In a constant volume calorimeter, `3.5` g of a gas with moleular 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 KJK^(-1)`, the numerical value for the enthalphy of combustion of the gas in KJ `mol^(-1)` is .

Answer

9

Explanation

`n=(3.5)/(28)`
`DeltaT = T_(2)- T_(1) = 298.45 - 298 = 0.45`
`C_(v)=2.5KJ K^(-1) = 2500 JK^(-1)`
`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) = 9030 J mol^(-1) = 9.030 KJ mol^(-1) = 9 KJ mol^(-1).`