The equilibrium expression can be evaluated even when a reaction mixture is not at equilibrium. The resulting quantity is the reaction quotient, $Q$.

For

$$a\mathrm A+b\mathrm B\rightleftharpoons c\mathrm C+d\mathrm D,$$

using the current concentrations gives

$$Q_c=\frac{[\mathrm C]^c[\mathrm D]^d}{[\mathrm A]^a[\mathrm B]^b}.$$

The algebraic form is the same as for $K_c$; the difference is conceptual. $Q_c$ describes the mixture now, while $K_c$ is the value that $Q_c$ has at equilibrium at the stated temperature.

Comparing them predicts the direction of net change:

• if $Q_c<K_c$, the mixture has too little product relative to equilibrium, so net reaction proceeds forward;
• if $Q_c>K_c$, it has too much product relative to equilibrium, so net reaction proceeds in reverse;
• if $Q_c=K_c$, the system is at equilibrium.

For example, suppose

$$\mathrm{A\rightleftharpoons B}$$

has $K_c=4.0$. If the current concentrations are

$$[\mathrm A]=0.50,\mathrm M,\qquad [\mathrm B]=0.50,\mathrm M,$$

then

$$Q_c=\frac{[\mathrm B]}{[\mathrm A]}=1.0.$$

Because $Q_c<K_c$, the system undergoes net forward reaction: A is consumed and B is produced until the quotient rises to 4.0.

If instead $[\mathrm A]=0.10,\mathrm M$ and $[\mathrm B]=0.80,\mathrm M$, then

$$Q_c=8.0>K_c,$$

so the net change is toward A.

The comparison $Q$ versus $K$ is more precise than memorizing that an equilibrium “shifts left” or “shifts right.” It identifies the direction required for the composition to recover the equilibrium relation after the mixture is prepared or disturbed.