Question

The standard free energy change $(ΔG°)$ for $50$ dissociation of $N_2O_4$ into $NO_2$ at $27°C$ and 1 atm pressure is $x\ J mol^{–1}$. The value of $x$ is _____. (Nearest Integer)
[Given : $R = 8.31\ JK^{–1} mol^{–1}$, $log\ 1.33 = 0.1239$, $ln\ 10 = 2.3$]

Answer 1

$N_2O_4 ⇋ 2NO_2$
$t = 0$ $1$ $0$
$t = t_{eq.}$ $1 – 0.5$ $2×0.5$

$P_{N_2O_4} = 0.33\ atm$
$P_{NO_2} = 0.66\ atm$
$K_p = \frac {(P_{NO_2})^2}{P_{N_2O_4}}$

$= \frac {(0.66)^2}{0.33}$
$= 1.33$
$△G = -RT\ ln K_p$
$△G = -0.831 \times 300 \times 2.3 \times log\ 1.33$
$△G ≃ 710\ Jmol^{-1}$

So, the correct answer is $710\ Jmol^{-1}.$