Question

One mole of N2O4 is enclosed in a 5 L container. At equilibrium, the container has 0.5 mole of N2O4. Equilibrium constant for decomposition of N2O4 [N2O4(g) \leftrightharpoons 2NO2(g)] is


a) 1

b) 2/5

c) 3

d) 1/5

Answer 1

answer : option (b) 2/5

explanation : decomposition of N2O4 is given as ....

$N_2O_4(g)\leftrightharpoons2NO_2(g)$

at t = 0, N2O4 = 1mol, NO2 = 0mol

at equilibrium, N2O4 = 1 - x = 0.5 ⇒x = 0.5

and NO2 = 2x = 0.5 × 2 = 1

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

= 4x²/(1 - x)V

putting, x = 0.5 and V = 5L

so, K = 4(0.5)²/(1 - 0.5)5

= 4 × 0.25/(0.5 × 5)

= 1/2.5

= 2/5

hence, equilibrium constant is 2/5

Answer 2

Explanation:

hope this will help you ☺️