Equilibrium Constant (Kc)

The chemical equilibrium is described mathematically in terms of the equilibrium constant, $K_c$. Consider a hypothetical reversible reaction $A+B\rightleftharpoons C+D$: since forward and reverse reactions occur simultaneously, rate equations can be written for each, using the law of mass action -- Rate$_{\text{forward}}=k_f[A][B]$ and Rate$_{\text{reverse}}=k_r[C][D]$. At equilibrium these two rates are equal, so $k_f[A][B]=k_r[C][D]$, which rearranges to $\dfrac{k_f}{k_r}=K_c=\dfrac{[C][D]}{[A][B]}$ -- this ratio, $K_c$, is the equilibrium constant. More generally, for $aA+bB\rightleftharpoons cC+dD$, $K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$. If the same equilibrium is instead written in the reverse sense, $cC+dD\rightleftharpoons aA+bB$, its equilibrium constant $K'_c=\dfrac{[A]^a[B]^b}{[C]^c[D]^d}=\dfrac{1}{K_c}$ -- the equilibrium constant of the reverse reaction is the reciprocal of the forward one. For the Haber-process equilibrium, $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$, $K_c=\dfrac{[NH_3]^2}{[N_2][H_2]^3}$; written in reverse, $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$, $K'_c=\dfrac{[N_2][H_2]^3}{[NH_3]^2}$. Although bot …