Question

If pcl5 is 80% dissociated at 523k .what is the vapour density of the equilibrium mixture at 523 k

Answer 1

Hey dear,

● Answer -
4.846 kg/m^3

● Explanation -
PCl5 dissociates upto 80% i.e. 0.8 .

Dissociation of PCl5 occurs as -
Reaction PCl5 --> PCl3 + Cl2
Equilibrium 0.8 => 0.2 0.2

At 523 K, density of PCl5 is calculated by-
d1 = 208 / 22.4 × 0.8 × 273 / 523
d1 = 3.877 g/L

At 523 K, density of PCl3 is calculated by-
d2 = 137 / 22.4 × 0.2 × 273 / 523
d2 = 0.638 g/L

At 523 K, density of Cl2 is calculated by-
d3 = 71 / 22.4 × 0.2 × 273 / 523
d3 = 0.331 g/L

Density of total mixture -
d = d1 + d2 + d3
d = 3.877 + 0.638 + 0.331
d = 4.846 kg/m^3

Therefore, vapour density of mixture is 4.846 kg/m^3.

Hope this helps...

Answer 2

Answer:

57.92

here's the answer and hope this helps you.