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)`, find the numerical value for the enthalpy of combustion of the gas in `kJ mol^(-1)`

Answer

9

Explanation

`n=(3.5)/(28)`
`DeltaT=T_(2)-T_(1)=298.45-298=0.45`
`C_(P)=C_(V)+R=2500+8.314=2508.314 JK^(-1)`
`Q_(P)=C_(P)DeltaT=1128.74 J`
`DeltaH=(Q_(P))/(n)=(1128.74)/(3.5//28)`
implies 9030 J `mol^(-1)`=9.030 kJ `mol^(-1)` = 9 kJ `mol^(-1)`.