Chemistry - Thermochemistry Question with Solution | TestHub

Given, 

${\mathrm{CH}}_4+2{\mathrm{O}}_2⟶{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}; \mathrm{\Delta H}=-210 \mathrm{kcal}/\mathrm{mol}$    $...\left(\mathrm{i}\right)$

${\mathrm{C}}_2{\mathrm{H}}_6+\frac{7}{2}{\mathrm{O}}_2⟶2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}; \mathrm{\Delta H}=-368 \mathrm{kcal}/\mathrm{mol}$    $...\left(\mathrm{ii}\right)$

On subtracting Eq. $\left(\mathrm{i}\right)$ from Eq $\left(\mathrm{ii}\right)$, we get

${\mathrm{CH}}_2+\frac{3}{2}{\mathrm{O}}_2⟶{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}; \mathrm{\Delta H}=-158 \mathrm{kcal}/\mathrm{mol}$

$\therefore$ Enthalpy of combustion of one ${\mathrm{CH}}_2 \mathrm{unit}$

$=-158 \mathrm{kcal}/\mathrm{mol}$

${\mathrm{\Delta H}}_{\mathrm{comb}}\left({\mathrm{C}}_{10}{\mathrm{H}}_{22}\right)={\mathrm{\Delta H}}_{\mathrm{comb}}\left({\mathrm{CH}}_4\right)+9\times {\mathrm{\Delta H}}_{\mathrm{comb}}\left({\mathrm{CH}}_2\right)$

$=-210+\left(9\times -158\right)$

$=-1632 \mathrm{kcal}$