Question

1 mole of A and 1 mole of B are taken in a 5 litre flask, 0.5 mole of C is formed in equilibrium of A + B ⇄ C + D.What is molar concentration of each species if the reaction is carried with 2 mole of A, 1 mole of B in a 5 litre flask at the same temperature.

Answer 1

A + B ⇄ C + D

at t = 0. 1 1 0 0

at equⁿ 1 - 0.5 1 - 0.5 0.5 0.5

so, equilibrium constant, K = (0.5)²/(1 - 0.5)² = 1

now again,

A + B ⇄ C + D

at t = 0, 2 1 0 0

at eqⁿ 2 - x 1 - x x x

so, equilibrium constant, K = 1 = x²/(2 - x)(1 - x)

or, x² = (2 - x)(1 - x)

or, x² = 2 - 2x -x + x²

or, 2 = 3x ⇒x = 2/3

so, mole of A = 2 - 2/3 = 4/3

hence, molar concentration of A = mole/volume = (4/3)/5 = 0.8/3 ≈ 0.2667M

mole of B = 1 - 2/3 = 1/3

so, molar concentration of B = (1/3)/5 = 0.2/3 = 0.0667M

mole of C = mole of D = 2/3

so, molar concentration of C = molar concentration of D = (2/3)/5 = 0.4/3 = 0.1333M