Explain dynamic chemical equilibrium and predict the effect of changes in concentration, pressure and temperature using Le Chatelier's principle

What this dot point is asking

A reversible reaction does not go to completion. In a closed system the forward reaction and the reverse reaction both keep happening, and eventually they happen at the same rate. At that point the concentrations of reactants and products stop changing, even though both reactions are still occurring. This is dynamic equilibrium: dynamic because the reactions never stop, equilibrium because the macroscopic properties (concentration, colour, pressure) are constant.

Three things must be true for equilibrium. The system must be closed (no matter enters or leaves), the temperature must be constant, and the reaction must be reversible, written with the equilibrium arrow as in $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. At equilibrium the forward rate equals the reverse rate, not the concentrations. Reactant and product concentrations are usually different from each other.

Le Chatelier's principle

Le Chatelier's principle predicts which way a system at equilibrium shifts when it is disturbed: the system responds in the direction that partially counteracts the imposed change. Use it to predict the qualitative effect of three changes.

Concentration

Adding a reactant shifts the position of equilibrium towards the products (the system consumes some of the added reactant). Removing a product also shifts towards products. Adding a product, or removing a reactant, shifts back towards reactants. Importantly, changing a concentration does not change $K_c$.

Pressure (for gases)

Increasing the pressure by decreasing the volume shifts equilibrium towards the side with fewer moles of gas, because that reduces the total pressure. Decreasing pressure shifts towards the side with more moles of gas. If both sides have equal moles of gas, a pressure change has no effect on position. Adding an inert gas at constant volume does not shift the equilibrium because it does not change the partial pressures of the reacting gases.

Temperature

This is the only change that alters the value of $K_c$. Treat heat as a reactant or product using the sign of $\Delta H$. For an exothermic forward reaction ($\Delta H$ negative), heat is a product, so increasing temperature shifts equilibrium towards reactants and decreases $K_c$. For an endothermic forward reaction, increasing temperature shifts towards products and increases $K_c$.

Catalyst

A catalyst speeds up the forward and reverse reactions equally. It makes equilibrium reached faster but does not shift the position and does not change $K_c$ or yield.

Worked example: the Haber process

Consider $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, $\Delta H = -92\ \text{kJ mol}^{-1}$.

Predict the effect of each change on the yield of ammonia.

1. Increase pressure. The left side has 4 mol of gas, the right has 2 mol. Higher pressure shifts towards fewer moles, so towards products. Yield of $\text{NH}_3$ increases.
2. Increase temperature. The forward reaction is exothermic, so heat behaves as a product. Adding heat shifts towards reactants. Yield decreases and $K_c$ decreases.
3. Remove ammonia as it forms. Removing a product shifts equilibrium towards products. Yield increases.
4. Add a catalyst. No shift in position. Yield is unchanged, but equilibrium is reached faster.

This explains the real industrial compromise: high pressure favours yield, but low temperature (which favours yield) makes the reaction too slow, so a moderate temperature plus a catalyst is used and ammonia is removed continuously.

When you answer these questions in the WACE examination, always state the direction of the shift and link it explicitly to the change. A sentence such as "the equilibrium shifts to the right because the system opposes the increase in reactant concentration" earns the reasoning marks; just writing "right" does not.