Effect of concentration

At equilibrium, the concentrations of reactants and products stay fixed. Adding more of a reactant or a product from outside disturbs this balance by raising that species' concentration -- and, according to Le Chatelier's principle, the system responds by shifting the equilibrium in the direction that consumes the added substance.

Consider $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$. Adding extra $H_2$ or $I_2$ disturbs the equilibrium; to minimise the stress, the system shifts the reaction in the direction that consumes the added $H_2$ and $I_2$ -- that is, more HI forms, and the equilibrium shifts to the right (forward direction), continuing until equilibrium is re-established. Removing HI (a product) likewise favours the forward reaction, for the same reason. Conversely, adding HI to the mixture raises $[HI]$, and the system responds by shifting in the reverse direction to consume the excess HI.

Worked example. At equilibrium, $[HI]=1$, $[H_2]=0.2$, $[I_2]=0.1$ M, so $K_C = \dfrac{1^2}{(0.2)(0.1)} = 50$. The equilibrium is disturbed by adding 0.1 M iodine, so immediately after the addition $[I_2]=0.2$ M while $[H_2]$ and $[HI]$ are momentarily unchanged. The system then shifts by some amount x: $[H_2]$ and $[I_2]$ both fall by x (to $0.2-x$), and $[HI]$ rises by 2x (to $1+2x$). The observed new $[HI] = 1.092$ M gives $2x = 0.092$, so $x = 0.046$ M, and therefore $[H_2]=[I_2]=0.2-0.046=0.154$ M. Checking the reaction quotient at this new state, $Q = \dfrac{(1.092)^2}{(0.154)^2} \approx 50$, which equals $K_C$ -- confirming the system has re-established equilibrium, with the added iodine driving the formation of additional HI (Fig. 8.5). …