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)`

In a constant volume calorimeter, `3.5 g` of a gas with molecular weig

1.	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» Number of moles of gas `= (3.5)/(28) = 0.125` Heat evolved = Heat capacity `xx DeltaT` `= 2.5 xx (298.45 - 298)` `= 2.5 xx 0.45 = 1.125 kJ` Heat evolved per mole `= (1.125)/(0.125) = 9 kJ`

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