The Law of Mass Action and the Equilibrium Constant $K_c$

While the idea of "equal opposing rates" explains why a reaction stops changing, it does not by

itself say what the actual equilibrium concentrations will be. That quantitative link was supplied by

Cato Guldberg and Peter Waage through the Law of Mass Action: at a given temperature, the rate of

a chemical reaction is directly proportional to the product of the "active masses" (in practice,

molar concentrations) of the reacting substances, each raised to a power equal to its stoichiometric

coefficient in the balanced equation.

Applying this law to both the forward and reverse directions of a general reversible reaction

$aA + bB \rightleftharpoons cC + dD$ gives a forward rate $r_f = k_f[A]^a[B]^b$ and a reverse rate

$r_r = k_r[C]^c[D]^d$. At equilibrium these two rates are equal, $r_f = r_r$, so

$k_f[A]^a[B]^b = k_r[C]^c[D]^d$. Rearranging,

The ratio of the two rate constants is itself a constant at a given temperature, called the

equilibrium constant, $K_c$ (the subscript $c$ signals that it is expressed in terms of molar

concentrations). This expression — products raised to their coefficients, divided by reactants raised

to their coefficients — is known as the Law of Chemical Equilibrium, and it holds for any

reversible reaction regardless of the mechanism by which equilibrium is actually reached (the

rate-based derivation above is a useful motivation, but $K_c$'s value depends only on the balanced

equation and the temperature, never on the pathway).

For example, for the industrially crucial ammonia synthesis,

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, the equilibrium-constant

expression is

A large value of $K_c$ (much greater than 1) signals that, at equilibrium, products dominate over

reactants — the reaction proceeds nearly to completion. A small value of $K_c$ (much less than 1)

signals the opposite — very little product forms before equilibrium is reached. $K_c$ therefore gives …