The value of K_p for the equilibrium reaction N_2 O_4(g) leftharpoons 2 NO_2(g) is 2 . The percentage dissociation of N_2 O_4(g) at a pressure of 0.5 ~atm is

Correct answer: D. 71 array llcc & N_2 O_4( g) leftharpoons 2 NO_2( g) & Initial moles & 1 & 0 & Moles of equil. & (1-α) & 2 α array (α= degree of dissociation ) Total number of moles at equil. aligned & =(1-α)+2 α & =(1+α) p_ N_2 O_ 4 & =((1-α))/((1+α)) p_ N O_2 & =(2 α)/((1+α)) K_P & = (p_ N_O )^2 p_ N_2 O_4 =(((2 α)/((1+α)) × P)^2)/(((1-α)/(1+α)) × P)=(4 α^2 P)/(1-α^2) aligned aligned & Given, K_P=2, P=0.5 atm & K_P=(4 α^2 P)/(1-α^2) & =(4 α^2 × 0.5)/(1-α^2) & α=0.707 0.71 & Percentage dissociation & =0.71 ×100=71 aligned

JEE Main 201219 MayOnline

Exam: JEE Main
Year: 2012
Date: 19 May
Mode: Online
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2012 · 19 May · Online · Chemistry · Chemical Equilibrium

The value of K_p for the equilibrium reaction N_2 O_4(g)...

The value of

K_p

for the equilibrium reaction

\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)

is 2 . The percentage dissociation of

\mathrm{N}_2 \mathrm{O}_4(g)

at a pressure of

0.5 \mathrm{~atm}

Options

Correct answer

D. 71

Step-by-step solution

\begin{array}{llcc} & \mathrm{N}_2 \mathrm{O}_4(\mathrm{ g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{ g}) \\ & \text{ Initial moles } & 1 & 0 \\ & \text{ Moles of equil. } & (1-\alpha) & 2 \alpha\end{array}

(\alpha=

degree of dissociation

)

Total number of moles at equil.

\begin{aligned}& =(1-\alpha)+2 \alpha\\ & =(1+\alpha)\\ p_{N_2 \mathrm{O}_{4}} & =\frac{(1-\alpha)}{(1+\alpha)} \timesP\\ p_{N O_2} & =\frac{2 \alpha}{(1+\alpha)} \timesP\\ K_P & =\frac{\left(p_{N_O}\right)^2}{p_{N_2 O_4}}=\frac{\left(\frac{2 \alpha}{(1+\alpha)} \times P\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right) \times P}=\frac{4 \alpha^2 P}{1-\alpha^2} \end{aligned}

\begin{aligned}& \text{ Given, } K_P=2, P=0.5 \mathrm{ atm}\\ & \therefore \quad K_P=\frac{4 \alpha^2 P}{1-\alpha^2}\\ & =\frac{4 \alpha^2 \times 0.5}{1-\alpha^2}\\ & \alpha=0.707 \approx0.71 \\ & \therefore \quad \text{ Percentage dissociation }\\ & =0.71 \times100=71 \end{aligned}

Practice Chemical Equilibrium on Quantrex Academy →

More from Chemical Equilibrium

Consider the following reactions in which all the reactants and products are present in… JEE Main 2026 Solid carbon, CaO and CaCO _3 are mixed and allowed to attain equilibrium at T K.… JEE Main 2026 In a closed flask at 600 K, one mole of X _2 Y _4 (g) attains equilibrium as given below… JEE Main 2026 One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to… JEE Main 2026 The reaction A(g) leftharpoons B(g) + C(g) was initiated with the amount ' a ' of A(g) .… JEE Main 2026 The values of pressure equilibrium constant recorded at different temperatures for the… JEE Main 2026 For the following reaction at 50° C and at 2 atm pressure, 2N_2O_5(g) leftharpoons… JEE Main 2026 At T(K) , the equilibrium constant of A_2(g) + B_2(g) leftharpoons C(g) is 2.7 × 10^ -5 .… JEE Main 2026 Full Chemical Equilibrium list All JEE Main PYQs