Question

Two moles of $PC{l}_5$ were heated to ${327}^{\circ }C$ in a closed two litre vessel and when equilibrium was achieved, $PC{l}_5$ was found to be 40% disociated into $PC{l}_3$ and $C{l}_2$. Calcutta the equilibrium constant $\left(K_c\right)$ for the reaction.

Answer 1

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

$2moles$ $-$ $-$

$2(1-\alpha )$ $2\alpha$ $2\alpha$ $\alpha =40$% $=0.4$

$V=2$lit

$\left[{PCl}_5\right]=\frac{2(1-\alpha )}{V}=\frac{2(1-0.4)}{2}=0.6mol/L$

$\left[{PCl}_3\right]=\left[{Cl}_2\right]=\frac{2\left(0.4\right)}{2}=0.4mol/L$

$K_C=\frac{0.4\times 0.4}{0.6}=0.27mol{L}^{-1}$