• # question_answer 15 moles of ${{H}_{2}}$ and 5.2 moles of ${{I}_{2}}$ are mixed and allowed to attain equilibrium at ${{500}^{o}}C$. At equilibrium, the concentration of $HI$ is found to be 10 moles. The equilbrium constant for the formation of $HI$ is  A) 50 B) 15 C) 100 D) 25

 [a]     ${{H}_{2}}$    +    ${{I}_{2}}$     $\rightleftharpoons$    $2HI$ 15             5.2               0 (155)      (5.25)         10 ${{K}_{C}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{10\times 10}{10\times 0.2}=50$