Chemical Energetics (7) Using Hess’s Law to determine Enthalpy Changes.

You have probably begun this part of the energetics topic using enthalpies of combustion to determine an enthalpy of formation in a reaction in organic chemistry.

The reaction will have been one that you cannot measure directly such as the formation of butane from carbon and hydrogen i.e.

4C (s)  +  5H₂ (g)    =     C₄H₁₀ (g)

Here the enthalpy of formation of butane is the heat evolved or taken in when one mole of butane is formed from its elements in their standard states under standard conditions.

You could pick any organic compound and carry out this calculation. 

Why not do that when I’ve been through this calculation and shown you how it is done?

The reaction cannot take place in practice since hydrogen and carbon do not react together directly. 

However it is fairly straightforward task to measure the standard enthalpies of combustion of the elements and compounds involved here.

The data is given to you in exam questions or you can find the values in a book of data or on the Internet.  

Substance          Formula             ΔH^o_combustion (kJ/mol)

Carbon               C(graphite)        –393.5

Hydrogen           H₂ (g)                 –285.8

Butane               C₄H₁₀(g)             –2876.5

The next thing you have to do is to build a Hess Cycle like this:

If you are wondering where the oxygen molecules have gone, I wouldn’t add them as it makes the cycle too cluttered and as the reactions are carried out using excess oxygen you can be sure there is enough to go round as it is in excess. 

Each letter stands for a heat of combustion or reaction.

Let’s look at each in turn. 

Be careful here because this is THE place mistakes are made.

A: This is the value we are going to calculate: the heat of formation of butane.

B: This is the heat change when 4 moles of carbon burn – notice the “4” in the equation.  So this value is  4* –393.5 = – 1574 kJ/mol

C: This is the heat change when 5 moles of hydrogen burn again notice the “5”in the equation.  So this value is 5*–285.8 = ­–1429 kJ/mol.

D: This is the heat change when one mole of butane burns so the value is –2876.5kJ/mol

We can now calculate A the enthalpy of formation of butane.

The enthalpy change A is equal to B  +  C  – D 

We must follow the flow of the arrows

So A  =  –1574  +   (– 1429)  –  (– 2876.5) 

You will notice that I have separated the values using brackets to avoid mixing up the signs.

If you are going to make a mistake it could well be here.

If we remove the brackets then the calculation becomes:

A  =  –1574  – 1429  + 2876.5  =  –126.5kJ/mol

You can then check your answer against the value in a data book or on the Internet.

So on the Internet we have –125.6 kJ/mol that’s from Wikipedia! (is that a misprint?)

On the NIST site we also have the value –125.6 kJ/mol here

So maybe Wikipedia is correct after all and it is my Nuffield book of Data that is out of date. 

You need now to test your self that you can carry out a similar calculation using Hess’s Law for a difference organic molecule.