$3.00 \, mol$ of $PCl_5$ kept in $1 \, L$ closed reaction vessel was allowed to attain equilibrium at $380 \, K$. Calculate the composition of the mixture at equilibrium. Given $K_c = 1.80$.

(N/A) The equilibrium reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$
Initial concentrations: $[PCl_5] = 3.0 \, M$,$[PCl_3] = 0 \, M$,$[Cl_2] = 0 \, M$
Let $x \, mol/L$ be the amount of $PCl_5$ dissociated at equilibrium.
Equilibrium concentrations: $[PCl_5] = (3.0 - x) \, M$,$[PCl_3] = x \, M$,$[Cl_2] = x \, M$
The equilibrium constant expression is: $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
Substituting the values: $1.8 = \frac{x^2}{3.0 - x}$
Rearranging into a quadratic equation: $x^2 + 1.8x - 5.4 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-1.8 \pm \sqrt{(1.8)^2 - 4(1)(-5.4)}}{2(1)}$
$x = \frac{-1.8 \pm \sqrt{3.24 + 21.6}}{2} = \frac{-1.8 \pm \sqrt{24.84}}{2} \approx \frac{-1.8 \pm 4.98}{2}$
Since $x$ must be positive: $x = \frac{3.18}{2} = 1.59 \, M$
Equilibrium composition:
$[PCl_5] = 3.0 - 1.59 = 1.41 \, M$
$[PCl_3] = 1.59 \, M$
$[Cl_2] = 1.59 \, M$