You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction

One mole of P...

Question

One mole of $P C l_{5} (g)$ dissociates and attains equilibrium after t seconds at $27^{\circ} C$ :
$P C l_{5} (g) ⇋ P C l_{3} (g) + C l_{2} (g)$
At equillibrium, the vapour density is found to be 100 and the volume of the system is 10.425 L at $27^{\circ} C$ .
The total pressure of the system at equilibrium will be atm.
$R = 0.082 L a t m m o l^{- 1} K^{- 1}$

Open in App

Solution

$P C l_{5} (g) ⇋ P C l_{3} (g) + C l_{2} (g)$
$I n t i a l 1 m o l e - -$
$A t e q (1 - α) α α$
$α = \frac{D}{d} - 1$
where D is the molar vapour density of $P C l_{5} = 104.25$
and $d$ is the vapour density after dissociation $= 100$
So, $α = 0.0425$
Total number of moles at equilibrium (n)
$= 1 + α = 1 + 0.0425 = 1.0425$
$P V = n R T$
$P = \frac{1.0425 \times 0.082 \times 300}{10.425} = 2.46 a t m$

Suggest Corrections

Similar questions

P C l_{5}

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

K_{p} = 1.78 a t m

P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g),

P C l_{5}, P C l_{3} a n d C l_{2}

K_{P}

B r_{2} (l) + C l_{2} (g) ⇌ 2 B r C l (g)

27^{\circ} C

k_{p} = 1 a t m

164 L

10

C l_{2}

27^{\circ} C

B r_{2} (l)

2.25 a t m

B r_{2} (l)

27^{\circ} C

0.25 a t m

[R = 0.082 L a t m m o l e^{- 1} K^{- 1}

= 80

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

P C l_{5}

K_{p}

K_{c}

P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)

P

P {C l}_{5}

x

P {C l}_{3}

Join BYJU'S Learning Program