Question

for the following equilibrium vant half factor is0.80 2A=A2 hence volume percentage of A2 in the total mixure

Answer 1

2A- A2

let 1 moles initially taken of 2A

At equilibrium let 2x moles dissociated

so left moles of 2A - 1 - 2x

Moles formed of A2 - x

Total moles = 1 - 2x + x = 1 - X

AS vant's hoff factor = 1 - x= 0.8

x= 0.2

So A2 - 0.2 moles

so it's 20% as volume percentage would be same