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

Temperature rise = `T_(2) - T_(1) - T_(1) = 298.45 - 298.0.45 K`
q = heat capacity `xx Delta T = 2.5 xx 0.45 = 1.125 kJ`
`implies "Heat produced per mole" = (1.125)/(3.5) xx 28 = 9 kJ`.