Predicting the direction of a reaction

Knowing $K_C$ tells us where equilibrium lies, but a reaction mixture is not always sitting at equilibrium. To find out which direction a given (non-equilibrium) mixture still needs to move, define the reaction quotient, Q, for a general reaction $xA + yB \rightleftharpoons lC + mD$ as

Q has exactly the same algebraic form as $K_C$, evaluated with whatever concentrations of reactants and products happen to be present at that instant -- concentrations that need not be the equilibrium ones. As the reaction proceeds, the concentrations of reactants and products keep changing, so Q keeps changing too, right up until the reaction reaches equilibrium -- at which point Q becomes numerically equal to $K_C$ and, once equilibrium is reached, stops changing any further. Comparing Q with $K_C$ at any instant therefore predicts the direction in which the reaction will move:

• If $Q = K_C$, the reaction is already at equilibrium (no further net change).
• If $Q > K_C$, the reaction proceeds in the reverse direction, i.e. toward the formation of reactants.
• If $Q < K_C$, the reaction proceeds in the forward direction, i.e. toward the formation of products.

Example 1. For $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$, $K_C = 48$ at 717 K. At a particular instant, $[H_2]=0.2$, $[I_2]=0.2$, $[HI]=0.6$ mol L$^{-1}$. $Q = \dfrac{(0.6)^2}{(0.2)(0.2)} = 9$. Since $Q(9) < K_C(48)$, the reaction proceeds in the forward direction.

Example 2. For $N_2O_4(g)\rightleftharpoons 2NO_2(g)$, $K_C = 0.21$ at 373 K. At a given time, $[N_2O_4]=0.125$, $[NO_2]=0.5$ mol dm$^{-3}$. $Q = \dfrac{(0.5)^2}{0.125} = 2$. Since $Q(2) > K_C(0.21)$, the reaction proceeds in the reverse direction, until Q falls back to 0.21. …