Question

Relation between enthalpy of formation and enthalpy of combustion

Answer 1

Formation is concerned with forming from the constituent elements while combustion is concerned with the change that might occur if you burn it. Sometimes, these two might have equivalent values but they are not the same thing.

In standard terms, enthalpy of formation is defined as the enthalpy change observed when elements combine to create one mole of a compound as measured in their standard states. Take the example of enthalpy of formation of NaCl(s):

Na(s) + 1212 Cl22(g) →→ NaCl(s)  ΔHff= -ve

The enthalpy change observed in the above reaction is the enthalpy of formation of NaCl(s). As you can see, a compound is being made from its constituent elements.

Na(s) + 1212 H22(g) + 1212O22(g) →→ NaOH(s)  ΔHff= -ve

The above example is the enthalpy of formation of NaOH(s). Of course, the above reaction is not possible directly so we take number of easy-to-perform steps which can be demonstrated in a Hess cycle.

Take the enthalpy of formation of CO22(g):

C(s) + O22(g) →→ CO22(g)  ΔHff= -ve …(a)

Now, enthalpy of combustion is concerned with combustion of substances till completion. In standard terms, enthalpy of combustion is defined as the enthalpy change observed when one mole of a substance is burned in excess oxygen as measured in their standard states. Of course, many things may not burn so this value may not exist for many other substances. For example:

Mg(s) + 1212O22(g) →→ MgO(s)  ΔHcc= -ve

The above is the enthalpy of combustion of Mg(s). Similarly:

C(s) + O22(g) →→ CO22(g)  ΔHcc= -ve

The above is the enthalpy of combustion of C(s). Note that the above and equation (a) is numerically equal. That is because formation of CO22(g) and combustion of C(s) is really the same thing. Similarly, formation of MgO(s) and combustion of Mg(s) is really the same thing as well. However, we should take care that they are not the same thing in many cases. Take the following:

CH33CH22OH(l) + 3O22(g) →→ 2CO22(g) + 3H22O(l)  ΔHcc= -ve

The above is the enthalpy of combustion of ethanol. We should know combustion of ethanol and formation of CO22(g) and H22O(l) is not the same thing. This is because we have to consider the enthalpy of formation of ethanol as well, which is:

2C(s) + 3H22(g) + 1212 O22(g) \rightarrow CH33CH22OH(l)  ΔHff= -ve

That is because enthalpy change is the sum of the enthalpy change of the products minus the sum of the enthalpy change of the reactants



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Answer 2

Short answer: Sometimes. But not really.

“Enthalpy of” whatever … is the enthalpy of the transformation (reaction) referred to by the “whatever”.

Enthalpy of formation specifically refers to the reaction where one mole of a compound is formed directly from its elements in their standard states. Note that this kind of enthalpy always refers to the product of the reaction. So, for example,

[1:] 12 N2 (g) + 32 H2 (g) → NH3 (g)

[2:] C (graphite) + O2 (g)→ CO2 (g)

are both standard reactions of formation because the product compounds (NH3 and CO2) are formed from their elements in standard states.

Enthalpy of combustion specifically refers to the reaction where one mole of the compound in question completely combines with oxygen gas. Unlike formation reactions, combustion is referring to one of the reactants (not the product).

Can you see that example #2 above is a standard reaction of combustion, too? One mole of C (graphite) is combining completely with oxygen gas. So the enthalpy of that reaction is both the enthalpy of formation of CO2 and the enthalpy of combustion of C (graphite).

But note from the above that the enthalpy of reaction #2 is not the enthalpy of formation and the enthalpy of combustion of the same substance. That can never be true (which is why I said “not really” in my short answer). It is the enthalpy of formation of something and the enthalpy of combustion of something else.