You visited us 20 times! Enjoying our articles? Unlock Full Access!

Reaction Quotient

PCl5 dissocia...

Question

$P C l_{5}$ dissociates according to the reaction $P C l_{5} ⇌ P C l_{3} (g) + C l_{2} (g) .$ At $523 K, K_{p} = 1.78 a t m .$ Find the density of the equilibrium mixture at a total pressure of 1 atm.

2.96 g/ml

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.51 g/ml

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.73 g/ml

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.44 g/ml

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 2.96 g/ml
Let at equilibrium, 1 mole each of $P C l_{3}$ and $C l_{2}$ are present.
The $x$ moles of $P C l_{5}$ is obtained from the equilibrium constant expression.
$K_{p} = \frac{1 \times 1}{x} = 1.78$
$\Rightarrow x = 0.56$
The average molecular mass of the gaseous mixture is as calculated below.
$\frac{0.56 (208.5) + 1 (137.5) + 1 (71)}{2.56} = 127$
The expression for the density is $ρ = \frac{P M}{R T} = \frac{1 \times 127}{0.082 \times 523} = 2.96 g / m l .$

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

250^{o} C

1

P {C l}_{5}

57.9

K_{p}

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

250^{o} C

K_{p}

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

1.80

523 K

g {l i t}^{- 1}

100 k P a

727^{0} C

S O_{3}

S O_{2}

O_{2}

S O_{3} (g) ⇌ S O_{2} (g) + 1 / 2 O_{2} (g)

727^{o}

1.23

S O_{3}

S O_{2}

O_{2}

S O_{3} (g) ⇌ S O_{2} (g) + 1 / 2 O_{2} (g)

0.9

Join BYJU'S Learning Program