The following reaction is performed at $298 \mathrm{~K}$. $2…

The following reaction is performed at $298 \mathrm{~K}$. $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) ightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$
The standard free energy of formation of $\mathrm{NO}(\mathrm{g})$ is $86.6 \mathrm{~kJ} / \mathrm{mol}$ at $298 \mathrm{~K}$. What is the standard free energy of formation of $\mathrm{NO}_{2}(\mathrm{~g})$ at $298 \mathrm{~K} ?\left(K_{p}=1.6 \times 10^{12}ight)$
  1. $86600-\frac{\ln \left(1.6 \times 10^{12}ight)}{\mathrm{R}(298)}$
  2. $0.5\left[2 \times 86,600-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)ight]$
  3. $\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)-86600$
  4. $86600+\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)$

Solution

$\Delta \mathrm{G}_{\mathrm{NOg}}^{\circ}=86.6 \mathrm{~kJ} / \mathrm{mol}=86600 \mathrm{~J} / \mathrm{mol}$
$\mathrm{G}_{\mathrm{NO}_{2}(\mathrm{~g})}^{\circ}=\mathrm{x} \mathrm{J} / \mathrm{mol}$
$\mathrm{T}=298, \mathrm{~K}_{\mathrm{p}}=1.6 \times 10^{12}$
$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{P}}$
Given equation, $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) ightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$
$\therefore \quad 2 \Delta \mathrm{G}_{\mathrm{NO}_{2}}^{\circ}-2 \Delta \mathrm{G}_{\mathrm{NO}}^{\circ}=-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)$
$2 \Delta \mathrm{G}_{\mathrm{NO}_{2}}^{^{\circ}}-2 \times 86600=-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)$
$2 \Delta \mathrm{G}_{\mathrm{NO}_{2}}^{^{\circ}}=2 \times 86600-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)$
$\Delta \mathrm{G}_{\mathrm{NO}_{2}}^{\circ}=\frac{1}{2}\left[2 \times 86600-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight]ight.$
$=0.5\left[2 \times 86600-\mathrm{R}(298) \ln \left(1.6 \times 10^{12}ight)ight]$

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more EQUILIBRIUM questions on Aicharya