The internal energy of a system is the total energy it contains, while its enthalpy is the sum of its internal energy and pressure times volume. You can't measure internal energy or enthalpy directly, but you can measure and calculate change in enthalpy. If pressure is constant and a system does no mechanical work, the change in enthalpy is just the heat absorbed or released by the reaction. Better yet, you don't even need to do an experiment to figure out what the change in enthalpy will be. You can calculate the change in enthalpy of a combustion reaction using the enthalpy of formation.
Things You'll Need
Write out the chemical equation for the combustion reaction and balance it. A combustion reaction is always similar to the following generic equation: Fuel + O2 ---> CO2 + H2O. If you need a refresher on how to balance equations of chemical reactions, click on the second link under the Resources section below.
Example: Propane has the molecular formula C3H8. Substituting it into the generic combustion equation yields the following: C3H8 + O2 ---> CO2 + H2O. This equation is not balanced, so to balance it you first multiply the number of CO2 on the right by 3 to balance the number of carbons. Next, you multiply the number of waters on the right by 4 to balance the number of hydrogens. This gives you the following: C3H8 + O2 ---> 3 CO2 + 4 H2O.
There are now 10 oxygens on the right side, so multiplying the O2 on the left side by 5 balances the oxygens, giving you the following final answer: C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
Find the "fH° of your fuel using the NIST page in the Resources. "fH° is the standard enthalpy of formation. With propane, for example, this number is -104.7 kilojoules per mole (kJ/mol).
Note that any element by itself in its most stable form at room temperature has a "fH° of 0. Consequently, you can ignore the "fH° of oxygen in the chemical equation.
Recall that the enthalpy of a reaction is equal to the sum of the "fH° of the products - the sum of the "fH° of the reactants. Note that the "fH° of CO2 is -393.51 kJ/mol, while the "fH° of water is -285.83 kJ/mol.
Multiply the "fH° of water by the number of water molecules in the balanced chemical equation. In the case of propane, for example, there are four molecules of water on the product side, so you multiply the "fH° of water to obtain the following: (-285.83 kJ/mol) x (4 moles) = -1143.32 kJ.
Multiply the "fH° of CO2 by the number of molecules of CO2 in the balanced equation. In the case of propane, for example, there are three molecules of CO2 on the product side, so you multiply the "fH° of CO2 to obtain the following: (-393.51 kJ/mol) x (3 moles) = -1180.53 kJ.
Add your results from Steps 6-7 to obtain the total "fH° of the products. In the case of propane, this is as follows: -1180.53 + -1143.32 = -2323.85 kJ.
Subtract the "fH° of your fuel from the result in the previous step. In the case of propane, you would have the following: -2323.85 kJ + 104.7 kJ = -2219.15 kJ / mol,
meaning that you should get roughly this amount of energy per mole of propane you burn.
Tips & Warnings
- NIST often lists the enthalpy of combustion if it has been determined experimentally. You can compare their results with your calculations to see how close you come to the true value. In the example, you were almost bang-on.
- "Chemical Principles: The Quest for Insight"; Peter Atkins, et al.; 2008
- University of Waterloo: Enthalpies of Reactions
- Photo Credit Jupiterimages/Photos.com/Getty Images
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