Question

$K_c$ for the synthesis of HI(g) from $H_2\left(g\right)$ and $I_2\left(g\right)$ is 50. The degree of dissociation of HI is:

• A. 0.1
• B. 0.14
• C. 0.22
• D. 0.93

Answer 1

Answer: (D)

0.93

The equilibrium reaction for the dissociation of HI is
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
$(2-2\alpha )\alpha \alpha$
Its equilibrium constant is reverse of the equilibrium constant for the formation of ammonia.
$50=\frac{{\alpha }^2}{(2-2\alpha {)}^2}$
Hence, $\alpha =0.93$