MCQ
For the reaction ${H_2}_{(g)} + {I_2}_{(g)}$ $\rightleftharpoons$ $2HI_{(g)}$  at $721\,K$ the value of equilibrium constant $({K_c})$ is $50$. When the equilibrium concentration of both is $0.5\,M$, the value of ${K_p}$ under the same conditions will be
  • A
    $0.002$
  • B
    $0.2$
  • $50$
  • D
    $50/RT$

Answer

Correct option: C.
$50$
(c) For the reaction ${H_2} + {I_2}$ $ \rightleftharpoons $  $2HI$

$\Delta n = 0$    So ${K_p} = {K_c}$ $\therefore 50.0$

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