Question

For the reaction $\sf{N_2 O_4}$ ⇌ $\sf{2NO_2}$ the degree of dissociation at equilibrium is 0.2 at 1 atm Then Kp is

Options

a) 1/2
b) 1/4
c) 1/6
d) 1/8

Hint :- Correct answer is 1/6​

Answer 1

Reaction : $\bf{N_2 O_4}$ ⇌ $\bf{2NO_2}$

At t = 0 ::

Moles of N₂O₄ → 1

Moles of NO₂ → 0

At equilibrium ::

Moles of N₂O₄ → (1 - α)

Moles of NO₂ → 2 × α = 2α

Total number of moles at equilibrium will be;

$\sf:\implies\:N_T=N_{N_2O_4}+N_{NO_2}$

$\sf:\implies\:N_T=(1-\alpha)+2\alpha$

$\sf:\implies\:N_T=1+\alpha$

Given that; α = 0.2

$\sf:\implies\:N_T=1+0.2$

$\bf:\implies\:N_T=1.2$

In order to find Kp of the reaction, first of all we have to calculate partial pressure of N₂O₄ and NO₂ at equilibrium.

Partial pressure of N₂O₄ ::

$\sf:\implies\:P_{N_2O_4}=\dfrac{N_{N_2O_4}}{N_T}\times P$

$\sf:\implies\:P_{N_2O_4}=\dfrac{1-\alpha}{1.2}\times 1$

$\sf:\implies\:P_{N_2O_4}=\dfrac{1-0.2}{1.2}\times 1$

$\sf:\implies\:P_{N_2O_4}=\dfrac{0.8}{1.2}\times 1$

$\bf:\implies\:P_{N_2O_4}=\dfrac{2}{3}\:atm$

Partial pressure of NO₂ ::

$\sf:\implies\:P_{NO_2}=\dfrac{N_{NO_2}}{N_T}\times P$

$\sf:\implies\:P_{NO_2}=\dfrac{2\alpha}{1.2}\times 1$

$\sf:\implies\:P_{NO_2}=\dfrac{2(0.2)}{1.2}\times 1$

$\sf:\implies\:P_{NO_2}=\dfrac{0.4}{1.2}\times 1$

$\bf:\implies\:P_{NO_2}=\dfrac{1}{3}\:atm$

Kp of the reaction is given by;

$\sf:\implies\:K_p=\dfrac{[P_{NO_2}]^2}{[P_{N_2O_4}]}$

$\sf:\implies\:K_p=\dfrac{(1/3)^2\:atm^2}{(2/3)\:atm}$

$\sf:\implies\:K_p=\dfrac{1}{3\times 2}\:atm$

$:\implies\:\underline{\boxed{\bf{\orange{K_p=\dfrac{1}{6}\:atm}}}}$

Answer 2

EXPLANATION.

Reaction,

⇒ N₂O₄ ⇄ 2NO₂.

Degree of dissociation = α = 0.2. at 1 atm.

As we know that,

N₂O₄ ⇄ 2NO₂.

1               0       at  t = 0.

1 - α         2α     at  t = equilibrium.

We can write Kp as,

$\sf \implies Kp = \dfrac{[P_{NO_2]^{2} }}{[P_{N_2O_4}]}$

Total no of moles = 1 - α + 2α = 1 + α.

As we know that,

⇒ P(N₂0₄) = 1 - α/1 + α x P.

Put the value in equation, we get.

⇒ P(N₂0₄) = 1 - 0.2/1 + 0.2 x 1.

⇒ P(N₂0₄) = 0.8/1.2.

⇒ P(N₂0₄) = 8/12 = 2/3.

⇒ P(NO₂) = 2α/1 + α x P.

Put the value in the equation, we get.

⇒ P(NO₂) = 2(0.2)/1 + 0.2 x 1.

⇒ P(NO₂) = 0.4/1.2.

⇒ P(NO₂) = 4/12 = 1/3.

As we know that,

Put the value in the expression of Kp, we get.

$\sf \implies Kp = \dfrac{[P_{NO_2]^{2} }}{[P_{N_2O_4}]} = \dfrac{\bigg(\dfrac{1}{3} \bigg)^{2} }{\dfrac{2}{3} }$

$\sf \implies Kp = \dfrac{[P_{NO_2]^{2} }}{[P_{N_2O_4}]} = \dfrac{\dfrac{1}{3} \times \dfrac{1}{3} }{\dfrac{2}{3} } = \dfrac{1}{9} \ \times \dfrac{3}{2} = \dfrac{1}{6}$

$\sf \implies Kp = \dfrac{[P_{NO_2]^{2} }}{[P_{N_2O_4}]} = \dfrac{1}{6}$

Hence, option [C] is correct answer.

                                                                                                                     

MORE INFORMATION.

Some other important points.

(1) = Kc unit = (moles/lit)^Δn.

(2) = Kp unit = (atm.)^Δn.

(3) = Total moles at equilibrium = [total initial moles + Δn(x)].

(4) = Time required to established equilibrium ∝ 1/Kc.

(5) = If in any heterogenous equilibrium solid substances is also present then its active mass & partial pressure is assumed to 1.