Question

$HI$ was heated in a sealed tube at ${400}^{o}C$ till the equilibrium was reached. $HI$ was found to be $22$% decomposed. The equilibrium constant for dissociation is:

• A. $1.99$
• B. $0.0199$
• C. $0.0796$
• D. $0.282$

Answer 1

Answer: (B)

$0.0199$

$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

Suppose, initially 2 moles of HI are present. $22\%$ or $2\times \frac{22}{100}=0.44$ moles will dissociate. $2-0.44=1.56$ moles of HI will be present.

$0.44$ moles of HI will form $0.44\times \frac{1}{2}=0.22$ moles of $H_2$ and $0.22$ moles of $I_2$.

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

$K_C=\frac{\left(0.22/V\right)\times \left(0.22/V\right)}{(1.56/V{)}^2}$

$K_C=0.0199$