Part 3: Why equilibrium shifts and when K changes

Use the equilibrium expression to distinguish the effect of changing temperature from changing concentration, pressure or catalyst.

A constant can stay fixed while composition changes

At fixed temperature, the equilibrium constant fixes a relationship among the equilibrium quantities, not the value of each concentration separately. If extra reactant is added, the current product-to-reactant expression usually falls below K. Net forward reaction then increases that ratio until the same K is restored.

Calling the current, possibly nonequilibrium ratio Q is a useful explanatory extension. Q < K gives net forward change; Q > K gives net reverse change. At equilibrium Q = K. You can apply the reasoning directly to the expression without introducing an extra symbol in an answer.

For H₂ + I₂ ⇌ 2HI with Kc = 8.00, suppose an instantaneous disturbance gives [HI]²/([H₂][I₂]) = 2.00. More HI forms until that ratio is 8.00 again. It is incorrect to say that the disturbance has made the equilibrium constant 2.00.

Compression changes all gaseous partial pressures together

Consider N₂ + 3H₂ ⇌ 2NH₃. If compression doubles every reacting partial pressure before reaction has adjusted, the numerator of its expression gains a factor of 2² and the denominator a factor of 2⁴. The ratio becomes Kp/4. Net forward reaction restores Kp, giving a greater equilibrium proportion of ammonia.

This explains the rule that compression at constant temperature favours the side with fewer moles of gas. For H₂ + I₂ ⇌ 2HI, the numerator and denominator gain the same factor, so ideal-gas composition does not shift. Count gas coefficients, not solid formula units.

An inert gas at constant volume increases total pressure but leaves each reacting partial pressure nRT/V unchanged, so no equilibrium shift is predicted. Adding inert gas at constant total pressure requires expansion; reacting partial pressures decrease and an equilibrium with unequal gas totals can shift. The constraint matters.

Temperature changes K itself

For a forward exothermic reaction, heating decreases K. At the new temperature, the previous composition now gives too large a product-to-reactant ratio relative to the smaller K, so net reverse reaction occurs. Cooling raises K and favours products.

For a forward endothermic reaction, heating increases K and the equilibrium composition adjusts toward products. Explain both changes: the new temperature changes the constant and the composition then satisfies that new value. Increasing temperature may accelerate both directions even when equilibrium product yield falls; rate and equilibrium yield answer different questions.

Suppose A ⇌ B is endothermic, with Kc = [B]/[A] increasing from 2.0 to 5.0 on heating. In a mixture with conserved [A] + [B] = 0.600 mol dm⁻³, at the first temperature [A] = 0.600/3 = 0.200 and [B] = 0.400. At the higher temperature [A] = 0.600/6 = 0.100 and [B] = 0.500 mol dm⁻³. The total is unchanged; the ratio changes.

Catalysts alter the approach, not the target

A catalyst provides an alternative reaction pathway with lower activation barriers for both directions. It does not change the reactant and product thermodynamic states, so it does not change K or the final equilibrium composition at fixed temperature. A catalysed mixture may contain more product after a short fixed time because it approaches equilibrium sooner.

For an exothermic industrial reaction, lower temperature can improve equilibrium yield while slowing the reaction. A catalyst allows a useful rate at a more favourable temperature. Increased pressure can improve yield where gas amount decreases, but equipment strength, compression energy, safety and cost still matter. Product separation and recycling improve overall conversion; they do not alter the stoichiometric atom economy of the reaction.

For a “predict and explain” response, state the written reaction direction, the condition changed, whether K changes, and the consequent composition change. Avoid saying “K moves right”: a numerical constant increases, decreases or stays fixed.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources