Answer:
5878 kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the calorimeter is zero.
Qcomb + Qcal = 0
Qcomb = -Qcal [1]
The heat absorbed by the calorimeter (Qcal) can be calculated using the following expression:
Qcal = Ccal . ΔT = 3.640 kJ/°C × 19.35°C = 70.43 kJ
where,
Ccal is the heat capacity of the calorimeter
ΔT is the difference in temperature
From [1],
Qcomb = - 70.43 kJ
Since this heat is measured at constant pressure, the molar enthalpy of combustion of cymene is:
[tex]\Delta Hcomb=\frac{-70.43kJ}{1.608g} .\frac{134.21g}{mol} =5878kJ/mol[/tex]
