Chemical systems rarely start at equilibrium. By comparing the reaction quotient (Q) to the equilibrium constant (K), you can predict the net direction a reversible reaction will proceed to re-establish equilibrium.

Core idea: Q vs K predicts the net change

What Q and K represent

Q is a “snapshot” of the system right now.

This figure illustrates a reversible system approaching equilibrium using $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, showing the mixture’s macroscopic change alongside a concentration-versus-time plot. The leveling of the curves represents the point where the composition stops changing (equilibrium), even though microscopic forward and reverse processes continue. This supports the idea that $Q$ describes the current composition, while equilibrium corresponds to the constant ratio defined by $K$ at that temperature. Source

To decide what happens next, compare that snapshot to the target ratio set by K.

K is not a measure of rate or time; it is the equilibrium position expressed as a ratio.

Mathematical form (why Q and K are directly comparable)

For gas-phase systems you may see $Q_p$ (using partial pressures); the comparison logic with K is identical as long as you compare like with like (i.e., $Q_c$ with $K_c$, or $Q_p$ with $K_p$).

Using Q vs K to decide shift direction

The three possible comparisons

This chart summarizes the three possible relationships between the reaction quotient and the equilibrium constant: $Q<K$, $Q=K$, and $Q>K$. It pairs each comparison with the corresponding net reaction direction needed to re-establish equilibrium (shift toward products or reactants). Source

• If $Q < K$: the mixture has too many reactants (or too few products) relative to equilibrium.

  • Net change: reaction proceeds forward overall to form more products until Q increases to K.

• If $Q > K$: the mixture has too many products (or too few reactants) relative to equilibrium.

  • Net change: reaction proceeds in reverse overall to form more reactants until Q decreases to K.

• If $Q = K$: the system is already at equilibrium.

  • Net change: no net shift (forward and reverse still occur, but cancel macroscopically).

These statements are exactly the syllabus requirement: Q less than K → forward; Q greater than K → reverse.

What “shifts” means (and what it does not mean)

• A predicted “shift” is a statement about net reaction progress, not about stopping one direction.

• When shifting:

  • Reactant concentrations/partial pressures decrease in a net forward shift and increase in a net reverse shift.

  • Product concentrations/partial pressures increase in a net forward shift and decrease in a net reverse shift.

• The system stops shifting when the ratio returns to Q = K.

Quick checks and common pitfalls

• Check you used the correct K: use $K_c$ with concentrations and $K_p$ with partial pressures.

• Use a balanced equation: exponents come from stoichiometric coefficients; the direction decision can be wrong if the reaction is not properly balanced.

• Interpret inequality carefully: “$Q$ smaller” means “ratio is smaller,” not “all concentrations are smaller.”

• Zero/very small terms: if the numerator is (effectively) zero, then Q is very small, so you should expect a forward net reaction to make products.