JEE Main 2022 · 25 Jun · Shift 1 · Chemistry Q86

← JEE Main 2022 · 25 Jun · Shift 1

ChemistryInteger+4 / -1Question 86 of 90

The standard free energy change ( $\Delta G^\circ$ ) for 50% dissociation of $\mathrm{N_2O_4}$ into $\mathrm{NO_2}$ at 27°C and 1 atm pressure is $-x$ J mol $^{-1}$ . The value of $x$ is ______. (Nearest Integer)

[Given : $R=8.31\ \mathrm{J\,K^{-1}mol^{-1}}$ , $\log1.33=0.1239$ , $\ln10=2.3$ ]

Your answer

Answer and worked solution

Official answer: 710

Equilibrium. From 1 mol $\mathrm{N_2O_4}$ , half dissociates: 0.5 mol $\mathrm{N_2O_4}$ and 1 mol $\mathrm{NO_2}$ , total 1.5 mol at 1 atm.

$K_p$ . $\dfrac{(2/3)^2}{1/3}=\dfrac43\approx1.33$ .

Free energy. $\Delta G^\circ=-2.3RT\log K=-2.3\times8.31\times300\times0.1239\approx-710$ J mol $^{-1}$ .

Answer: 710

Topics:Equilibrium and Gibbs energy Law of mass action and Kc, Kp

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JEE Main 2022 · 25 Jun · Shift 1, Chemistry question 86