NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
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 m o l - 1 is
Correct answer
9
Step-by-step solution
Energy released by combustion of 3.5 g gas = 2.5 × ( 298.45 - 298 ) k J Energy released by 1 mole of gas = 2.5 × 0.45 3.5 28 = 9 k J m o l - 1