Q26 / 90 · JEE Main 2022 (25 Jun Shift 1) · Chemistry

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

Correct answer: 710

Solution

$N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$

$1 - α 2 α$

$K_{p} = \frac{4 α^{2} p}{1 - α^{2}} = \frac{4 \times {(0.5)}^{2} \times 1}{1 - {(0.5)}^{2}} = \frac{1}{0.75}$

$K_{p} = \frac{4}{3}$

${ΔG}^{o} = - 2.303 RT \log K_{p}$

$= - 2.3 \times 8.31 \times 300 [\log \frac{4}{3}] = - 2.3 \times 8.31 \times 300 \times 0.1239$

$= - 710 J$

Hence, the correct answer is 710.