In a closed vessel,$PCl_{5(g)}$ is obtained by the chemical reaction between $PCl_{3(g)}$ and $Cl_{2(g)}$. If the equilibrium concentrations in this vessel of $PCl_3$,$Cl_2$,and $PCl_5$ at $500 \ K$ are $1.59 \ M$,$1.59 \ M$,and $1.41 \ M$ respectively,then find the equilibrium constant $K_c$ for the reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

(A) The equilibrium constant $K_c$ for the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$ is given by the expression:
$K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]}$
Given equilibrium concentrations:
$[PCl_3] = 1.59 \ M$
$[Cl_2] = 1.59 \ M$
$[PCl_5] = 1.41 \ M$
Substituting these values into the expression:
$K_c = \frac{1.41}{1.59 \times 1.59} = \frac{1.41}{2.5281} \approx 0.558 \ M^{-1}$