In a constant volume calorimeter, 3.5g of a gas with molecular mass 28 was burnt in excess oxygen at 298.0 K. the temperature of the calorimeter was found to increase from 298.0K to 298.45K, due to the combustion process. Given, that the heat capacity of the calorimeter is `2.5kJ*K^(-1)`, what will be the value of enthalpy of combustion of the gas?

Explanation

For a combustion reaction carried out in a constant volume calorimeter, the amount of liberated heat is given by the relation, `q=C_(cal)xxDeltaT`
`C_(cal)=2.5kJ*K^(-1) and DeltaT=(298.45-298)K=0.45K`
thus, `q=2.5xx0.45kJ=1.125kJ`
For 3.5g off the gas, the number of moles =`(3.5)/(28)=0.125`
Thus, the burning of 0.125 mol of the given gas liberates 1.125 kJ of heat.
Hence, the enthalpy of combustion of the gas is `(1.125)/(0.125)kJ*mol^(-1)=9kJ*mol^(-1)`.