Question

For the reaction, N2O3 = NO +NO2, the
value of equilibrium constant Kp at fixedtemperature is 4. What will be the amount of
dissociation at same temperature and 5
atmospheric pressure ?
(1)1/3
(2)2/3
(3)7/9
(4)2/4​

Answer 1

Answer:

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?

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 ⇒ αNO2 ⇒ αso, n = (1 - α) + α + α = 1 + αthen, 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 = 4so, 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

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Answer 2

Answer:

Equilibrium constant for the given reaction can be written as

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