Reaction quotient Q and predicting direction: WACE Year 12 Chemistry

What this dot point is asking

The equilibrium constant $K_c$ describes the ratio of product to reactant concentrations only when a system is at equilibrium. But a real mixture may not start at equilibrium. The reaction quotient $Q$ lets you test where a system sits relative to equilibrium and predict the direction of net change.

For the general reaction

both have the form

The difference is that $K_c$ uses equilibrium concentrations and $Q$ uses the actual concentrations at the moment you are interested in.

Comparing Q with Kc

The system always shifts in the direction that brings $Q$ towards $K_c$.

• $Q < K_c$: there are too few products relative to reactants. The forward reaction is favoured; the reaction proceeds to the right, making more product until $Q$ rises to equal $K_c$.
• $Q > K_c$: there are too many products. The reverse reaction is favoured; the reaction proceeds to the left, consuming product until $Q$ falls to equal $K_c$.
• $Q = K_c$: the system is already at equilibrium and there is no net change.

Why this is useful

The reaction quotient lets you do three things the equilibrium constant alone cannot. First, predict the direction of reaction for any starting mixture. Second, decide whether adding more of a substance will cause net forward or reverse change. Third, it underpins the solubility comparison ($Q$ versus $K_{sp}$) used to predict precipitation, so the same logic carries across the whole equilibrium topic.

Handling gases, solids and the units

When the question gives amounts in moles, always divide by the volume to get concentrations before substituting, exactly as for $K_c$ itself; the only time you can skip this is when the total moles of gas are equal on both sides, so the volume cancels. Pure solids and pure liquids are left out of $Q$ just as they are left out of $K_c$, because their effective concentration is constant. For a heterogeneous equilibrium such as $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, the quotient is simply $Q = [\text{CO}_2]$, and comparing it with $K_c$ predicts whether more carbonate decomposes or more carbon dioxide is reabsorbed.

Q for solubility: a preview

The same comparison drives precipitation prediction in the solubility topic. There the quotient is called the ionic product, written with the same ion-concentration form as $K_{sp}$. If the ionic product exceeds $K_{sp}$ the solution is supersaturated and a precipitate forms (the system shifts to consume ions); if it is below $K_{sp}$ no precipitate forms. Recognising that "$Q$ versus $K$" is a single unifying idea across gaseous equilibria, acid ionisation and solubility is one of the higher-order links WACE rewards.

Connecting to Le Chatelier

The reaction quotient gives a quantitative version of Le Chatelier's principle. When you add reactant, $Q$ drops below $K_c$, so the system shifts forward; when you add product, $Q$ rises above $K_c$, so it shifts in reverse. The two ideas describe the same behaviour, but $Q$ lets you put numbers to the prediction.