Entropy (5) Discussion of ΔG = – RTlnK and thermodynamic vs kinetic control

Edexcel A level Chemistry (2017)

Topic 13B: Entropy:

Here are the remaining learning objectives to do with free energy change ΔG:

13B/20. To be able to use the equation ΔG = −RT ln K to show that reactions which are feasible in terms of ΔG have large values for the equilibrium constant and vice versa.

13B/21. To understand why a reaction for which the ΔG value is negative may not occur in practice.

13B/22. To know that reactions that are thermodynamically feasible may be inhibited by kinetic factors.

Free energy change (ΔG) and equilibrium constant (K)

Well here’s a thing if a reaction is spontaneous and feasible it’s likely to go.

But how far will it go? 

Will it go to completion?

What do we mean by “going to completion”?

The relationship between free energy change and equilibrium constant is this

ΔG = −RT ln K

Peter Atkins calls this equation one of the most important equations in chemical thermodynamics.

From is equation we can predict the value of the equilibrium constant for any reaction from its thermodynamic data.

So let’s see if we can show you how to calculate the equilibrium constant for this reaction at 25^oC or 298K.

N₂ (g)   +     3H₂ (g)     ⇌    2NH₃ (g)

If

ΔG = −RT ln K

Then rearranging

therefore

since 2 moles of ammonia form and we have rationalised the units multiplying up the Gibbs energy value by 1000

this then gives

lnK  =  e^13.318

therefore

K  =  6.08  ×  10⁵  dm⁶.mol^—2

So when the free energy change is large and negative the natural logarithm of the equilibrium constant is large and positive.

e.g.  consider now the combustion of magnesium in oxygen

2Mg(s)    +    O₂(g)  ⟶    2MgO(s)

where ΔH^o_reaction   =   —1203.4 kJ.mol^—1

and ΔS_system   =   —216 J.mol^—1.K^—1

so that at 298K:

if 

ΔG  =   ΔH —TΔS_system

then

ΔG  =   — 1,203,400 —   298  ×  —216  =   —1139kJ.mol^—1

and therefore

ln K =   —ΔG/RT

so  

lnK    =   1139× 1000/8.314×298   =  459.7

so

K  =  e^459.7

which is very large and positive!!

(the calculator on my iphone produced an error statement when I tried the last part of this calculation!!)

This result confirms the previous conclusion from the formation of ammonia at 298K

Let’s summarise the implications of the values of ΔG on the feasibility of a reaction.

Thermodynamic vs kinetic control

Now all this discussion of the feasibility of a chemical reaction because of the sign and value of the free energy change is all well and good but all the factors that enable a reaction to proceed have not been considered.

The values of free energy change, total entropy change and the equilibrium constant are used to predict the feasibility of a reaction.

The standard reduction potential E^⦵   can be used to predict the feasibility of redox reactions in aqueous solution but these four thermodynamic factors do not include any reference to the rate at which the reaction proceeds.

The study of the thermodynamic dimensions of a reaction does not involve the measure of the activation energy, E_a of the reaction or the size of the rate constant k.

In other words, all the thermodynamic factors might suggest to us that a reaction is feasible but it could be just too slow to happen in any reasonable time we have because of its very high activation energy!!

These four thermodynamic values say nothing about how fast the reaction will go.