$PCl_5$,$PCl_3$,and $Cl_2$ are at equilibrium at $500 \ K$ with concentrations $[PCl_3] = 1.59 \ M$,$[Cl_2] = 1.59 \ M$,and $[PCl_5] = 1.41 \ M$. Calculate $K_c$ for the reaction:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

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(1.79) The equilibrium constant expression for the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is given by:
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
Substituting the given equilibrium concentrations into the expression:
$K_c = \frac{(1.59 \ M)(1.59 \ M)}{1.41 \ M}$
$K_c = \frac{2.5281}{1.41} \approx 1.79$

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