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Unit 6

6.8 Enthalpies of Formation

3 min readaugust 29, 2020

Anika P

Dalia Savy

Dylan Black


What is an Enthalpy of Formation?

The standard enthalpy of formation (ΔHf) or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements in their most stable state at standard conditions (25°C and 1atm). Essentially, it is the amount of energy it takes to form a compound.
For example, the ΔHf of CO2 can be represented as the ΔH for the following reaction: C + O2 --> CO2. Just break down the product😊! If the product was CO, the equation would be C + 1/2O2 --> CO.
For an element, the standard enthalpy of formation is zero.
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Image Courtesy of Google Sites

Using ΔHf to Calculate ΔHrxn

There is a simple formula to calculate ΔHrxn from ΔHfs:
ΔHrxn = ΣnΔHf(prod) - ΣmΔHf(reactants).
n and m are the stoichiometric coefficients for each reactant and product.
Note that the heat of formation formula is products - reactants, but the bond dissociation energy formula is reactants - products. Figure out a way to ensure you don't confuse the two. Luckily, the heat of formation formula is on the reference table🎉.

Example Problem

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When first doing these questions, try to write down the formula to help you memorize it🧠.
ΔHrxn = ΣnΔHf(prod) - ΣmΔHf(reactants)
Then, simply plug values into the formula to calculate your answer. ΔHrxn = ([8 * -393.5] + [10 * -241.8]) - ([2 * -147.3] + [13 * 0]) = -5271.4 kJ
n and m are seen in this formula by the constants. There are 8 carbon dioxide molecules in the products, this is why -393.5 kJ/mol is multiplied by 8. Same goes for every other compound in this reaction✍️.

Example Problem #2

Use standard enthalpies of formation to find the ΔH of reaction for the following reaction:
CH4(g) + 2O2 (g) --> CO2(g) + 2 H2O (l)
SubstancekJ/mol
CH4 (g)-74.8
O2 (g)0
CO2 (g)-393.5
H2O (g)-241.8
H2O (l)-285.8
Let's write down the formula first!
ΔHrxn = ΣnΔHf(prod) - ΣmΔHf(reactants)
Now, plug in the values but be careful with which number you use for H2O. The table gives you two different standard enthalpies of formation, one for water vapor and one for liquid water. Be sure to use the correct one; the AP creators love to play tricks on us😉. ΔHrxn = ([2 * -285.8] + [1 * -393.5]) - ([1 * -74.8] + [2 * 0]) = -890.3 kJ/mol

Practice AP Question

The following question is from the 2014 AP Chemistry Exam - #6 part c.
In a separate experiment, the student measures the enthalpies of combustion of propene and vinyl chloride. The student determines that the combustion of 2.00 mol of vinyl chloride releases 2300 kJ of energy, according to the equation below.
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Image Courtesy of College Board

Using the table of standard enthalpies of formation below, determine whether the combustion of 2.00 mol of propene releases more, less, or the same amount of energy that 2.00 mol of vinyl chloride releases. Justify your answer with a calculation.
The balanced equation for the combustion of 2.00 mol of propene is 2 C3H6(g) + 9 O2(g) → 6 CO2(g) + 6 H2O(g).
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-wYq3WOwNn5ZX.JPG?alt=media&token=608509d7-25c8-446e-a5bd-96ab8b408d59

Image Courtesy of College Board

Reading all these words tells us that we have to figure out which reaction releases more energy. But in order to compare the two, we must have both heat of reactions, and we only have the heat of reaction for the combustion of vinyl chloride😔.
Seeing standard enthalpies of formation, you should think of ΔHrxn = ΣnΔHf(prod) - ΣmΔHf(reactants) right away, now let's plug in the values!
ΔHrxn = ([6 * -242] + [6 * -394]) - ([9 * 0] + [2 * 21]) = -3858 kJ/mol
Now all you have to do is compare -2300 to -3858 and write a statement that says the combustion of 2.00 mol of propene releases more energy than the combustion of 2.00 mol of vinyl chloride.
You must show your calculations and substitutions into the formula in order to get full credit, especially since the question said "Justify your answer with a calculation."

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