Question

$HI$ was heated in a sealed tube at ${440}^{o}C$ till the equilibrium was reached. $HI$ was found to be $22\%$ decomposed. The equilibrium constant for the dissociation of $HI$ is: $[2HI\rightleftharpoons H_2+I_2]$

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

Answer 1

Answer: (B)

$0.0796$

$2HI\stackrel{K}{\rightleftharpoons }{H}_2+I_2$

At $t=0$ $100$ $0$ $0$

At equilibrium $100-22=78$ $22$ $22$

$K=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}=\frac{22\times 22}{{78}^2}$

$\Rightarrow K=0.0796$