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

`:' C_(P)-C_(v)=R`
`C_(p)=C_(v)+R`
`C_(p)=2500+8.314=2508.314 JK^(-1)`
(`:' C_(v)=2.5 kJ`)
`q_(p)=C_(p)xxDeltaT ( :. DeltaT=0.45)`
`=2508.314xx0.45=1128.74 J`
Now `DeltaH=(q_(p))/n=1128.74/(3.5//28)~~9.0299 k J mol^(-1)`