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Q. $PCl_5$, $PCl_3$ and $Cl_2$ are at equilibrium at $500\,K$ with concentration $2.1\,M\, PCl_3$, $2.1\,M \,Cl_2$ and $1.9 \,M \,PCl_5$. The equilibrium constant for the given reaction is
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

Equilibrium

Solution:

$K_{c}=\frac{\left[PCl_{3}\right]\left[Cl_{2}\right]}{\left[PCl_{5}\right]}$
$=\frac{2.1\times2.1}{1.9}$
$=2.32$