Thermodynamics And Thermochemistry Result 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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Answer:
Correct Answer: 15. $(9 \mathrm{~kJ})$
Solution:
- Temperature rise $=T_2-T_1=298.45-298=0.45 \mathrm{~K} $
$ q=\text { heat capacity } \times \Delta T=2.5 \times 0.45=1.125 \mathrm{~kJ} $
$ \Rightarrow \text { Heat produced per mole }=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}$
BETA
