Question

4 mol of $HI$ is taken in a lit closed vessel and heated to equilibrium. At equilibrium the concentration of $H_2$ is one mol ${lit}^{-1}$. The equilibrium constant is:

• A. $4$
• B. $0.5$
• C. $2$
• D. $0.25$

Answer 1

The correct option is D $0.25$

$2HI\rightleftharpoons H_2+I_2$

$4$ $-$ $-$

$4-2x$ $x$ $x$ given $x=1$

$K_C=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}=\frac{1\times 1}{4}=0.25$