Question

If one-third of HI decomposes at a particular temperature $K_c$ for the reaction : $2HI\rightleftharpoons H_2+I_2$ is:

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{4}$
• D. $\frac{1}{16}$

Answer 1

The correct option is D

$\frac{1}{16}$

$\alpha =\frac{1}{3}$

$\begin{array}{cccccc}& & & & & \\ & 2HI& \rightleftharpoons & H_2& +& I_2\\ \text{Initial:}& \text{1 mole}& & 0& & 0\\ \text{at equilibrium}& 1(1-\alpha )& & \frac{\alpha }{2}& & \frac{\alpha }{2}\\ \text{at equilibrium}& 1-\frac{1}{3}=\frac{2}{3}& & \frac{1}{6}& & \frac{1}{6}\end{array}$

$\therefore K_c=\frac{\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)}{{\left(\frac{2}{3}\right)}^2}=\frac{1}{16}$