Q. In the reaction, $H_{2} + I_{2} \rightleftharpoons 2HI$ In a 2 I flask 0.4 moles of each $H_{2}\ and\ I_{2}$ are taken. At equilibrium 0.5 moles of $HI$ are formed. What will be the value of equilibrium constant $K_{c}$?
A) 20.2
B) 25.4
C) 0.284
D) 11.1
Explanation
(d)
$H_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ + I_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \rightleftharpoons 2HI$
Initial 0.4 0.4 0
At equilibrium 0.4-0.25 0.4-0.25 0.05
=0.15 =0.15
$K_{c} = \frac{\lbrack HI\rbrack^{2}}{\left\lbrack H_{2} \right\rbrack\lbrack I_{2}\rbrack}$
$= \frac{\left( \frac{0.50}{2} \right)^{2}}{\left( \frac{0.15}{2} \right)\left( \frac{0.15}{2} \right)} = \frac{0.5 \times 0.5}{0.15 \times 0.15} = 11.11$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.