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# 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.

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)    Δr1 (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.

From Hess’ law, we have, Δr1 = Δr2 + Δr3

Lets explore Δr2 + Δr3 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,

Δr2 = – [sum of Std enthalpies of formation of reactants].

Δr2 = ΣΔr (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,

Δr3 = – [sum of Std enthalpies of formation of products].

Δr3 = ∑ Δr (prod)

So now, we can write:

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

Δr1 = f(CO2) + 2 Δf(H2O)] – [Δf(CH4) + 2 Δf(O2)]

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

Δr1 = [(-393.5 kJ) + 2 (-285.8 kJ)] – [(-74.7 kJ) + 2 (0)]

Δr1 = -890.4 kJ

In general, we can write:

Δr = ∑p Δr(prod) – ∑rΔr(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)?

Δr = ∑ Δf(prod) – ∑ Δf(reactants)

Δr = [4 Δf(NO) + 6 Δf(H2O)] – [4 Δf(NH3) + 5 Δf(O2)]

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

Δr = ΣΔf(prod) – ΣΔf(reactants)
Δr = [5/2Δf(B2O3) + 9/2 Δf(H2O)] – [Δf(B5H9) + 6 Δf (O2)]
Δr = [5/2 (-1272.77) + 9/2 (-285.83)] – [(73.2) + 6 (0)]
Δr = -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)

Solve for the only unknown:

ΔfHº (C6H6,l) = 49 kJ/mol