Determine the heat of combustion of ethyl alcohol
(C2H6O).
the right answer is -1238.43 kj/mol. Help with process thank you
2C2H6O + 6O2 ==> 4CO2 + 6H2O
heat comb = (n*dHf CO2 + n*dHf H2O) - (2*dHf ethanol) = ?
dHf = heat formation = delta Hf. These are in the appendix in your text/notes.
To determine the heat of combustion of ethyl alcohol (C2H6O), you need to use the concept of enthalpy of combustion.
The enthalpy of combustion (ΔHcomb) is defined as the heat energy released when one mole of a substance undergoes complete combustion in excess oxygen. In this case, ethyl alcohol will combust and produce carbon dioxide (CO2) and water (H2O) as the products.
To calculate the heat of combustion, you need to follow these steps:
1. Write the balanced chemical equation for the combustion reaction:
C2H6O + O2 → CO2 + H2O
2. Determine the molar masses of each compound involved in the reaction:
Molar mass of C2H6O: 2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
Molar mass of CO2: 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Molar mass of H2O: 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
3. Calculate the change in enthalpy (ΔH) by using the molar masses and the balanced equation:
ΔHcomb = [ΔHf(CO2) + ΔHf(H2O)] - [ΔHf(C2H6O) + ΔHf(O2)]
ΔHf(CO2) = -393.5 kJ/mol (standard enthalpy of formation)
ΔHf(H2O) = -285.8 kJ/mol (standard enthalpy of formation)
ΔHf(C2H6O) = -277.7 kJ/mol (standard enthalpy of formation)
ΔHf(O2) = 0 kJ/mol (oxygen is in its standard state)
ΔHcomb = [-393.5 + (-285.8)] - [-277.7 + 0]
= -678.8 kJ/mol + 277.7 kJ/mol
= -401.1 kJ/mol
4. Convert the heat of combustion from grams to moles by using the molar mass of ethyl alcohol:
-401.1 kJ/mol / 46.07 g/mol = -8.703 kJ/g
Therefore, the heat of combustion of ethyl alcohol (C2H6O) is approximately -8.703 kJ/g or -1238.43 kJ/mol (rounded to two decimal places).