Q. On a given condition, the equilibrium concentration of $HI,\ H_{2}\ and\ I_{2}$are 0.80, 0.10 and 0.10 mol/L. The equilibrium constant for the reaction, $H_{2} + I_{2} \rightleftharpoons 2HI$, will be
A) 8
B) 16
C) 32
D) 64
Explanation
(d)
$H_{2} + I_{2} \rightleftharpoons 2HI$
$\lbrack HI\rbrack = 0.80,\ \left\lbrack H_{2} \right\rbrack = 0.10,\ \left\lbrack I_{2} \right\rbrack = 0.10$
$K_{c} = \frac{\lbrack HI\rbrack^{2}}{\left\lbrack H_{2} \right\rbrack\left\lbrack I_{2} \right\rbrack} = \frac{0.80 \times 0.80}{0.10 \times 0.10} = 64$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.