NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice
The vapour density of N 2 O 4 at a certain temperature is 30. What is the percentage dissociation of N 2 O 4 at this temperature ?
Options
- A53.3 %
- B106.6 %
- C26.7 %
- DNone of these
Correct answer
A. 53.3 %
Step-by-step solution
The reaction is N 2 O 4 ⇌ 2 NO 2 mol. wt. of N 2 O 4 = 14 x 2 + 16 x 4 = 92 ∴ Vapour density (D) of N 2 O 4 = 92/2 = 46 We know that, α = D - d n - 1 d = 4 6 - 3 0 2 - 1 3 0 = 1 6 3 0 where, D = Initial vapour density of reactant d = vapour density of equilibrium mixture α = degree of dissociation ∴ α = 0.533 = 53.3 %