Question

Find the total pressure to terms of $K_p$ for 50% dissociation of $PC{l}_5$ at ${250}^{\circ }C$.

• A. $K_P$
• B. $\frac{3}{2}{K}_P$
• C. $3K_P$
• D. $\frac{1}{2}{K}_P$

Answer 1

The correct option is C $3K_P$

$PC{l}_5\underset{\rightleftharpoons }{\left(C\right)}PC{l}_3+C{l}_2$
$\begin{array}{ccccc}T=0& 1& 0& 0& \text{(Iniitial moles)}\\ T=eq.& (1-\frac{1}{2})& \frac{1}{2}& \frac{1}{2}& \text{(moles at equlibrium)}\\ & =0.5& 0.5& 0.5& \end{array}$
Total moles at equilibrium
= 0.5 + 0.5 + 0.5 = 1.5
$\therefore K_p=\frac{P_{PC{l}_2}.P_{C{l}_2}}{P_{PC{l}_5}}$
$=\frac{(\frac{0.5}{1.5}\times p)(\frac{0.5}{1.5}\times P)}{(\frac{0.5}{1.5}\times P)}(\because P=\text{Total pressure})$
$\Rightarrow K_p=\frac{1}{3}P$
$\Rightarrow P=3\text{Kp}$