A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat capacity of 490 J/K, and the calorimeter contained 730 g of water. Burning 4.40 g of ethanol, resulted in a rise in temperature from 16.8 °C to 20.5 °C. Calculate the enthalpy of combustion of ethanol, in kJ/mol. (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.) Enthalpy of combustion = kJ/mol

The enthalpy of combustion of ethanol in kJ/mol is - 1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write:

qcal = Ccal * ΔT = 490 J/K * 276.7 K = 135,583 J = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

Now that we have the heat of combustion, we need to calculate the molar heat.

Because qsystem = qrxn + qcal and qrxn = - qcal, the heat change of the reaction is - 135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol; therefore, we can write the conversion factor as 135.58 kJ / 4.40 g.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is:

molar heat of combustion = - 135.58 kJ/4.40 g x 46.07 g / 1 mol = - 1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is - 1419.58 kJ/mol.