Chemical Energetics (8) Using enthalpy changes of formation to determine enthalpy changes of reaction.

These types of calculation are another way of using Hess’s Law. 

In these calculations there is no need to carry out an experiment. 

Enthalpy changes of reaction can be calculated for hypothetical reactions and those reactions that cannot be carried out directly.

Let’s use a typical reaction as a example first:

Here is the equation for the thermal decomposition reaction of sodium hydrogen carbonate

2NaHCO₃(s)    =   Na₂CO₃(s)    +   H₂O(l)  +   CO₂(g)

The next thing we need to do is find the standard enthalpies of formation of the substances involved in the equation. 

Here they are:

	ΔHfo   kJ/mol
2NaHCO3(s)	2 × –950.8
Na2CO3(s)	–1130.7
H2O(l)	–285.8
CO2(g)	–393.5

Next it is best to construct a Hess Cycle in this example though you will see that in the end a Hess Cycle will not be necessary every time you do this kind of a calculation. 

Here it is:

  

So the next thing to do is to calculate each enthalpy change:

1     is of course the value we are looking for.

2     Is found by multiplying the enthalpy of formation of sodium hydrogen carbonate by 2 (2 moles used in the equation above)  i.e. 2 ×  –950.8  =  –1901.6 kJ/mol.

3     Is found by adding together the values for the three products i.e.

–1130.7  +   –285.8  +   –393.5   =   –1810 kJ/mol.

Next we apply Hess’s law so that 1=3–2 as the enthalpy change must be the same whichever route is taken. 

ΔH_r^o   [2NaHCO₃(s)]  =   –1810 – (–1901.6)  =   + 91.6 kJ/mol

where ΔH_r^o  [2NaHCO₃(s)]  refers to the thermal decomposition of two moles of sodium hydrogen carbonate.

Therefore we can conclude that the energy required to decompose two moles of sodium hydrogen carbonate is 91.6 kJ. 

As this is a thermal decomposition we note that the value is endothermic as expected. 

From this example and many others we can draw a fairly firm conclusion in the form of a general application of Hess’s law to the problem of determining the standard enthalpy of reaction from standard enthalpies of formation.

Or we can apply this law to the effect that

ΔH^o_reaction     =    ΔH_f^o  [Products]  —  ΔH_f^o  [Reactants]