Question

.


For the reaction N2O4⇄2NO2
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

Answer 1

Answer:

1/4

$thank$

Answer 2

Step-by-step explanation:

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.

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

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

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

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.