MCQ
At a given temperature, the equilibrium constant for reaction $PC{l_5}_{(g)}$ $\rightleftharpoons$ $PC{l_3}_{(g)} + C{l_2}_{(g)}$ is $2.4 \times {10^{ - 3}}$. At the same temperature, the equilibrium constant for reaction $PC{l_3}_{(g)} + C{l_2}_{(g)}$ $\rightleftharpoons$ $PC{l_5}_{(g)}$ is
  • A
    $2.4 \times {10^{ - 3}}$
  • B
    $ - 2.4 \times {10^{ - 3}}$
  • $4.2 \times {10^2}$
  • D
    $4.8 \times {10^{ - 2}}$

Answer

Correct option: C.
$4.2 \times {10^2}$
(c) Reaction is reversed. Hence

$K = \frac{1}{{(2.4 \times {{10}^{ - 3}})}} = 4.2 \times {10^2}$

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