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QJEEM22Ch364a: JEE Main 2022 Chemistry - Class 11 JEEM22Ch364a Solution: The standard free energy change ( Δ G ∘ ) for 50% dissociation…

The standard free energy change (ΔG∘) for 50% dissociation of N2O4 into NO2 at 27∘C and 1 atm pressure is − x J mol−1. The value of x is ___________. (Nearest Integer)

[Given : R = 8.31 J K−1 mol−1, log 1.33 = 0.1239 ln 10 = 2.3]

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JEE Main 2022 (Online) 25th June Morning Shift Chemistry - Chemical Equilibrium Question 36 English Explanation
kP=(11.5×1)2(0.51.5×1)=10.75=10075\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{1}{1.5} \times 1\right)^2}{\left(\frac{0.5}{1.5} \times 1\right)}=\frac{1}{0.75}=\frac{100}{75}

=1.33=1.33

ΔG0=−RTℓnkP\Delta \mathrm{G}^0=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{P}}

=−8.31×300×ln⁡(1.33)=−710.45 J/mol=-8.31 \times 300 \times \ln (1.33)=-710.45 \mathrm{~J} / \mathrm{mol}

=−710 J/mol=-710 \mathrm{~J} / \mathrm{mol}

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