What is the enthalpy change for the given reaction if enthalpies of formation of ${\mathrm{Al}}_2{\mathrm{O}}_3$ and ${\mathrm{Fe}}_2{\mathrm{O}}_3$ are $-1670 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and $-834 \mathrm{kJ} {\mathrm{mol}}^{-1}$ respectively ? ${\mathrm{Fe}}_2{\mathrm{O}}_3+2\mathrm{Al}\to {\mathrm{Al}}_2{\mathrm{O}}_3+2\mathrm{Fe}$

An athlete is given $100 \mathrm{g}$ of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ for energy. This is equivalent to $1800 \mathrm{kJ}$ of energy. The $50\%$ of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is $\mathrm{\_\_\_\_\_g}$ (Nearest integer) Assume that there is no other way of consuming stored energy.

Given : The enthalpy of evaporation of water is $45 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Molar mass of $\mathrm{C},\mathrm{H}\&\mathrm{O}$ are $12.1$ and $16 \mathrm{g} {\mathrm{mol}}^{-1}$.

The change in enthalpy $\left[\Delta \mathrm{H}\right]$ in ${\mathrm{kJmol}}^{-1}$ for the reaction is $\mathrm{Mg}+2 \mathrm{F}\to {\mathrm{MgF}}_2$

Given: $\mathrm{EA}$ of $\mathrm{F}=328 \mathrm{kJ} {\mathrm{mol}}^{-1},{\mathrm{IE}}_1$ of $\mathrm{Mg}=737 \mathrm{kJ} {\mathrm{mol}}^{-1},{\mathrm{IE}}_2$ of $\mathrm{Mg}=1451 \mathrm{kJ} {\mathrm{mol}}^{-1}$

The Born-Haber cycle for $\mathrm{KCl}$ is evaluated with the following data:
${\mathrm{\Delta }}_f{\mathrm{H}}^{\mathrm{\Theta }}\text{ for }\mathrm{KCl}=-436.7{\mathrm{kJmol}}^{-1};{\mathrm{\Delta }}_{\mathrm{sub}}{\mathrm{H}}^{\mathrm{\Theta }}\text{ for }\mathrm{K}=89.2{\mathrm{kJmol}}^{-1};$

${\mathrm{\Delta }}_{\text{ionization }}{\mathrm{H}}^{\mathrm{\Theta }}\text{ for }\mathrm{K}=419.0\mathrm{kJ}{\text{ mol }}^{-1};{\mathrm{\Delta }}_{\text{electron gain }}{\mathrm{H}}^{\mathrm{\Theta }}\text{ for }{\mathrm{Cl}}_{\left(\mathrm{g}\right)}=-348.6{\mathrm{kJmol}}^{-1}$

${\mathrm{\Delta }}_{\text{bond }}{\mathrm{H}}^{\mathrm{\Theta }}\text{ for }{\mathrm{Cl}}_2=243.0{\mathrm{kJmol}}^{-1}$
The magnitude of lattice enthalpy of $\mathrm{KCl}$ in ${\mathrm{kJmol}}^{-1}$ is (Nearest integer)

For one mole of NaCl(s) the lattice enthalpy is

${\mathrm{Na}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\stackrel{+108.4 \mathrm{kJ}/\mathrm{mol}}{⟶}{\mathrm{Na}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\stackrel{+495.6 \mathrm{kJ}/\mathrm{mol}}{⟶}{\mathrm{Na}}_{\left(\mathrm{g}\right)}^{+}+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\stackrel{121 \mathrm{kJ}/\mathrm{mol}}{⟶}$

${\mathrm{Na}}_{\left(\mathrm{g}\right)}^{+}+\mathrm{Cl}\left(\mathrm{g}\right)\stackrel{-348.6 \mathrm{kJ}/\mathrm{mol}}{⟶}{\mathrm{Na}}_{\left(\mathrm{g}\right)}^{+}+{\mathrm{Cl}}_{\left(\mathrm{g}\right)}^{-}\stackrel{{\mathrm{\Delta H}}^0 \mathrm{altice} }{⟶}\mathrm{Nacl}$ (solid) $\underset{-411.2 \mathrm{kJ}/\mathrm{mol}}{\leftarrow }{\mathrm{Na}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

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)