Question

At a certain temperature, only 50% $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is:

• A. 1
• B. 3
• C. 0.5
• D. 0.25

Answer 1

The correct option is A 1

$2HI\rightleftharpoons H_2+I_2$

$t=0$ $N$ $-$ $-$

$t=eq$ $N-2y$ $y$ $y$

Given $N-2y=50$% of $N=\frac{N}{2}$

$y=N/4$

$K_C=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}=\frac{{\left(N/4\right)}^2{\left(\frac{N}{4}\right)}^2}{{\left(\frac{N}{2}\right)}^2}=\frac{4}{16}=0.25$