Question

What will be the expression of K_p for the given reaction if the total pressure inside the vessel is P and degree of dissociation of the reactant is α? The reaction: N₂O₄ (g) ⇌ 2NO₂

• A. \(\frac{4α^2P}{(1+α^2)}\)
• B. \(\frac{4α^2P}{(1-α^2)}\)
• C. \(\frac{α^2P}{(1-α^2)}\)
• D. \(\frac{α^2}{(1-α)}\)

Hint:

For N2O4 ⇌ 2NO2, total moles at equilibrium = 1 - α + 2α = 1+α.

Answer 1

The Correct Option is B

Reaction	N2O4 (g) ⇌ 2NO2
Moles at t = 0	1	0
Moles at equilibrium	1-α	2α

Total moles at equilibrium = 1 - α + 2α = 1+α
\(p_{N_2 O_4}=\left(\frac{1-a}{1+a}\right) P ;\)
\(p_{NO_2}=\left(\frac{2a}{1+a}\right)P\)
\(K_{P}=\frac{p_{NO_2}}{p_{N_2O_4}}=\frac{\left(\frac{2a}{1+a}\right)^{2} p^{2}}{\left(\frac{1-a}{1+a}\right)p}\)
\(=\frac{4a^{2}p}{1-a^{2}}\)

Answer 2

The correct option is B.

\(A→2B\) moles

\(K_p = K_n [\frac{P}{n}]^{\Delta ng}\)  P⟶ Total pressure, n= Total moles, \(n=1−α+2α=1+α\)

\(K_p = \frac{[nB]^2}{n_A} [\frac{P}{n}]^{2-1}\)

\(K_p = \frac{[2\alpha]^2}{[1-\alpha]} [\frac{P}{1+\alpha}]\)

\(K_p = \frac{4\alpha^2P}{1-\alpha^2}\)

The expression of K_p for the given reaction if the total pressure inside the vessel is P and the degree of dissociation of the reactant is \(\frac{4α^2P}{(1-α^2)}\) for the reaction: N₂O₄ (g) ⇌ 2NO₂