The standard free energy change ( \(\Delta \mathrm{G}^{\circ}\) ) for \(50 \%\) dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) into \(\mathrm{NO}_{2}\) at \(27^{\circ} \mathrm{C}\) and 1 atm pressure is \(-\mathrm{x} \mathrm{J} \mathrm{mol}^{-1}\). The value of x is \(\_\_\_\_\) . (Nearest Integer) [Given : \(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 1.33=0.1239 \ln 10=2.3]\)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Chemical Thermodynamics

The standard free energy change (

Δ G^{\circ}

) for

50%

dissociation of

N_{2} O_{4}

into

NO_{2}

2 7^{\circ} C

and 1 atm pressure is

- xJ mol^{- 1}

. The value of x is

____

. (Nearest Integer)
[Given :

R = 8.31 J K^{- 1} mol^{- 1}, lo g 1.33 = 0.1239 ln 10 = 2.3]

Answer: 710

For the equilibrium

N_{2} O_{4} ⇌ 2 NO_{2}

with 50% dissociation:

Initial: 1 mol N₂O₄
At equilibrium: 0.5 mol N₂O₄, 1 mol NO₂
Total moles = 1.5

Partial pressures:

K_{P} = \frac{( 2/3 ) ^{2}}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} = 1.33

Using

Δ G^{\circ} = - R T ln K_{P}

Δ G^{\circ} = - 8.31 \times 300 \times ln (1.33)

= - 8.31 \times 300 \times 0.1239 \times 2.3

= - 710.45 J/mol

Therefore, x = 710.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer