Haber Equilibrium: Three Proposed Changes That Do Not All Work
In a rigid 2.0 dm³ steel vessel held at 700 K, an equilibrium mixture contains 0.50 mol N₂, 1.00 mol H₂ and 1.00 mol NH₃ according to N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Under these conditions Kc = 8.0 when concentrations are expressed in mol dm⁻³. A technician proposes three changes to increase the equilibrium amount of ammonia: (i) raise the temperature to 800 K; (ii) inject 0.50 mol of argon at constant volume; (iii) inject 0.50 mol of argon at constant pressure, letting the vessel expand at 700 K. A supervisor approves all three on the grounds that each raises the collision rate or the total pressure inside the vessel.
1. Verify Kc from the stated composition, then calculate the reaction quotient after 0.50 mol of N₂ is added and 0.50 mol of NH₃ is removed at constant volume, and predict the direction of the net change.
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Check Kc. Concentrations: [N₂] = 0.50/2.0 = 0.25; [H₂] = 1.00/2.0 = 0.50; [NH₃] = 1.00/2.0 = 0.50 mol dm⁻³. Kc = [NH₃]²/([N₂][H₂]³) = (0.50)²/(0.25 × 0.50³) = 0.25/(0.25 × 0.125) = 0.25/0.03125 = 8.0, consistent with the stated value.
After the changes (volume still 2.0 dm³): N₂ = 1.00 mol → 0.50; H₂ = 1.00 mol → 0.50; NH₃ = 0.50 mol → 0.25 mol dm⁻³. Q = (0.25)²/(0.50 × 0.50³) = 0.0625/(0.50 × 0.125) = 0.0625/0.0625 = 1.0.
Prediction. Q < Kc (1.0 < 8.0), so the net change is forward, producing more NH₃, until Q rises to 8.0. Removing product lowers Q by a factor of 4 here, which outweighs the effect of adding reactant, so the mixture does not simply 'settle back' at the old composition.
2. Assess proposal (i), raising the temperature to 800 K, including what happens to Kc and to the rate at which equilibrium is reached.
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The forward reaction is exothermic (ΔH = −92 kJ mol⁻¹), so the reverse reaction is endothermic. Raising the temperature drives the equilibrium in the endothermic direction, i.e. backwards, so the equilibrium amount of NH₃ falls and Kc decreases (for an exothermic reaction Kc falls as T rises; the van 't Hoff relation gives this without needing a new Kc value).
It is true that the rate constants for both directions increase and that equilibrium is reached faster - that is the supervisor's valid point. But rate and yield are different quantities: a higher temperature trades yield for speed.
So proposal (i) fails its stated aim. The supervisor's argument that 'a higher collision rate means more ammonia' confuses kinetics with thermodynamics: the position of equilibrium depends only on temperature here, not on how quickly it is attained. Industrially, 700 K is already a compromise between yield (favoured by low T) and rate (favoured by high T).
3. Assess proposals (ii) and (iii), explaining any difference between adding argon at constant volume and at constant pressure.
Read the worked answer
(ii) Argon at constant volume: no effect. Argon is inert and does not appear in Q. Because the vessel volume is fixed, the concentrations (and partial pressures) of N₂, H₂ and NH₃ are unchanged by adding argon, so Q is still 8.0 = Kc. Total pressure rises, but total pressure is not the variable in the equilibrium expression - only the partial pressures of the species in the reaction matter. No shift occurs.
(iii) Argon at constant pressure: shifts backwards. To keep the pressure constant the vessel must expand, so every species is diluted and all partial pressures fall by the same factor α < 1. Writing Q_p = p(NH₃)²/(p(N₂)p(H₂)³), the numerator scales as α² and the denominator as α⁴, so Q scales as α⁻² > 1: Q becomes greater than K. The system responds by shifting towards the side with more gaseous moles, the reactants (left), and the NH₃ amount falls. This is the same conclusion Le Chatelier's principle gives for dilution of a reaction with Δn_gas = −2.
So of the three proposals, none increases the ammonia yield: (i) decreases it, (ii) does nothing, (iii) decreases it. Only compressing the mixture or removing NH₃ would help.