Consider the reaction,

$4{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right),\mathrm{ }{∆}_{\mathrm{r}}\mathrm{H}=-111\mathrm{ }\mathrm{kJ}.$

If ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)$ is formed instead of ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)$ in the above reaction, the ${∆}_{\mathrm{r}}\mathrm{H}$ value will be

(Given, $∆\mathrm{H}$ of sublimation for ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)$is 54 $\mathrm{k}\mathrm{J}\mathrm{ }\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$)

For complete combustion of ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)⟶2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$, the amount of heat produced as measured in bomb calorimeter, is $1364.47 \mathrm{kJ} {\mathrm{mol}}^{-1}$ at $25°\mathrm{C}$. Assuming ideality the enthalpy of combustion, ${\mathrm{\Delta }}_{\mathrm{C}}\mathrm{H}$ reaction will be :

$\left(\mathrm{R}=8.314 \mathrm{kJ} {\mathrm{mol}}^{-1}\right)$

Oxidising power of chlorine in an aqueous solution can be determined by the parameters indicated below:

$\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\stackrel{\frac{1}{2}{\mathrm{\Delta }}_{\mathrm{dis}}{\mathrm{H}}^{\ominus }}{⟶}\mathrm{Cl}\left(\mathrm{g}\right)\stackrel{{\mathrm{\Delta }}_{\mathrm{eg}}{\mathrm{H}}^{\ominus }}{⟶}{\mathrm{Cl}}^{-}\left(\mathrm{g}\right)\stackrel{{\mathrm{\Delta }}_{\mathrm{hyd}}{\mathrm{H}}^{\ominus }}{⟶}{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

The energy involved in the conversion of

$\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)$ to ${\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

(using the data, ${\mathrm{\Delta }}_{\mathrm{diss}}{\mathrm{H}}_{{\mathrm{Cl}}_2}^{\ominus }=240 \mathrm{kJ} {\mathrm{mol}}^{-1}$, ${\mathrm{\Delta }}_{\mathrm{eg}}{\mathrm{H}}_{\mathrm{Cl}}^{\ominus }=-349 \mathrm{kJ} {\mathrm{mol}}^{-1}, {\mathrm{\Delta }}_{\mathrm{hyd}}{\mathrm{H}}_{{\mathrm{Cl}}^{-}}^{\ominus }=-381 \mathrm{kJ} {\mathrm{mol}}^{-1}$) will be:

On the basis of the following thermochemical data:

${\mathrm{\Delta H}}_{\mathrm{f}}°\left({\mathrm{H}}_{\mathrm{aq}}^{+}\right)=0$

$\begin{array}{l}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right); \mathrm{\Delta H}°=57.32\mathrm{kJ}\\ {\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); \mathrm{\Delta H}°=-286.20\mathrm{kJ}\end{array}$

The value of the enthalpy of formation of ${\mathrm{OH}}^{-}$ ions at $25°\mathrm{C}$ is: