Chemistry - Chemical Equilibrium Question with Solution | TestHub

$2\mathrm{ICl}\rightleftharpoons {\mathrm{I}}_2+{\mathrm{Cl}}_2$ Initial $0.600$ At eqm : $0.6-2xxx$ $\begin{array}{rl}& K_c=0.14=\frac{\left[{\mathrm{I}}_2\right]\left[{\mathrm{Cl}}_2\right]}{[\mathrm{ICl}{]}^2}=\frac{x^2}{(0.6-2x{)}^2}\\ & 0.37=\frac{x}{0.6-2x}\Rightarrow 0.222-0.74x=x\\ & \Rightarrow 1.74x=0.222\therefore x=0.128\mathrm{M}\end{array}$