Question

Which of the following changes would decrease the dissociation of $PC{l}_5$ in the reaction? $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

• A. Decrease in pressure
• B. Increase in pressure
• C. Addition of an inert gas at constant volume
• D. Addition of an inert gas at constant pressure

Answer 1

Answer: (A), (D)

Addition of an inert gas at constant pressure
Decrease in pressure

${PCl}_{5\left(g\right)}\rightleftharpoons {PCl}_{3\left(g\right)}+{Cl}_{2\left(g\right)}$

To increase dissociation of ${PCl}_5$ we should make $Q<K_C$ which drives forward.

$K=Q_C=\frac{P^{{PCl}_3}\times P^{{Cl}_2}}{P^{{PCl}_5}}$

No if we decrease total pressure $Qi$ decreases because numerator proportional to $P^2$ where as denominator proportional to $P^1$. Hence reaction moves forward (increases dissociation)

(or)

According Lechatlier principle decrease in pressure drives reaction towards more moles of gas.