In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess of oxygen at 298 k. The temperature of the calorimeter was found to Increase from 298 k to 298.45 k due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 KJ/K. The Numerical value for the enthalpy of combustion of the Gas in KJ /Mole is
9
Given, Cv = 2.5 kJ/K = 2500 J/K
ΔT = T2 − T1 = 298.45 − 298 = 0.45K.
ΔH, due to combustion of 3.5g of gas
= Cv × Δ T = 2500 × 0.45 = 1125J
1mole = 28. (Molwt of gas = 28)
Hence, ΔH due to combustion of 1mole of gas
= 11253.5 × 28 = 9000 J = 9 kJ/Mol