NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice
K p for the reaction P C l 5 g ⇌ P C l 3 g + C l 2 g at 250 o C is 0 .82 . Calculate the degree of dissociation at given temperature under a pressure of 5 atm . What will be the degree of dissociation if the equilibrium pressure is 10 atm at same temperature.
Options
- A27.5%
- B23%
- C35.5%
- D40%
Correct answer
A. 27.5%
Step-by-step solution
K P = x 1 + x P × x 1 + x P 1 - x 1 + x P K P = x 2 1 - x 2 P = 0.82 x 2 = 0.82 5.82 x = 0.375 Now new pressure is 10 K P = y 2 × P 1 - y 2 , 0.82 = y 2 1 - y 2 × 10 y = 0.275 , y = 27.5 %