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 increases 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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n = 
Δ T = T 2 – T 1 = 298.45 – 298 = 0.45
C V = 2.5 kJ k –1 = 2500 JK –1
C P = C V + R = 2500 + 8.314 = 2508.314 JK –1
Q P = C P Δ T = 1128.74 J
Δ H = 
= 9030 J mol –1 = 9.030 KJ mol –1 = 9 KJ mol –1 .
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