Question

If one-third of $HI$ decomposes at a particular temperature, $K_c$ for the equilibrium is:

$HI\rightleftharpoons H_2+I_2$

• A. 1/16
• B. 1/4
• C. 1/6
• D. 1/2

Answer 1

Answer: (B)

1/4

Given 1/3 $HI$ decomposes to give 1/6 of $H_2$ and $I_2$ each.
$2HI\rightleftharpoons H_2+I_2$
At equilibrium:
$K_c=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$
$\to K_c=\frac{\frac{1}{3}\times \frac{1}{3}}{\frac{2}{3}\times \frac{2}{3}}$

$K_c=\frac{1}{4}$