Question

For the reaction n2o3=no+no2, the value of equilibrium constant kp at fixed temperature is 4. What will be the amount of dissociation at same temperature and 5 atmospheric pressure?

Answer 1

equilibrium constant for the given reaction can be written as $k_p=\frac{P_{NO}\times P_{NO_2}}{P_{N_2O_3}}$

let dissociation constant is α.

we have to find out partial pressure of each of the components.

from reaction ,at equilibrium,

N2O3 ⇒(1 - α)

NO ⇒ α

NO2 ⇒ α

so, n = (1 - α) + α + α = 1 + α

then, partial pressure of N2O3 = P(1 - α)/(1 + α)

partial pressure of NO = Pα/(1 + α)

partial pressure of NO2 = Pα/(1 + α)

now, Kp = {Pα/(1 + α) × Pα/(1 + α) }/{P(1 - α)/(1 + α)}

or, Kp = Pα²/(1 - α²)

putting P = 5 and Kp = 4

so, 4 = 5α/(1 - α²)

or, 4(1 - α²) = 5α²

or, 4 - 4α² = 5α²

or, 9α² = 4

or, α² = 4/9 = (2/3)²

or, α = 2/3

hence, amount of dissociation is 2/3

Answer 2

Answer:

2/3

Explanation:

I hope this will help u dude