Chemical Energetics (6) Introducing Hess's Law

Measuring enthalpy change using Hess’s Law.

How can we measure the enthalpy change of a reaction that we can’t actually carry out in the lab?

The enthalpy change of this reaction cannot be easily measured in the laboratory:

MgSO₄(s)   +    7H₂O(l)   =   MgSO₄.7H₂O(s)

white powder                       clear crystals

The reason is that the process is very difficult to control precisely. 

However, what we can do is measure the enthalpy of solution of both clear magnesium sulphate crystals and white anhydrous magnesium sulphate powder. 

Using Hess’s Law we can then work out the enthalpy change above. 

What is Hess’s Law?

Hess’s Law is a form of the First Law of Thermodynamics (there are three laws of thermodynamics the Zeroth Law, the First Law and the Second Law!!). 

The First Law of Thermodynamics can be stated in several different ways e.g. the mass and energy content of the Universe is constant or energy can be neither created nor destroyed. 

Hess’s Law puts the Frist Law of Thermodynamics like this: The energy change from reactants to products in a reaction is independent of the route taken to produce those products. 

Here is a typical diagram you’ll find on the Internet that is supposed to illustrate Hess’s Law:

This diagram needs some explanation.

The enthalpy change ΔH₁ carries the same value as that of ΔH₂ and ΔH₃ combined or ΔH₄, ΔH₅ and ΔH₆ combined.

That is the implication of Hess’s Law. 

Applying Hess’s law

Let’s apply Hess’s law to our problem the determination of the enthalpy of reaction for the hydration of anhydrous magnesium sulphate powder. 

Let’s measure the heat of solution of both salts, the hydrated salt and the anhydrous salt.

Using 0.0250mol of each salt then the reactions require slightly different amounts of water to produce the same concentration of solutions since the hydrated salt contains a given number of moles of water in its crystals

In each case we measure the greatest temperature change.

The anhydrous salt dissolves exothermically but the hydrated salt dissolves endothermically.

The Hess cycle will look like this:

So ΔH₁ =  ΔH₂–ΔH₃

The negative sign reverses the value of ΔH₃ so that the two routes are equivalent to each other. 

Here is a typical set of results:

For the enthalpy of solution of the hydrated crystals (MgSO₄ .7H₂O):

Mass of water 41.85g

Amount of crystals: 0.0250mol

Temperature change: –1.4^oC

Applying   E = m c ΔT

Energy taken in   =   41.85g  *  4.18J/g/^oC  *  1.4^oC

So energy taken in  =   +240J 

Therefore enthalpy of solution ΔH₃   =   +240J            =    +9600J

                                                                0.0250mol

For the enthalpy of solution of the anhydrous salt  (MgSO4(s))

Mass of water 45.00g

Amount of crystals: 0.0250mol

Temperature change: 11.3^oC

Applying   E = m c ΔT

Energy taken in   =   45.00g  *  4.18J/g/^oC  *  11.3^oC

So energy taken in  =   2120J 

Therefore enthalpy of solution ΔH₂   =   –2120J          =    –84800J

                                                                0.0250mol

So if  ΔH₁ =  ΔH₂–ΔH₃   then

  ΔH₁   =   –84800 – 9600   =    – 94400 J/mol  or –94.4kJ/mol