IIT-JEE 2009 · Paper 2 · Chemistry Q12

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ChemistryInteger+4 / -1Question 12 of 57

In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K $^{-1}$ , the numerical value for the enthalpy of combustion of the gas in kJ mol $^{-1}$ is

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Answer and worked solution

Official answer: 9

Heat released. $q=C\Delta T=2.5\times0.45=1.125$ kJ.

Per mole. $n=\dfrac{3.5}{28}=0.125$ mol, so $\dfrac{1.125}{0.125}=9$ kJ mol⁻¹.

At constant volume this is strictly $\Delta U$ ; the question takes its numerical value (9) as the enthalpy of combustion, neglecting the small $\Delta n_gRT$ correction.

Answer: 9

Topics:Calorimetry

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IIT-JEE 2009 · Paper 2, Chemistry question 12