Question

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$. If 1 mole of $PC{l}_5$ be put in a container of volume V litre and at equilibrium, x moles of it was decomposed, find its $K_p$ and $K_c$ at equilibrium pressure of P atm.

• A. $K_c=\frac{x^2}{(1-x)V}$ and $K_P=\frac{x^{2}P}{1-x^2}$
• B. $K_c=\frac{x^4}{(1-x)V}$ and $K_P=\frac{x^{4}P}{1-x^2}$
• C. $K_c=\frac{x}{(1-x)V}$ and $K_P=\frac{xP}{1-x^2}$
• D. None of the above

Answer 1

The correct option is A $K_c=\frac{x^2}{(1-x)V}$ and $K_P=\frac{x^{2}P}{1-x^2}$

The given reaction is $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$K_P=\frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}and{K}_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$
$\begin{array}{cccccc}& PC{l}_5& \rightleftharpoons & PC{l}_3& +& C{l}_2\\ \text{Initial moles}& 1& & 0& & 0\\ \text{Moles at equilbrium}& 1-x& & x& & x\\ \text{Concentration at equilbrium}& \frac{1-x}{V}& & \frac{x}{V}& & \frac{x}{V}\end{array}$
Total moles at equilbrium$=1-x+x+x=1+x$
$K_c=\frac{{\left(\frac{x}{V}\right)}^2}{\left(\frac{1-x}{V}\right)}=\frac{x^2}{(1-x)V}$
Partial pressure of $P_{PC{l}_3}=X_{PC{l}_3}{P}_{eqb},P_{PC{l}_5}=X_{PC{l}_5}{P}_{eqb},P_{C{l}_2}=X_{C{l}_2}{P}_{eqb}$
$P_{eqb}=P,X_{PC{l}_3}=\frac{x}{1+x},X_{C{l}_2}=\frac{x}{1+x},X_{PC{l}_5}=\frac{1-x}{1+x}$

so, $K_P=\frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}$
$K_P=\frac{{\left(\frac{x}{1+x}P\right)}^2}{\left(\frac{1-x}{1+x}\right)P}=\frac{x^{2}P}{1-x^2}$