The standard enthalpies of formation of ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ $,{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ and glucose (s) at $25^{\text{o}}C$ are $-\text{400 kJ}/\text{mol}$, $-\text{300 kJ}/\text{mol}$ and $-\text{1300 kJ}/\text{mol}$, respectively. The standard enthalpy combustion per gram of glucose at $25^{\text{o}}C$ is 

(${\mathrm{\Delta H}}_{\mathrm{f}}^0=-1300\mathrm{kJ}{\mathrm{mol}}^{-1}$ for Glucose)

($\mathrm{U}$ : internal energy, $\mathrm{S}:$ entropy, $\mathrm{p}:$ pressure, $\mathrm{V}:$ volume, $\mathrm{R}$ : gas constant) 

(Given: molar heat capacity at constant volume, ${\mathrm{C}}_{\mathrm{V},\mathrm{m}}$ of the gas is $\frac{5}{2}\mathrm{R}$)

Give your answer as the nearest integer.

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in $\mathrm{K}$) at which the reduction of cassiterite by coke would take place.

$298 \mathrm{K}: {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{SnO}}_2\left(\mathrm{s}\right)\right)=-581.0 \mathrm{kJ} {\mathrm{mol}}^{-1}, {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)=-394.0 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{S}}^0\left({SnO}_2\left(\mathrm{s}\right)\right)=56.0 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}, {\mathrm{S}}^0\left(\mathrm{Sn}\right(\mathrm{s}\left)\right)=52.0 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}$

${\mathrm{S}}^0\left(\mathrm{C}\right(\mathrm{s}\left)\right)=6.0 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}, {\mathrm{S}}^0\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)=210.0 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

Assume that the enthalpies and the entropies are temperature independent.

The surface of copper gets tarnished by the formation of copper oxide. ${\mathrm{N}}_2$ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the ${\mathrm{N}}_2$ gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:

$2\mathrm{Cu}\left(\mathrm{s}\right)+\mathrm{ }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\to \mathrm{ }{\mathrm{Cu}}_2\mathrm{O}\left(\mathrm{s}\right)\mathrm{ }+\mathrm{ }{\mathrm{H}}_2\left(\mathrm{g}\right)$

${\mathrm{P}}_{{\mathrm{H}}_2}$ Is the minimum partial pressure of ${\mathrm{H}}_2$ (in bar) needed to prevent the oxidation at $1250\mathrm{K}$. The magnitude of value of $\ln \left({\mathrm{P}}_{{\mathrm{H}}_2}\right)$ is ____.

(Given: total pressure = 1 bar, R (universal gas constant) $=\mathrm{ }8\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{K}}^{-1}\mathrm{ }{\mathrm{mol}}^{-1},\ln \left(10\right)\mathrm{ }=\mathrm{ }2.3.\mathrm{ }\mathrm{Cu}\left(\mathrm{s}\right)\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{Cu}}_2\mathrm{O}\left(\mathrm{s}\right)\mathrm{ }$ are mutually immiscible.

At $1250\mathrm{ }\mathrm{K}:\mathrm{ }2\mathrm{Cu}\left(\mathrm{s}\right)+\frac{1}{2}\mathrm{ }{\mathrm{O}}_2\left(\mathrm{g}\right)\to \mathrm{ }{\mathrm{Cu}}_2\mathrm{O}\left(\mathrm{s}\right);\mathrm{ }∆{\mathrm{G}}^{⊝}\mathrm{ }=\mathrm{ }-\mathrm{ }78,000\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{mol}}^{-1}$

${\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to \mathrm{ }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right); ∆{\mathrm{G}}^{⊝}\mathrm{ }=\mathrm{ }-\mathrm{ }1,78,000\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{mol}}^{-1};$) 

Round off the answer up to the nearest integer.