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]

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. 

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

Chemical Energetics (5) Measuring the enthalpy change of reaction (ii)

This new blog looks at a common experiment that is often used to show you how to calculate an enthalpy of reaction.

The reaction often used is the redox reaction between zinc powder and copper sulphate solution. 

Zn(s)  +   CuSO₄ (aq)    =    Cu(s)  +   ZnSO₄ (aq)

Zinc powder is added gradually to copper sulphate solution in a polystyrene cup (calorimeter).

Polystyrene minimises heat loss from the reaction mixture especially if a lid is added to the cup.

Zinc is added gradually to prevent excess frothing of the mixture and possibly frothing over of the reaction mixture. 

The temperature of the mixture is taken every 10-20s then the zinc is added and a graph of temperature against time is plotted. (see below)   

A colour change from blue solution (due to Cu²⁺ ions) to colourless solution (due to Zn²⁺ ions) is observed since excess zinc is used. 

The reaction mixture is stirred during the addition of the zinc to ensure all the zinc that can react does so and no zinc remains coated in copper.

A brown coloured finely divided solid appears in the solution at the end of the reaction.  

Let’s have a look at a typical set of results from this experiment that you can find in the internet.

Here it is:

So for the Brits among you “styrofoam” is Americanese for expanded polystyrene. 

A “single replacement reaction” is, I think, another Americanese for the displacement or better redox reaction that occurs since zinc being a more reactive metal than copper displaces it from the solution. 

Experimental Analysis

Let’s examine the quantities used to see which reagent is in excess. 

Above you can see that in the chemical equation 1 mole of zinc combines exactly with 1 mole of copper ions. 

Reacting quantities

How many moles of zinc is used in this experiment?

We are told the mass of powdered zinc was 1.3g so the number of moles was 1.3/65 or 0.02moles.

How many moles of copper ions were used in this experiment?

We are told again that 100ml of 0.1M copper sulphate solution was used.

So moles copper ions =  0.1  *100  =  0.01 moles copper ions

                                            1000

So in this reaction the zinc is in excess.

With the zinc in excess we can be sure that all the copper ions will be reduced to copper atoms. 

The enthalpy change will be determined by the amount of copper ions available in the reaction i.e. the limiting molar quantity.

Calculation of enthalpy change

How much energy was evolved during the reaction?

Here we use the equation with which you ought to be familiar that is:

E = m c ΔT

Therefore E  =  mass of solution used *  specific heat capacity of water * temp change

Therefore  E  =  100g   *  4.18 J/g/oC  *   11^oC 

[I don’t think we can get a temp change better than that (2 sig figs) given the lack of fine detail in the graph above]

So    E   =   4598J

But this number of Joules of energy are produced from 0.1moles copper ions

Therefore the enthalpy change of reaction is 4598J   =   45980J 

                                                                                      0.1mol

or –46kJ/mol

This is a poor result when you consider that the enthalpy change is accepted as  –217kJ/mol

You need to ask why this value for the enthalpy change is so low!!

The answer you ought to come up with is probably to do with poor insulation of the reaction in the cup.  May be no lid was used.

And may be the zinc was poor quality and some was already oxidised so not enough actual zinc atoms went into the reaction with copper ions. 

Here is another set of actual results for you to determine the enthalpy of reaction for yourself:

25.00cm³ of 1.00M copper(II) sulphate solution was placed in a polystyrene cup.  About 6g of powdered zinc were added which is an excess of zinc.  A temperature rise of 50.6^oC was recorded.

Can you show that the enthalpy change measured in this experiment is

 –212kJ/mol?

.