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 K to 298 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K- 1 , the value for the enthalpy of combustion of the gas 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 K to 298 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K- 1 , the value for the enthalpy of combustion of the gas is (A) 90 kJ mol- 1 (B) 9 kJ mol- 1 (C) 18 kJ mol- 1 (D) 6 kJ mol- 1

Tardigrade Admin · Accepted Answer

Energy released by combustion of 3.5 g gas =2.5× (.298.45-298) kJ Energy released by 1 mole of gas =(2.5 × 0.45/3.5 / 28)=9 kJmol- 1

90 kJ $m o l^{- 1}$

9 kJ $m o l^{- 1}$

18 kJ $m o l^{- 1}$

6 kJ $m o l^{- 1}$

Energy released by combustion of 3.5 g gas

$= 2.5 \times (298.45 - 298) k J$

Energy released by 1 mole of gas $= \frac{2.5 \times 0.45}{3.5/28} = 9 k J m o l^{- 1}$

90 kJ mol−1

9 kJ mol−1

18 kJ mol−1

6 kJ mol−1

Energy released by combustion of 3.5 g gas =2.5×(298.45−298)kJ Energy released by 1 mole of gas =3.5/282.5×0.45​=9kJmol−1

90 kJ $m o l^{- 1}$

9 kJ $m o l^{- 1}$

18 kJ $m o l^{- 1}$

6 kJ $m o l^{- 1}$

Energy released by combustion of 3.5 g gas

$= 2.5 \times (298.45 - 298) k J$

Energy released by 1 mole of gas $= \frac{2.5 \times 0.45}{3.5/28} = 9 k J m o l^{- 1}$