Question

2.0 mol of $PC{l}_5$ were introduced in a vessel of 5.0 L capacity at a particular temperature. At equilibrium, $PC{l}_5$ was found to be 35% dissociated into $PC{l}_3$ and $C{l}_2$. The value of $K_C$ for the reaction is-

• A. 1.89
• B. 0.377
• C. 0.75
• D. 0.075

Answer 1

The correct option is D 0.075

35% of $PC{l}_5$ was dissociated. The number of moles of $PC{l}_5$ dissociated $=\frac{2\times 35}{100}=\mathrm{0.7.}$
The number of moles of $PC{l}_5$ left undissociated $=2-0.7=1.3mol.$
The equilibrium concentrations are $\left[PC{l}_5\right]=\frac{1.3}{5}M,\left[PC{l}_3\right]=\frac{0.7}{5}M,\left[C{l}_2\right]=\frac{0.7}{5}M.$
The expression for the equilibrium constant $K=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{\left(\frac{0.7}{5}\right)\left(\frac{0.7}{5}\right)}{\left(\frac{1.3}{5}\right)}=\mathrm{0.075.}$