For the gas-phase equilibrium… | JEE Advanced 2024 P2

For the gas-phase equilibrium N2O4(g)rightleftharpoons 2NO2(g), the degree of dissociation of N2O4 is 0.20 at 300 K and total pressure 1.0 atm, and…

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For the gas-phase equilibrium

\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)}

, the degree of dissociation of

\mathrm{N_2O_4}

0.20

300\ \mathrm{K}

and total pressure

1.0\ \mathrm{atm}

, and

0.50

350\ \mathrm{K}

and total pressure

2.0\ \mathrm{atm}

. Assuming

\Delta H^\circ

is independent of temperature, and

R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}

, the value of

\Delta H^\circ

for the reaction is closest to

Show answer and solution

Answer: (A) $48.4\ \mathrm{kJ\,mol^{-1}}$

Start with 1 mol

\mathrm{N_2O_4}

: at equilibrium

(1-\alpha)

mol

\mathrm{N_2O_4}

and

2\alpha

mol

\mathrm{NO_2}

, total

(1+\alpha)

. Then

K_p=\dfrac{(2\alpha P/(1+\alpha))^2}{(1-\alpha)P/(1+\alpha)}=\dfrac{4\alpha^2P}{1-\alpha^2}

. At 300 K:

K_1=\dfrac{4(0.04)(1)}{0.96}=0.1667

. At 350 K:

K_2=\dfrac{4(0.25)(2)}{0.75}=2.667

. Van't Hoff:

\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac1{300}-\dfrac1{350}\right)

. With

K_2/K_1=16

\ln16=2.773

and the bracket

=4.762\times10^{-4}\ \mathrm{K^{-1}}

, so

\Delta H^\circ=\dfrac{8.314\times2.773}{4.762\times10^{-4}}\approx4.84\times10^4\ \mathrm{J}=48.4\ \mathrm{kJ\,mol^{-1}}

. Using

P=1

atm in the second case (ignoring the pressure change) gives 36.3 kJ, a typical error.

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake