Thermodynamics and Thermochemistry 2 Question 15
15. 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
(2009)
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Solution:
- Temperature rise $=T_{2}-T_{1}=298.45-298=0.45 \mathrm{~K}$
$q=$ heat capacity $\times \Delta T=2.5 \times 0.45=1.125 \mathrm{~kJ}$
$\Rightarrow$ Heat produced per mole $=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}$
