Answer:
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= <u>135,583 J</u> = 135.58 kJ
Note we expressed the temperature change in K, because the heat capacity is written in K.
<u>
Now that we have the heat of combustion, we need to calculate the molar heat. </u>
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 <u>conversion factor as 135.58 kJ/ 4.40 g</u>.
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.
