In a constant volume calorimeter, $3.5 \mathrm{~g}$ of a gas with molecular weight $=28$ was burnt in excess…
In a constant volume calorimeter, $3.5 \mathrm{~g}$ of a gas with molecular weight $=28$ was burnt in excess oxygen at $298.0 \mathrm{~K}$. The temperature of the calorimeter was found to increases from $298.0 \mathrm{~K}$ to $298.45 \mathrm{~K}$ due to the combustion process. Given, that the heat capacity of the calorimeter is $2.5$ $\mathrm{kJ} \mathrm{K}^{-1}$, the numerical value for the enthalpy of combustion of the gas in $\mathrm{kJ} \mathrm{mol}^{-1}$ is
Solution
The temperature rise is : $\Delta T=T_2-T_1=298.45-298=0.45 \mathrm{~K}$
This indicates that heat produced from combustion of $3.5 \mathrm{~g}$ of compound rises temperature of calorimeter by $0.45 \mathrm{~K}$.
Heat produced $=0.45 \mathrm{~K} \times 2.5 \mathrm{k} \mathrm{JK}^{-1}=1.125 \mathrm{~kJ}$
$\Rightarrow$ Heat produced from $28 \mathrm{~g}$ of compound $(1.0 \mathrm{~mol})=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}$