JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Chemical Thermodynamics

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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}

Answer: 9

Solution

Heat released = heat capacity × temperature change =

2.5 \times 0.45 = 1.125

kJ for 3.5 g.

Number of moles =

\frac{3.5}{28} = 0.125

mol.

Enthalpy of combustion per mole =

\frac{1.125}{0.125} = 9

kJ mol⁻¹.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer

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