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 kJmol -1 is

Question

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 kJmol -1 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy release at constant volume due to combustion of 3.5 gm of a gas =2.5 × 0.45 Hence energy released due to the combustion of 28 gm (i.e., 1 mole) of a gas =2.5 × 0.45 × (28/3.5)=9 kJmol -1

Energy release at constant volume due to combustion of 3.5gm of a gas =2.5×0.45 Hence energy released due to the combustion of 28gm (i.e., 1 mole) of a gas =2.5×0.45×3.528​=9kJmol−1

Energy release at constant volume due to combustion of $3.5 g m$ of a gas $= 2.5 \times 0.45$
Hence energy released due to the combustion of $28 g m$ (i.e., 1 mole) of a gas
$= 2.5 \times 0.45 \times \frac{28}{3.5} = 9 k J m o l^{- 1}$