Q. Equivalent amounts of $H_{2}\ and\ I_{2}$ are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to $H,\ the\ K_{c}$ at this temperature is
A) 64
B) 16
C) 0.25
D) 4
Explanation
(a)
$H_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ + \ \ \ I_{2}\ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ \ 2HI$
1 1 0
(1-0.8) (1-0.8) $2 \times 0.8$
=0.2 =0.2 =1.6
$K_{c} = \frac{\lbrack HI\rbrack^{2}}{\left\lbrack H_{2} \right\rbrack\left\lbrack I_{2} \right\rbrack}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1.6 \times 1.6}{0.2 \times 0.2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{c} = 64$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.