Question

In the reaction,
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_3\left(g\right)$ the amounts of $PC{l}_5,PC{l}_3$ and $C{l}_2$ are $2$ mole each at equilibrium and the total pressure is $3$ atmospheres. The equilibrium constant, $K_p$ is:

• A. $1$ atm
• B. $2$ atm
• C. $3$ atm
• D. $1.5$ atm

Answer 1

The correct option is A $1$ atm

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)222\text{Total moles}=6$

$\text{Total pressure}=3atm.$

$P_{{PCl}_5}=P_{PC{l}_3}=P_{C{l}_2}=\left(\frac{2}{6}\right)\times 3$

and$K_p=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}=\frac{\frac{2}{6}\times 3\times \frac{2}{6}\times 3}{\frac{2}{6}\times 3}=1atm$