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)`

Options

A. 8
B. 7
C. 9
D. 6

Answer

C

Explanation

Heat produced from the combustion of `3.5 g` of the gas can be calculated as follows:
`q = C Delta T`
`= (2.5 kJ K^(-1)) (298. 45 K - 298 K)`
`= 1.125 kJ`
Enthalpy of combustion of the gas will be equal to heat Produced when 1 mol of gas undergoes combustuion. According to unitary method,
`Delta_(C ) H = (1.125 kJ)/(3.5 g) xx 28 g mol^(-1)`
`= 9 kJ`