Standard Enthalpies of Formation

The Standard Enthalpy of formation is defined as the enthalpy change of a standard formation reaction.     We saw the definition of the reaction type earlier.  Now let’s explore how we can use these enthalpy change values.

Here are a few reactions that can be defined as standard formation reactions and whose enthalpy changes are useable as standard enthalpies of formation.

C(s,graphite) + O2(g) → CO2	ΔfHº (CO2) = –393.5 kJ/mol
H2(g) + 1/2 O2 → H2O(l)	ΔfHº (H2O) = –285.8 kJ/mol
C(graphite) + 2H2(g) → CH4(g)	ΔfHº (CH4) = –74.7 kJ/mol

Note that the phase of standard state carbon is solid graphite, not just solid.  That’s because there are more than one solid phase of carbon, for example, diamond is solid carbon.  In the cases where there are more than one allotrope of an element, the most stable one is defined to be the standard state.

Question: What will be the standard enthalpy of formation of an element in its standard state [like O₂(g)]?

Answer: Zero. Consider the reaction that is being described in the question.  It would look like this:  O₂(g) → O₂(g).  Clearly there is no reaction as the product and the reactant are the same so no enthalpy change has actually occurred.  We can generalize this discussion to say:  the standard enthalpy of formation of any element in its standard state is zero.

Now let’s use these with Hess’ law to determine the reaction enthalpy for the following reaction.

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(l)    Δ_rHº₁ (combustion of methane)

According to Hess’ law, we could chose to break down the reactants to their constituent elements in their standard state and then reconstruct them into the products.

CH4(g) + 2 O2(g) → C(s) + 2H2(g) + 2O2(g)	ΔrHº2
C(s) + 2H2(g) + 2O2(g) → CO2(g) + 2H2O(l)	ΔrHº3

From Hess’ law, we have, Δ_rHº₁ = Δ_rHº₂ + Δ_rHº₃

Lets explore Δ_rHº₂ + Δ_rHº₃ a bit further.

In step 2, we broke the reactants (methane and oxygen) into their constituent elements at standard state.  This is the reverse of the formation of these reactants so the total enthalpy change is simply the negative of the sum of the standard enthalpy of formation for the reactants (multiplied, of course by the appropriate coefficients to give us the correct stoichiometry of the desired reaction).  Thus,

Δ_rHº₂ = – [sum of Std enthalpies of formation of reactants].

Δ_rHº₂ = ΣΔ_rHº (react)
In step 3, we took elements in their standard state and formed the products.  Thus, the  total enthalpy change is simply the sum of the standard enthalpy of formation for the products (multiplied, of course by the appropriate coefficients to give us the correct stoichiometry of the desired reaction). Thus,

Δ_rHº₃ = – [sum of Std enthalpies of formation of products].

Δ_rHº₃ = ∑ Δ_rHº (prod)

So now, we can write:

For the combustion of methane we’re dealing with here, we can write

Δ_rHº₁ = [Δ_fHº(CO₂) + 2 Δ_fHº(H₂O)] – [Δ_fHº(CH₄) + 2 Δ_fHº(O₂)]

We can look up the values for Heat of formation in any standard reference book.

Δ_rHº₁ = [(-393.5 kJ) + 2 (-285.8 kJ)] – [(-74.7 kJ) + 2 (0)]

Δ_rHº₁ = -890.4 kJ

In general, we can write:

Δ_rHº = ∑_p Δ_rHº(prod) – ∑_rΔ_rHº(react)

Note that the coefficients n used here are unitless because they represent mole ratios, not numbers of moles. This allows the final enthalpy change values to have units of kJ/mol, as they must have.  Note that if one or more of the reactants or products are not in their standard state, we simply can remove the superscript ‘not’ from the final enthalpy change symbol and the same equation will still work.

This procedure is merely a special application of Hess’ Law whereby we are adding the enthalpies of formation in a way that gives us the overall enthalpy change.

Let’s try a few examples to get some practice with these calculations:

What is the standard enthalpy change for

4 NH₃(g) + 5 O₂(g) → 4NO(g) + 6 H₂O(l)?

Δ_rHº = ∑ Δ_fHº(prod) – ∑ Δ_fHº(reactants)

Δ_rHº = [4 Δ_fHº(NO) + 6 Δ_fHº(H₂O)] – [4 Δ_fHº(NH₃) + 5 Δ_fHº(O₂)]

Δ_rHº = [4 (90.25 kJ/mol) + 6 (-285.83 kJ/mol)] – [4 (-46.11 kJ/mol) + 5 (0)]
Δ_rHº = -1169.54 kJ/mol

Now here’s another one.

B₅H₉(g) reacts exothermically with O₂ to form B₂O₃(s) and water. What is the standard enthalpy change for the reaction of 1 mol of B₅H₉(g)?

balance this for 1 mol B5H9	B5H9(g) + O2(g)	B2O3(s) + H2O(l)
1 B5H9(g) + 6 O2(g)	5/2 B2O3(s) + 9/2 H2O(l)

Δ_rHº = ΣΔ_fHº(prod) – ΣΔ_fHº(reactants)
Δ_rHº = [5/2Δ_fHº(B₂O₃) + 9/2 Δ_fHº(H₂O)] – [Δ_fHº(B₅H₉) + 6 Δ_fHº (O₂)]
Δ_rHº = [5/2 (-1272.77) + 9/2 (-285.83)] – [(73.2) + 6 (0)]
Δ_rHº = -4541.4 kJ/mol

And another: 1 mol of benzene burns in air at standard state conditions and gives off 3267 kJ of heat. What is the enthalpy of formation of C₆H₆?

C₆H₆(‘) + 7.5O₂(g) → 6 CO₂(g) + 3 H₂O(l)

ΔcombHº	= -3267 kJ/mol
ΔcombHº	= ΣΔfHº(prod) – ΣΔfHº(reactants)
-3267 kJ/mol	= [6 ΔfHº (CO2,g)  + 3 ΔfHº (H2O,l)] – [ΔfHº (C6H6,l) + 7.5 ΔfHº(O2,g)]
Substitute in values from data table.
-3267 kJ/mol	= [6 (-393.5)  + 3 (-285.83)] – [ΔfHº(C6H6,l) + 7.5 (0)]

 Solve for the only unknown:

Δ_fHº (C₆H₆,l) = 49 kJ/mol