Question

A 50 litre flask containing one mole of $N_2$ and one mole of $PC{l}_5$ is connected to a 50 litre flask containing 2 moles of $PC{l}_5$, both vessels are heated to ${227}^{\circ }C$. The equilibrium pressure is found to be 2.05 atm. Assuming ideal behaviour and for the reaction $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$ correct option(s) is/are

• A. degree of dissociation of $PC{l}_5$ is $\frac{1}{3}$
• B. $K_p$ for the reaction is 0.205 atm
• C. Partial pressure of $N_2$ in each flask is 0.41 atm
• D. mole fraction of $N_2$ in each flask is 0.1

Answer 1

Answer: (A), (B), (C)

The correct options are
A degree of dissociation of $PC{l}_5$ is $\frac{1}{3}$
B $K_p$ for the reaction is 0.205 atm
C Partial pressure of $N_2$ in each flask is 0.41 atm

$N_2+PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

Moles at t=0 1 3 0 0
Moles 1 (3-x) x x

Total moles present at equilibrium = 4 + x
Given total pressure at equilibrium = 2.05
Total volume = 100 Litre

PV=nRT

$\therefore 2.05\times 100=(4+x)\times 0.082\times 500$

$\therefore x=1.0$

Degree of disassociation,$\alpha$ for $PC{l}_5$

$=\frac{\text{Moles dissociated}}{\text{Total moles present}}=\frac{1}{3}=0.3333=33.33\%$

Also $K_P=\frac{x^2}{(3-x)}{\left[\frac{P}{\sum n}\right]}^{\Delta n_g}=\frac{x^2}{(3-x)}\times \frac{2.05}{(4+x)}(\Delta n_g=1)$

$=\frac{(1{)}^2\times 2.05}{(3-1)(4+1)}$

$K_P=0.205$ atm