Question

Phosphorus pentachloride dissociates as follows in a closed reaction vessel.

PCl5(g) PCl3(g) + Cl2(g)

If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be :

• A. $\left( \frac{x}{x + 1} \right)P$
• B. $\left( \frac{2x}{1 - x} \right)P$
• C. $\left( \frac{x}{x + 1} \right)P$
• D. $\left( \frac{x}{1 - x} \right)P$

Answer 1

Answer: (A)

$\left( \frac{x}{x + 1} \right)P$

Answer 2

Given : PCI5(g) PCI3(g) + CI2(g)

t = 0 1 0 0

teq 1–x x x

Total number of moles = 1 – x + x + x = 1 + x

Thus partial pressure of$PCl_{3} = \left( \frac{x}{1 + x} \right)P.$