Question

For the synthesis of $HI$ (g) from $H_2\left(g\right)$ and $I_2\left(g\right)$ equilibrium constant $K_c$ is $50$. The degree of dissociation of $HI$ is:
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
(Value of $\sqrt{\frac{1}{50}}=0.1414$)

• A. $0.54$
• B. $0.48$
• C. $0.79$
• D. $0.22$

Answer 1

The correct option is D $0.22$

Given, $K_c=50$ for $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
equilibrium constant is reversed for the formation of $HI$ i.e. $K_c=\frac{1}{50}$
$2HI\rightleftharpoons H_2+I_2$
$1-2\alpha$ $\alpha$ $\alpha$ -
$\frac{1}{50}=\frac{{\alpha }^2}{(1-2\alpha {)}^2}$
Taking square root both sides.
$\sqrt{\frac{1}{50}}=\frac{\alpha }{(1-2\alpha )}0.1414=\frac{\alpha }{(1-2\alpha )}0.1414-0.2828\alpha =\alpha \alpha =\frac{0.1414}{1.2828}=0.11$
Hence, degree of dissociation is $0.11$.
According to the balanced chemical reaction, two moles of $HI$ is formed, $\alpha =0.11\times 2=0.22$.