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 process. Given that the heat capacity of the calorimeter is 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 `"mol"^(-1)` is

Answer

`9 kJ mol^(-1)`

Explanation

`q=ms DeltaT=2.5xx0.45=1.125 kJ rArr DeltaE=-1.125/3.5xx28=-9" kJ/mole"`
`|DeltaE|=9" kJ/mole"`
[Since `DeltaH=DeltaE+Deltan_(g) RT` to determine `DeltaH` from `DeltaE` we must know `Deltan_(g)` (i.e. Chemical reaction or Compound) which is not given therefore having no other option answer is 9]