Chemistry - Thermodynamics - II Question with Solution | TestHub

$\Delta {\text{G}}_{\text{reac}}^{\circ }=-{\text{2.303 RT log K}}_{\text{P}}$

$=-{\text{RT ln K}}_{\text{P}}$

$=-\text{R}\left(298\right)\text{ln }\left(\text{1.6}\times 10^{12}\right)$

$\Delta {\text{G}}_{\text{reac}}^{\circ }=2\Delta {\text{G}}_{\text{f}\left({\text{NO}}_2\right)}^{\circ }-2\Delta {\text{G}}_{\text{f}\left(\text{NO}\right)}^{\circ }$

$\Rightarrow -\text{R}\left(298\right)\text{ln }\left(\text{1.6}\times {10}^{12}\right)=2\Delta {\text{G}}_{\text{f}\left({\text{NO}}_2\right)}^{\circ }-2\times \text{86.6}\times 10^3$

$\Rightarrow 2\Delta {\text{G}}_{\text{f}\left({\text{NO}}_2\right)}^{\circ }=-\text{R}\left(298\right)\text{ln}\left(\text{1.6}\times 10^{12}\right)+2\times \text{86,600}$

$\Delta {\text{G}}_{\text{f}\left({\text{NO}}_2\right)}^{\circ }=\text{86,600}-\frac{\text{R}\left(298\right)}{2}\text{ln}\left(\text{1.6}\times 10^{12}\right)$

$=\text{0.5}\left[2\times \text{86,600}-\text{R}\left(298\right)\text{ln}\left(\text{1.6}\times 10^{12}\right)\right]$