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in Chemical thermodynamics by (64.7k points)

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 mol-1 is 

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Temperature rise = T2 – T1 = 298.45 – 298 = 0.45K

q = heat-capacity × ∆T = 2.5 × 0.45 = 1.125 kJ 

⇒Heat produced per mol = 1.125/3.5 x 28 = 9 kJ

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