Question

A reaction at equilibrium involving $2moles$ each of $P{Cl}_5,P{Cl}_3,{Cl}_2$ is maintained at ${250}^{o}C$ and a total pressure of $3atm$. The value of $K_p$ is:

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

Answer: (D)

$1$

The given reaction is :-

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

At eqm : $2$ $2$ $2$

(moles)

Total no. of moles $=6$, Total pressure $=3atm$

$\therefore K_P=\frac{p_{{PCl}_3}.p_{{Cl}_2}}{p_{{PCl}_5}}=\frac{(\frac{2}{6}\times 3)(\frac{2}{6}\times 3)}{(2/6\times 3)}=1$