Q. 15 moles of $H_{2}\ and\ 5.2\ moles\ of\ I_{2}$ are mixed and allowed to attain equilibrium at 500$℃$. At equilibrium, the concentration of $HI$ is found to be 10 moles. The equilibrium constant for the formation of $HI$ is

A) 50
B) 15
C) 100
D) 25
Explanation

(a)
$H_{2}\ \ \ \ \ \ + \ \ \ \ \ \ \ I_{2}\ \ \ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ 2HI$
15 5.2 0
(15-5) (5.2-5) 10
Equilibrium constant
$\left( K_{c} \right) = \frac{\lbrack HI\rbrack^{2}}{\left\lbrack H_{2} \right\rbrack\lbrack I_{2}\rbrack} = \frac{10 \times 10}{10 \times 0.2} = 50$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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