Question

0.16 g of methane was subjected to combustion at ${27}^{\circ }C$ in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by ${0.5}^{\circ }C$. If the heat of combustion of methane at constant volume and constant pressure is $x$ kJ. The thermal capacity of the calorimeter system is $17.7kJ{K}^{-1}$.
Then $-x$ (nearest integer value) is:

$[R=8.314J{K}^{-1}mo{l}^{-1}]$.

Answer 1

$Heatgivenbyburningof0.16gmethane=heattakenupbycalorimeter=17.7\times 0.5=8.85kJ$

$\therefore Heatgivenbyburningof16gmethane=\frac{8.85\times 16}{0.16}=885kJ$

The burning is at constant volume in bomb calorimeter

$\therefore \Delta U=-885kJmo{l}^{-1}$

$\because \Delta H=\Delta U+\Delta nRT$

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O\left(l\right)$;

$\therefore \Delta n=-2$

$\therefore \Delta H=-885+(-2)\times 8.314\times {10}^{-3}\times 300$

$=-889.98\approx 890kJmo{l}^{-1}$

Hence, $x$ is 890