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. 

Intermolecular forces (4) Solid structures of simple molecular lattices: Iodine and Ice

OCR A level H432 Chemistry A (from 2017)

Learning objectives

2.2.2 Intermolecular forces

(n) explanation of the solid structures of simple molecular lattices, as covalently bonded molecules attracted by intermolecular forces, e.g. I2, ice 


(o) explanation of the effect of structure and bonding on the physical properties of covalent compounds with simple molecular lattice structures including melting and boiling points, solubility and electrical conductivity. 


Solid structures of simple molecular lattices

Simple molecular lattices are very common especially in organic chemistry.

Think of sugar molecules, alkane molecules or alcohol molecules: when they freeze they exist in simple molecular lattices.

Two examples that come from inorganic chemistry are Iodine and ice.

Iodine I₂(s)

Iodine is a grey, shiny solid at room temperature and pressure.

With seven electrons in its outer electron energy level, an atom of iodine is covalently bonded to another to form a diatomic molecule I_2.

Such an arrangement means that each iodine atom possesses a full outer energy level of eight electrons.

The bond between these two atoms is a σ bond where the electron density sits along the axis joining the two iodine atoms.

The bond is formed by the end on overlap of two atomic orbitals gving a molecular orbital that is complete with two electrons.

The picture to the left  is an attempt at a representation of this covalent bond.

A weak induced dipole—induced dipole force holds the iodine molecules in the molecular lattice since both atoms have the same electronegativity and so do not possess permanent bond polarity.

The weak intermolecular force is called by its generic name in the diagram above as a van der Waals force.

The final picture attempts to show the arrangement of iodine molecules in the crystal lattice as they are close packed together.

Ice H₂O(s)

Ice is a hard, colourless solid at temperatures below 0^oC or +273K.

It is quite brittle and breaks easily to produce sharp shards that can cut through skin.

Ice is composed of two elements of significantly different electronegativity:  oxygen (3.44) and hydrogen (2.20. 

So ice molecules are polar covalent molecules.

The covalent σ bonds that exist between oxygen and hydrogen are polar with the oxygen atom pulling the bonding electron pairs towards itself. 

Because of the covalent bond polarity molecules hydrogen bonds are found between molecules of water in ice.

There are two hydrogen bonds per water molecules that mean the structure of ice is very similar to that of diamond.

Compare the two structures here:

                                                         Diamond   

                                                                Ice

Two hydrogen bonds per molecule are possible because the molecule of water has two lone pairs on the oxygen atom.

The intermolecular forces in ice are permanent dipole forces and the much weaker induced dipole forces that we find exist between all molecules at some level. 

Substance	Structure	Bonding	Solubility in water	Mp and bp /oC	Electrical conductivity
Iodine	Diatomic molecular Lattice	Covalent	Partially soluble	387K 457K	no
Ice	Simple molecular lattice	Polar covalent	Fully miscible	273K 373K	yes

Entropy (4) Using ΔG Gibbs free energy to predict reaction feasibility

Edexcel A level Chemistry (2017)

Topic 13B: Entropy:

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

13B/18. To know that the balance between the entropy change and the enthalpy change determines the feasibility of a reaction and is represented by the equation

ΔG = ΔH – TΔSsystem

13B/19.  To be able to use the equation ΔG = ΔH − TΔSsystem to:

i) predict whether a reaction is feasible 


ii) determine the temperature at which a reaction is feasible

Using the Gibbs Free energy to predict reaction feasibility

If we want to predict the feasibility of a chemical change we can look at the entropy changes in the system and in the surroundings.

And if the sum of these two entropy changes is positive then we can conclude that the chemical change is feasible. 

We saw here in the analysis of the combustion of magnesium in oxygen that the sum of the entropy change in the system and surrounding was positive.

Defining ΔG, the Gibbs Free Energy change.

If we look at the Gibbs Free Energy change the problem is simplified because in the 19^th century J.W.Gibbs showed that we can combine the two entropy calculations into one.

We have seen that

ΔS_total =  ΔS_system   +    ΔS_surroundings

And if we want to use this equation we have to do two calculations to find both ΔS_system and ΔS_surroundings  

And if we find that    ΔS_total  > 0   then the chemical change is spontaneous and feasible.  

We can also say that

and if we insert this expression into the expression for ΔS_total  we obtain

Here the total entropy change is determined in terms of the properties   of the system alone i.e. ΔS_system ,  ΔH and T.

All we need to remember is that this relationship only holds true if the temperature and pressure of the process are constant. 

Now if we multiply through by —T we obtain:

—TΔS_total  = —TΔS_system  +  ΔH

and this leads to a definition of the Gibbs Free Energy change since

ΔG  =   ΔH —TΔS_system

And also that

ΔG  =  —TΔS_total

Implications of ΔG

From this it is possible to conclude that the Free Energy Change is proportional to the total change in entropy in both system and surroundings.

But there is a difference in sign between ΔG and ΔS_total

Whereas a spontaneous change meant that  ΔS_total   >   0 now , providing temperature and pressure remain constant in the process, ΔG <  0  for a spontaneous chemical change. 

So in a spontaneous chemical change at constant temperature and pressure the Gibbs Free Energy falls.        

But again this is only significant because the total entropy is increasing!!

The other thing we get from the Gibbs Free Energy change is that its value is a measure of the total non-expansion work that we can obtain from any chemical change.  More about that under redox equilibria.

Reaction Feasibility and ΔG

So the equation above:

                                          ΔG  =   ΔH —TΔS_system

shows us that the feasibility of a reaction depends upon the values of ΔH and ΔS for the system.   

Here’s how we can see if a reaction is feasible using ΔG:

ΔH (enthalpy change of the system)	ΔSsystem	ΔG (free energy change of the system)	Feasibility of the reaction
Negative	positive	Always negative	Reaction is feasible
Positive	negative	Always positive	Reaction is never feasible
Negative	negative	Negative at low temps	Feasible at low temps
Positive	positive	Negative at high temps	Feasible at high temps.

So most exothermic reactions are, as we suspected, feasible and spontaneous.

And even if the system becomes more ordered the positive entropy change in the surroundings more than compensates for the negative value in the system and ensures that the reaction is spontaneous.

We can see also that enthalpy (ΔH) more than entropy (ΔS_system) contributes to ΔG.

And we can also use

ΔG  =   ΔH —TΔS_system

to find out the temperature at which a reaction becomes spontaneous.

Question:

At what temperature does the decomposition of magnesium carbonate become spontaneous and feasible?

Equation:                             MgCO₃     ⟶  MgO    +   CO₂

First find out the enthalpy change in the system:

ΔH_reaction    =     (—601.7  +   —393.5 )—  —1095.8

ΔH_reaction    =     +100.6 kJmol^—1

The enthalpy change is endothermic.

Second find out what is the entropy change in the system?

ΔS_system   =     (+26.9  +   +213.6 )— +65.7

ΔS_system   =     +174.8  J.mol^—1. K^—1

The entropy change in the system suggests that the reaction is feasible.

If

ΔG  =   ΔH —TΔS_system

Then rearrange this equation to find out the temperature  T when ΔG is zero i.e. when the reaction becomes spontaneous

So if now 

0  =   ΔH —TΔS_system

Then adding TΔS_system to each side gives

TΔS_system =   ΔH

And therefore

T =   ΔH/ΔS_system

So

T =   100.6  × 1000/174.8  =  576K

So this confirms what we already know from experience that heating magnesium carbonate decomposes it and in fact if the temperature rises to 576 K or above (302^oC) the substance decomposes spontaneously.