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What is a dynamic equilibrium, and how do changing conditions shift it?

Reversible reactions and dynamic equilibrium, Le Chatelier's principle and the effects of changing concentration, pressure and temperature, the effect of a catalyst on equilibrium, and the equilibrium constant Kc.

A CCEA Life and Health Sciences answer on chemical equilibria: reversible reactions and dynamic equilibrium, Le Chatelier's principle for changes in concentration, pressure and temperature, the effect of a catalyst, and the equilibrium constant Kc.

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  1. What this dot point is asking
  2. Dynamic equilibrium
  3. Le Chatelier's principle
  4. The equilibrium constant Kc
  5. Examples in context
  6. Try this

What this dot point is asking

CCEA wants you to explain reversible reactions and dynamic equilibrium, apply Le Chatelier's principle to predict the effect of changing concentration, pressure and temperature, state the effect of a catalyst on equilibrium, and write and calculate the equilibrium constant Kc. This is the theory that explains the operating conditions chosen for industrial processes such as the Haber and Contact processes.

Dynamic equilibrium

A reversible reaction can go in both directions, shown by the reversible arrow. If you start with reactants, the forward reaction is fast at first and slows as reactants are used up; the backward reaction starts slow and speeds up as products build. When the two rates become equal, the system is at equilibrium and the concentrations no longer change. Equilibrium can only be reached in a closed system (nothing added or escaping), because if a product escaped the system could never balance.

Le Chatelier's principle

To apply the principle, identify what changed and reason how the system opposes it. For pressure changes, count the moles of gas on each side and shift towards the side with fewer to reduce pressure. For temperature changes, identify which direction is endothermic (it absorbs the added heat) and shift that way; raising temperature therefore favours the endothermic direction and lowers the yield of an exothermic forward reaction. These rules let chemists choose conditions that maximise the yield of the desired product.

The equilibrium constant Kc

For a reaction, the equilibrium constant Kc is written as the concentrations of the products multiplied together, each raised to the power of its balancing number, divided by the same for the reactants. For the general reaction where a moles of A react with b moles of B to give c moles of C and d moles of D, the expression is the product concentrations over the reactant concentrations, each raised to its coefficient. Kc has a fixed value at a given temperature: a large Kc means the equilibrium lies towards the products, and a small Kc means it lies towards the reactants. Changing concentration or pressure shifts the position of equilibrium but does not change Kc; only a change in temperature changes the value of Kc.

Examples in context

Example 1. The Haber process. Ammonia forms by the exothermic reaction of nitrogen and hydrogen, with fewer gas moles on the product side. Le Chatelier's principle predicts a high pressure and a low temperature would maximise yield, but a very low temperature gives an unworkably slow rate, so a moderate temperature is used as a compromise, with an iron catalyst to reach equilibrium faster.

Example 2. Carbon dioxide in the blood. Carbon dioxide and water are in equilibrium with carbonic acid, which forms hydrogen ions. When carbon dioxide is exhaled and removed at the lungs, the equilibrium shifts to oppose the loss, releasing carbon dioxide from the blood. This biological equilibrium shows Le Chatelier's principle operating in the body, linking chemistry to human physiology.

Try this

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

  • Cue. In a closed system, the forward and backward reactions occur at equal rates, so concentrations stay constant while both reactions continue.

Q2. A forward reaction is endothermic. Predict the effect of increasing temperature on the yield of product, with a reason. [2 marks]

  • Cue. Yield increases; raising temperature favours the endothermic (forward) direction, opposing the rise.

Q3. State the effect of adding a catalyst on the position of an equilibrium. [1 mark]

  • Cue. No effect on the position; it only speeds the approach to equilibrium.

Exam-style practice questions

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

CCEA AS 36 marksFor the reaction N2(g) + 3H2(g) reversible 2NH3(g), the forward reaction is exothermic. Use Le Chatelier's principle to predict and explain the effect on the yield of ammonia of (i) increasing the pressure and (ii) increasing the temperature.
Show worked answer →

Each part needs the prediction plus a Le Chatelier explanation referring to the relevant property.

Increasing the pressure: there are 4 moles of gas on the left and 2 moles on the right. Le Chatelier's principle says the equilibrium shifts to oppose the increase in pressure, so it shifts to the side with fewer gas moles, the right. This increases the yield of ammonia.

Increasing the temperature: the forward reaction is exothermic. Le Chatelier's principle says the equilibrium shifts to oppose the rise in temperature by favouring the endothermic direction, which is the reverse reaction. So the equilibrium shifts left and the yield of ammonia decreases.

A linking point: this is why the Haber process uses a high pressure (good for yield) but only a moderate temperature (a compromise between yield and a workable rate).

Markers reward counting gas moles and shifting to fewer moles for pressure, identifying the endothermic direction for temperature, and the correct effect on yield in each case.

CCEA AS 35 marksWrite the expression for the equilibrium constant Kc for the reaction H2(g) + I2(g) reversible 2HI(g), and calculate Kc when the equilibrium concentrations are 0.20 mol per cubic decimetre of hydrogen, 0.20 mol per cubic decimetre of iodine and 1.6 mol per cubic decimetre of hydrogen iodide.
Show worked answer →

Write Kc as products over reactants, each raised to its balancing number, then substitute.

The expression is:

Kc=[HI]2[H2][I2]K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}

Substituting the equilibrium concentrations:

Kc=(1.6)2(0.20)(0.20)=2.560.04=64K_c = \frac{(1.6)^2}{(0.20)(0.20)} = \frac{2.56}{0.04} = 64

Because the powers cancel (2 on top, 1 plus 1 on the bottom), Kc has no units here.

Markers reward the correct Kc expression with the square on HI, correct substitution, the value 64, and the comment that there are no units in this case.

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