JEE MAIN - Chemistry (2022 - 25th June Morning Shift - No. 19)
The standard free energy change ($$\Delta$$G$$^\circ$$) for 50% dissociation of N2O4 into NO2 at 27$$^\circ$$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]
Answer
710
Explanation
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$\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{1}{1.5} \times 1\right)^2}{\left(\frac{0.5}{1.5} \times 1\right)}=\frac{1}{0.75}=\frac{100}{75}$
$=1.33$
$\Delta \mathrm{G}^0=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{P}}$
$=-8.31 \times 300 \times \ln (1.33)=-710.45 \mathrm{~J} / \mathrm{mol}$
$=-710 \mathrm{~J} / \mathrm{mol}$
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