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IBDP Chemistry -Reactivity 3.2 Electron transfer reactions - IB Style Questions For SL Paper 2 -FA 2025

Question

Ammonia is manufactured on an industrial scale using the following reversible reaction:
\( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \)
(a) Calculate the value of the equilibrium constant, \(K\), when \(30.0\ \text{mol}\) of \( \text{N}_2 \), \(40.0\ \text{mol}\) of \( \text{H}_2 \) and \(6.17\ \text{mol}\) of \( \text{NH}_3 \) are present at equilibrium in a reaction vessel of volume \(55.0\ \text{dm}^3\) at \(450^\circ\text{C}\).
(b) The volume of the reaction container is suddenly reduced while the temperature is kept constant. Sketch a graph showing how the concentration of nitrogen varies with time until a new equilibrium is reached.

Most-appropriate topic codes (IB Chemistry 2025):

Reactivity 3.2: The equilibrium law — part (a)
Reactivity 3.2: Le Châtelier’s principle and changes in conditions — part (b)
▶️ Answer/Explanation

(a)
1. Equilibrium concentrations:
\( [\text{N}_2] = \frac{30.0}{55.0} = 0.545\ \text{mol dm}^{-3} \)
\( [\text{H}_2] = \frac{40.0}{55.0} = 0.727\ \text{mol dm}^{-3} \)
\( [\text{NH}_3] = \frac{6.17}{55.0} = 0.112\ \text{mol dm}^{-3} \)

2. Equilibrium expression:
\( K = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \)

3. Substitution and calculation:
\( K = \frac{(0.112)^2}{(0.545)(0.727)^3} = 0.0599 \)
\(\boxed{0.0599}\)

(b)

• When the volume is decreased, the concentration of \( \text{N}_2 \) increases suddenly due to compression.
• The system then shifts to the right, towards the side with fewer gas molecules, causing \( [\text{N}_2] \) to decrease gradually.
• A new equilibrium is reached where the nitrogen concentration is higher than the original equilibrium value but lower than the immediate post-compression value.
Sharp increase → gradual decrease → new constant level above the initial concentration.

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