NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice
At a definite temperature, the equilibrium constant for a reaction, A + B ⇌ 2 C , was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :
Options
- A9 1 1
- B1 1 1
- C2 1 1
- D7 1 1
Correct answer
A. 9 1 1
Step-by-step solution
A + B ⇌ 2 C 1 1 0 1 - x 1 - x 2 x ∴ 8 1 = 4 x 2 1 - x 2 or 2 x 1 - x = 9 ∴ x = 9 1 1 Thus, A + B ⇌ 2 C 2 1 1 2 1 1 18 1 1 ( Total mole = 2 1 1 + 2 1 1 + 18 1 1 = 22 1 1 ) ∴ Mole fraction of C= 18 1 1 2 2 1 1 = 9 1 1