Entropy (2) Determination of ΔStotal for the reaction: magnesium burning in air

Edexcel A level Chemistry (2017)

Topic 13B: Entropy:

Here are the learning objectives to do with calculation of ΔS_total:

13B/16. be able to calculate the entropy change for the system, ΔSsystem , in a reaction, given the entropies of the reactants and products

13B/17. be able to calculate the entropy change in the surroundings, and hence ΔStotal , using the expression:

13B/15. understand that the total entropy change in any reaction is the entropy change in the system added to the entropy change in the surroundings, shown by the expression:

ΔStotal = ΔSsystem + ΔSsurroundings

Determination of ΔS_total  for the reaction magnesium burning in air

First, let’s calculate the entropy change in the system when magnesium burns in air:

We need the standard molar entropies (at 298K) of the elements and compounds involved.

The element molar entropies are per mole of atoms so the oxygen value is for half a mole of oxygen molecules because oxygen exists as diatomic molecules.

Magnesium:                S^o [Mg(s)] =  +32.7 Jmol^—1K^—1

Oxygen:                       S^o [½_(g)]=  +102.5 Jmol^—1K^—1

Magnesium Oxide      S^o [MgO (s)] =  +26.9 Jmol^—1K^—1

Equation:           2Mg(s)   + O₂(g)          ⟶    2MgO(s)

Entropies:          2×+32.7     2×+102.5           2×+26.9

_ΔSsystem  =  S^o_products  —   S^o  _reactants

ΔS_system =    2  × +26.9    —    (2×  +32.7 +  2× +102.5)

ΔS_system = —216.6 Jmol^—1K^—1

But what does this result mean?

The fact that the entropy of the system is negative means that the system decreases in entropy or that it moves from a disordered to a more ordered state because there is a fall in the number of moles of substances from 3 to 2 and a move in state from solid and gas to just solid.

If only the change in the entropy of the system determined whether the reaction proceeded or not then it would not go.

But we must not only consider the change in entropy of the system but also the change of entropy in the surroundings.

Entropy changes in the surroundings because the reaction is really exothermic—the energy causes a change in the energy quanta of all the molecules in the surroundings

These molecules become more disordered if the reaction is exothermic. 

The greater the exothermic character of the reaction the greater the entropy change.

So we have to change the sign of the enthalpy change putting a negative sign before the relationship.

But the higher the temperature of the surroundings the less effect a given enthalpy change has at that temperature.

This thinking suggests that entropy change varies with the inverse of temperature

Therefore

So to calculate ΔS_surroundings, first you have to calculate the enthalpy change required.

If

2Mg(s)    +    O₂(g)  ⟶    2MgO(s)

then we need the enthalpy change of formation of magnesium oxide at 298K

ΔH^o_f [MgO(s)]   =   —601.7 kJ.mol^—1

And so

ΔH^o_reaction   =   2   ×  —601.7 kJ.mol^—1

Don’t forget to rationalise the units of entropy (J.mol^—1.K^—1) with enthalpy (kJ.mol^—1) so the enthalpy change of reaction needs multiplying by 1000.

Then add in the temperature change in Kelvins.

 =   +4038 J.mol^—1.K^—1

Now we can calculate the total entropy change ΔS_total

ΔS_total   =   ΔS_system   +   ΔS_surroundings

Now we have already calculated

ΔS_system   =   —216 J.mol^—1.K^—1

                           ΔS_surroundings =  + 4038 J.mol^—1.K^—1

So

ΔS_total   =   —216.6    +   +4038

=  +3821 J.mol^—1.K^—1

And the meaning of this positive value is that the reaction ought to proceed at room temperature and go to completion and of course we see that it does. 

In other words, the reaction is spontaneous.

What we don’t know from this value however is just how fast the reaction is going to be. 

To see that you have understood this analysis you could see what the total entropy change is for this reaction:

CuCO₃(s)    ⟶  CuO(s)    +   CO₂(g)