Question

The value of $K_C$ for the reaction $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$ is $1\times {10}^{-4}$ At a given time, the composition of the reaction mixture is $\left[HI\right]=2\times {10}^{-5}mol,\left[H_2\right]=1\times {10}^{-5}mol$ and $\left[I_2\right]=1\times {10}^{-5}$ mol In which direction will the reaction proceed?

Answer 1

$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

$K_c=1\times {10}^{-4}$

i.e., at equilibrium, $Q_c=K_c=1\times {10}^{-4}$

Where $Q_c$ is reaction quotient, which is the ratio of concentration of product to the concentration of reactant not necessarily at equilibrium.

$K_c=\left(Conc{.}_{products}\right)x/\left(Conc{.}_{reactants}\right)y$

$K_c=\frac{{10}^{-10}}{4\times {10}^{-10}}=\frac{1}{4}=0.25$

$=2.5\times {10}^{-1}$

At this point, where

$\left[HI\right]=2\times {10}^{-5}$,

$\left[H_2\right]=1\times {10}^{-5}$ ,

$\left[I_2\right]=1\times {10}^{-5}$

$Q_c>K_c$

$\therefore$ The reaction will proceed in backward direction so as to re-attain the equilibrium according to Le Chateliers principle.