressure of PCl_5 if the partial pressures of PCl_3 and Cl_2 are both 0.20 Pa at equilibrium.
🔍 Worked example — Exam-style
In a reaction A_(g) + B_(g) leftharpoons 2C_(g), the total pressure is 1.0 atm at equilibrium. The mole fraction of C is 0.6. Calculate the partial pressure of A.
⚠️ Common errors
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🎯 Exam question
Key ideas — Calculating quantities at equilibrium
The partial pressure of a gas in a mixture is its mole fraction multiplied by the total pressure: p_A = x_A × P_total, where the mole fraction x_A is the moles of gas A divided by the total moles of gas present. Given the equilibrium amounts of each species (often found from initial amounts and the amount reacted), the mole fractions, partial pressures and hence K_p (or the concentrations and hence K_c) can all be calculated, as shown in the worked example.
🧑🏫 Teacher's explanation
Imagine you are in a crowded room where people are constantly moving between two different areas: a lounge and a kitchen. Even if people are moving back and forth, the total number of people in the lounge stays the same because for every person entering, one person leaves. This is 'dynamic equilibrium'. In chemistry, molecules are doing the same thing in a closed container; they react forward and backward at the exact same rate.
📖 Definitions
🔍 Worked example — Easy
A mixture of 2.0 mol of N_2(g) and 3.0 mol of H_2(g) is placed in a container at a total pressure of 100 kPa. Calculate the partial pressure of N_2(g).
🔍 Worked example — Standard
For the reaction N_2(g) + 3H_2(g) leftharpoons 2NH_3(g), the equilibrium partial pressures are p_N_2 = 0.20 bar, p_H_2 = 0.30 bar, and p_NH_3 = 0.10 bar. Calculate the value of K_p.
🔍 Worked example — Exam-style
A 10.0 dm^3 vessel contains 4.0 mol of A and 6.0 mol of B at equilibrium. The reaction is A_(g) + B_(g) leftharpoons 2C_(g). If the total pressure is 2.0 atm, find the partial pressure of C if the equilibrium concentration of C is 0.5 mol dm^-3.
⚠️ Common errors
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🎯 Exam question
Key ideas — The Haber process and the Contact process
The Haber process, N_2(g) + 3H_2(g) leftharpoons 2NH_3(g), Δ H = -92\ mathrmkJ\ mol^-1, uses a compromise of about 450 °C and 200 atm with an iron catalyst: high pressure favours the forward reaction (4 mol of gas becomes 2 mol), but very high pressure is costly to generate and maintain, while a moderately low temperature would increase the equilibrium yield (the forward reaction is exothermic) but is not used because it would make the rate of reaction too slow; the catalyst speeds up attainment of equilibrium without affecting its position. The Contact process, 2SO_2(g) + O_2(g) leftharpoons 2SO_3(g), Δ H = -196\ mathrmkJ\ mol^-1, similarly uses a compromise temperature of about 450 °C with a vanadium(V) oxide catalyst, and a pressure only slightly above atmospheric, since the equilibrium yield is already high enough at this pressure that the extra cost of higher pressure is not justified.
🧑🏫 Teacher's explanation
Imagine you are walking on a treadmill. You are moving forward, but the belt is moving backward at the exact same speed. To an observer, you stay in the same spot, even though you are working hard. This is 'dynamic equilibrium': the forward and backward reactions are happening at the same rate, so the concentrations of reactants and products don't change, even though the molecules are constantly reacting.
📖 Definitions
🔍 Worked example — Easy
In the Haber process, ceN2(g) + 3H2(g) <=> 2NH3(g), Δ H = -92 kJ mol^-1. If the temperature is increased, which way does the equilibrium shift and why?
🔍 Worked example — Standard
For the reaction ceN2(g) + 3H2(g) <=> 2NH3(g), the equilibrium constant K_c is 0.50 at a certain temperature. If the concentration of ceN2 is 0.20 mol dm^-3, ceH2 is 0.40 mol dm^-3, and ceNH3 is 0.10 mol dm^-3, calculate the value of K_c to verify the data.
🔍 Worked example — Exam-style
In the Contact process, ce2SO2(g) + O2(g <=> 2SO3(g), Δ H = -196 kJ mol^-1. A reaction vessel contains 0.50 mol of ceSO2, 0.50 mol of ceO2, and 0.50 mol of ceSO3 in a 2.0 dm^3 container. Calculate the equilibrium constant K_c for this reaction.
⚠️ Common errors
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🎯 Exam question
Worked example
Question. 1.00 mol of N_2O_4(g) was allowed to reach equilibrium with NO_2(g) in a closed vessel at constant temperature: N_2O_4(g) leftharpoons 2NO_2(g). At equilibrium, 0.60 mol of N_2O_4 remained, and the total pressure was 200 kPa. Calculate K_p for this reaction, stating its units. [4]
🔍 Worked example — Easy
In the reaction A(g) leftharpoons B(g), the equilibrium constant K_c is 4.0. If the concentration of A is 0.50 mol dm^-3 at equilibrium, calculate the concentration of B.
🔍 Worked example — Standard
For the reaction ceN2O4(g) <=> 2NO2(g), the total pressure is 100 kPa. At equilibrium, the mole fraction of ceN2O4 is 0.4. Calculate the partial pressure of ceNO2.
🔍 Worked example — Exam-style
1.00 mol of ceN2O4(g) was allowed to reach equilibrium with ceNO2(g) in a closed vessel at constant temperature: ceN2O4(g) <=> 2NO2(g). At equilibrium, 0.60 mol of ceN2O4 remained, and the total pressure was 200 kPa. Calculate K_p for this reaction, stating its units.
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Practice
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