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

Moles of $PC{l}_5$ dissociated $=\frac{2\times 35}{100}=0.7$
Moles of $PC{l}_5$ left undissociated = 2 - 0.7 = 1.3 mol
$\left[PC{l}_5\right]=\frac{1.3}{5}M,\left[PC{l}_3\right]=\frac{0.7}{5}M,\left[C{l}_2\right]=\frac{0.7}{5}M$
$K=\frac{\left[PC{l}_3\right]}{\left[PC{l}_5\right]}\left[C{l}_2\right]=\frac{\left(\frac{0.7}{5}\right)\left(\frac{0.7}{5}\right)}{\left(\frac{1.3}{5}\right)}=0.075$