Explain dynamic equilibrium, write equilibrium constant expressions, and predict shifts using Le Chatelier's principle.

What this dot point is asking

TASC expects you to describe equilibrium at the particle level, write and use the equilibrium constant expression, compare $Q$ with $K_c$, and predict shifts when a system is disturbed.

Dynamic equilibrium

A reversible reaction can proceed forwards and backwards. Early on the forward rate is high because reactant concentration is high. As products build up, the reverse rate rises. Eventually the two rates become equal and the concentrations of all species stop changing. This is dynamic equilibrium: both reactions still occur, but there is no net change. It is reached only in a closed system and can be approached from either direction.

The equilibrium constant

For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:

The reaction quotient Q

$Q$ has the same form as $K_c$ but uses concentrations at any instant. Comparing $Q$ with $K_c$ predicts the direction of net reaction:

• $Q < K_c$: too few products, so the forward reaction is favoured.
• $Q > K_c$: too many products, so the reverse reaction is favoured.
• $Q = K_c$: the system is at equilibrium.

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts to partly counteract the disturbance:

• Concentration: adding a reactant (or removing a product) shifts the position right; removing a reactant shifts it left.
• Pressure (gases): increasing pressure (decreasing volume) shifts towards the side with fewer gas moles; adding inert gas at constant volume has no effect because partial pressures of the reacting gases are unchanged.
• Temperature: heating favours the endothermic direction. For an exothermic forward reaction ($\Delta H < 0$), heating shifts the system left and lowers $K_c$.
• Catalyst: speeds up forward and reverse rates equally, so equilibrium is reached faster but the position and $K_c$ are unchanged.

Recognising equilibrium experimentally

You cannot see equilibrium directly, but several observations point to it. Macroscopic properties such as colour, pressure or pH become constant and stay constant in a closed system. The reaction can be approached from either direction and reaches the same final state for given conditions. A classic demonstration uses the brown gas $\text{NO}_2$ in equilibrium with colourless $\text{N}_2\text{O}_4$: warming the tube deepens the brown colour (favouring the endothermic dissociation to $\text{NO}_2$) and cooling lightens it, a visible Le Chatelier shift.

The industrial compromise

Real processes balance yield against rate and cost. The Haber process for ammonia is exothermic and goes from more to fewer gas moles, so high yield favours low temperature and high pressure. But a low temperature makes the rate impractically slow, so industry uses a moderate temperature (around $400\ ^\circ\text{C}$ to $450\ ^\circ\text{C}$), a high pressure, an iron catalyst, and continuous removal of ammonia to keep shifting the position forward. This compromise sacrifices some equilibrium yield to achieve an economical rate, a pattern repeated across industrial chemistry.

When you answer a Le Chatelier question, always state the change, the direction of the shift, and the effect on the named species or on $K_c$. Markers reward that three-part structure.