Question

At certain temperature, 50% of $HI$ is dissociated into $H_2$ and $I_2$. Then the equilibrium constant is:

• A. 1
• B. 3
• C. 0.5
• D. 0.25

Answer 1

The correct option is D 0.25

Let intial $\left[HI\right]=2X$ M. 50% of its decomposes. Hence, $2X\times \frac{50}{100}=X$ M will decompose.

	$HI$	$H_2$	$I_2$
Initial concentration (M)	2X	0	0
Change in concentration (M)	-X	0.5X	0.5X
Equilibrium concentration (M)	$2X-X=X$	0.5X	0.5X

$K=\frac{\left[H_2\right]\left[I_2\right]}{\left[HI\right]}=\frac{0.5X\times 0.5X}{X^2}=0.25$