Doc Brown's A Level Chemistry  Advanced Level Theoretical Physical Chemistry – AS A2 Level Revision Notes – Basic Thermodynamics

GCE Thermodynamics–thermochemistry sub–index links below

Part 3: ΔS Entropy Changes and ΔG Free Energy Changes

 3.4 ΔG, Gibbs free energy changes and reaction feasibility

Part 3.4 Introduces the Gibbs free energy equation for determining the feasibility of a reaction. The ideas and equation are applied to any chemical changes and more specifically the extraction of metals (e.g. feasibility of carbon reduction of metal oxide) and the direction of chemical change in an electrical cell (i.e. a 'battery' of two half–cells and their reactions). How to calculate the minimum temperature for reaction feasibility is described and how to obtain enthalpy change and entropy change from a graph of free energy versus temperature.

Energetics–Thermochemistry–Thermodynamics Notes INDEX

 

3.4a An Introduction to the Gibbs free energy expression

All advanced level syllabuses require a knowledge of entropy concepts and simple calculations and some specifications require a 'working knowledge' of the Gibbs free energy (ΔG) expressions.

Before dealing with free energy changes (ΔG) a quick recap on the calculations so far before I explain what free energy changes are.

In order for a reaction to be feasible the overall entropy change  ΔStot must be >0 and the overall free energy change ΔGsys must be <0.

If ΔGsys is ~0 it is likely to be a significant equilibrium i.e. not significantly on the right or the left of an equilibrium situation.

For a reaction the overall entropy change is given by:

ΔSθsys= ΣSθproducts – ΣSθreactants       and     ΔSθsurr = –ΔHθsys–reaction/T

giving the overall entropy change equation:  ΔSθtot = ΔSθsys +  ΔSθsurr

This has been dealt with in section 3.3b ΔS, Entropy changes and the feasibility of a chemical change

Before getting into free energy change calculations a little digression.

If the criteria for a feasible reaction is ΔStot >0 and ΔG<0 what is ΔG? and what is the connection between them?

PLEASE NOTE that you do not need to be able to derive the Gibbs free energy equation but I think it is important you realise there is a direct connection with the entropy expressions and that delta G expressions have not been pulled out of a hat!

From above and substituting: ΔSθtot = ΔSθsys –ΔHθsys/T, then multiplying through by T gives

TΔSθtot = TΔSθsys –ΔHθsys, then changing signs throughout gives

–TΔSθtot = –TΔSθsys + ΔHθsys

–TΔSθtot is called the Gibbs Free Energy change denoted by the letter G,

so giving the familiar free energy equation: ΔGθsys = ΔHθsys – TΔSθsys

A working knowledge of which is required by some courses.

 

3.4b Calculating free energy changes for chemical reactions – applications

Using the free energy equation: ΔGθsystem/reaction = ΔHθsystem/reaction – TΔSθsystem/reaction

The free energy can be thought of as energy that is available to do useful work.

A reaction is feasible if ΔGθ < 0  (its the parallel to feasibility when ΔSθtot > 0)

Rearrangements of : ΔGθ = ΔHθ – TΔSθ

ΔHθ = ΔGθ + TΔSθ

ΔSθ = (ΔHθ – ΔGθ) / T

T = (ΔHθ – ΔGθ) / ΔSθ

Note: To calculate temperature point of 'first' feasibility, at which ΔGθ = 0,

you argue that at this point: ΔHθsys – TΔSθsys = 0, so T = ΔHθsys / ΔSθsys

If you know the entropy change and enthalpy change of the system, you can then calculate whether any reaction is feasible at any given temperature by using the equation

ΔGθ = ΔHθ – TΔSθ, though this does assume that enthalpy and entropy changes do not change with temperature, unfortunately this is not the case, BUT it does give a fair indication of feasibility;

Units note: Remember T is in K, ΔH and ΔG are usually given in kJmol–1,

but S and ΔS are usually in JK–1mol–1, so its probably best to convert ΔS into kJ (that's what I usually do).

In the case of cells the free energy equation is:   ΔGθsys =  –neEθF (see section 3.4c)

which gives the electrical energy available.

The free energy change of a feasible reaction, ΔGθsys, must be <0, effectively the net cell p.d. must be positive.

If ΔGθsys is close to zero, then you are likely to be dealing with a significant equilibrium situation.

and remember however feasible a reaction might be thermodynamically, it does not necessarily mean it will happen spontaneously because of kinetic limitations.

For more details on this last point see 3.6 Kinetic stability versus thermodynamic instability for a more detailed discussion

and "KINETICS" section 6.


How can you calculate the temperature at which the free energy changes for two different reactions are the same?

ΔG = ΔH – TΔS, therefore for two reactions

ΔG1 = ΔH1 – TΔS1   and   ΔG2 = ΔH2 – TΔS2

ΔH1 – TΔS1 = ΔH2 – TΔS2

TΔS2 – TΔS1 = ΔH2 – ΔH1

∴ T(ΔS2 – TΔS1) = ( ΔH2 – ΔH1)

∴ T = ( ΔH2 – ΔH1) / (ΔS2 – ΔS1)


Introduction to the examples of free energy calculations and the linked concepts

ΔGθsystem/reaction = ΔHθsystem/reaction – TΔSθsystem/reaction  is a bit cumbersome, so for simplicity, in the table below, I've just used the main symbols involved in the Gibbs free energy equation.

ΔG = ΔH – TΔS    and a chemical reaction is truly thermodynamically feasible if ΔG < 0

All the data (from standard data books) for a question I've tabulated under the equation for convenience, but a handy way of preceding the enthalpy, entropy and free energy calculations.

I've worked everything out from first principles, so in an exam you might be given some data which I've worked out, but no matter, its all good practice in following the conceptual and numerical argument for free energy calculations.

The are four possible scenarios depending whether the enthalpy and entropy changes are positive and negative.

(1) ΔH positive (endothermic), ΔS positive (entropy increases)
ΔG   =

free energy change

ΔH

enthalpy change

– TΔS

– temp (K) x entropy change

  positive, endothermic ΔS positive, entropy increases

Not feasible at low temperature, the  –TΔS term can't outweigh the positive ΔH term, BUT, at high temperature the –TΔS term becomes sufficiently negative to outweigh the positive ΔH term and give a negative ΔG free energy overall, so the higher the temperature the more feasible the reaction.

 
(2) ΔH positive (endothermic), ΔS positive (entropy increases)
ΔG    = ΔH – TΔS
  positive, endothermic ΔS negative, entropy decreases

Since – – = +, the –TΔS term will always be positive, and, since ΔH is also positive, ΔG is always positive (can't be < 0), and so the reaction can NEVER be feasible.

 
(3) ΔH positive (endothermic), ΔS positive (entropy increases)
ΔG   = ΔH – TΔS
  negative, exothermic ΔS positive, entropy increases

Since both ΔH and –TΔS terms are always negative, ΔG is always < 0, so the reaction is ALWAYS feasible.

 
(4) ΔH positive (endothermic), ΔS positive (entropy increases)
ΔG   = ΔH – TΔS
  negative, exothermic ΔS negative, entropy decreases

Since – – = +, at low temperatures the –TΔS term is not too positive to outweigh the negative ΔH and the reaction is feasible, BUT at high temperatures the –TΔS term (– – = +) becomes to positive (– – = +) and outweighs the negative ΔH, so ΔG becomes positive and the reaction becomes less and less feasible.

 

Examples of calculations using the Gibbs free energy equation

Examples (1) to (4) match the 'concept' analysis above, after that its whatever came into my head!

In the questions, the following definitions apply (to reduce repetition!)

ΔHθf = standard enthalpy of formation of a compound at 298K, 1 atm./101kPa pressure in kJmol–1

(for elements in their normal stable standard state ΔHθf = 0).

ΔGθf = standard free energy of formation of a compound at 298K, 1 atm./101kPa pressure in kJmol–1

(for elements in their normal stable standard state ΔGθf = 0).

Sθf = standard entropy energy content at 298 K, 1 atm./101kPa pressure in JK–1mol–1

(standard entropy can never be zero!)

In calculating the free energy change at temperatures other than 298K it is assumed the enthalpy and entropy of the species does not change, however, this is not true, so any such calculations are an approximation, but the general trends in feasibility are quite correctly deduced.


 

(1) ΔH positive (endothermic), ΔS positive (entropy increases)

ΔG   = ΔH – TΔS
  positive, endothermic ΔS positive, entropy increases

Not feasible at low temperature, the  –TΔS term can't outweigh the positive ΔH term, BUT, at high temperature the –TΔS term becomes sufficiently negative to outweigh the positive ΔH term and give a negative ΔG free energy overall, so the higher the temperature the more feasible the reaction.

Example (1) The thermal decomposition of lithium carbonate

Thermodynamic data Li2CO3(s) ====> Li2O(s) + CO2(g)
ΔHθformation (kJmol–1) –1216   –596   –394
ΔGθformation (kJmol–1) –1133   –560   –395
Sθspecies (JK–1mol–1) 90.4   39.2   214

To calculate the enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = (–596 –394) – (–1216) = +226 kJmol–1

To calculate the free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = (–560 –395) – (–1133) = +178kJmol–1

To calculate the entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = (39.2 + 214) – (90.4) = +162.8 JK–1mol–1 (0.1628 kJK–1mol–1)

Substituting into the Gibbs free energy change equation

ΔGθsys = ΔHθsys – TΔSθsys

ΔGθsys = +226 – 0.1628T kJmol–1

To calculate the temperature at which the decomposition of lithium carbonate becomes feasible

When ΔGθsys = 0, you have reached the point of first feasibility

Therefore, at this point: ΔHθsys – TΔSθsys = 0

So T = ΔHθsys / ΔSθsys = 226 / 0.1628 = 1388 K (1115oC)

and the decomposition will occur at any temperature above 1388 K, below 1388 it becomes less and less feasible, negligible product (theoretically!).

Suppose you given a series of free energies at various temperatures for a given reaction?

How do you calculate the enthalpy change and entropy change?

ΔGθsys = ΔHθsys – TΔSθsys

y = mx + c, is the classic algebraic equation for a linear graph connecting two variables x and y, slope = m, c = a constant.

This can be applied to the Gibbs free energy equation

y = ΔGθsys, gradient = m = – ΔSθsys, x = T(K), constant = c = ΔHθsys

So, by plotting a graph of ΔGθsys versus T, you can derive ΔHθsys and TΔSθsys

I actually used the free energy equation to generate the perfect data for the graph below, so that we are now working as if the starting point was ΔG data and not ΔH, S or ΔS data.

You can see from the graph, the point of reaction feasibility, when ΔG is =<0, is around 1400K or higher temperatures.

With the help of Excel software and applying the y = mx + c logic:

From the gradient: m = –ΔS = –0.1628, ΔS = 0.163 kJK–1mol–1  or +163JK–1mol–1

From the constant c, ΔH = +226 kJmol–1


 

(2) ΔH positive (endothermic), ΔS positive (entropy increases)

ΔG    = ΔH – TΔS
  positive, endothermic ΔS negative, entropy decreases

Since – – = +, the –TΔS term will always be positive, and, since ΔH is also positive, ΔG is always positive (can't be < 0), and so the reaction can NEVER be feasible.

Example (2) The 'synthesis' of ethene from its elements

Thermodynamic data C(s) + 2H2(g) ====> C2H4(g)
ΔHθformation (kJmol–1) 0   –0   +52.3
ΔGθformation (kJmol–1) 0   0   +68.1
Sθspecies (JK–1mol–1) 5.7   2 x 131   219

To calculate the enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = +52.3 – (0 + 0) = +52.3 kJmol–1

To calculate the free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = +68.1 – (0 + 0) = +68.1 kJmol–1

To calculate the entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = (219) – {(5.7 + (2 x 131)} = –48.7 JK–1mol–1  (–0.0487 JK–1mol–1)

Substituting into the Gibbs free energy equation for other temperatures

ΔGθsys = ΔHθsys – TΔSθsys

ΔGθsys = +52.3 – (–0.0487T) = 52.3 + 0.0487 kJmol–1

Both terms on the right are always positive, so ΔG can never be negative, so the reaction is never feasible at any temperature!

Suppose you given a series of free energies at various temperatures for a given reaction?

How do you calculate the enthalpy change and entropy change?

The arguments are given in detail in Example (1) above, so I'm just starting from the given graph of free energy versus temperature from now on.

With the help of Excel software and applying the y = mx + c logic:

From the gradient: m = = –ΔS = 0.0487, ΔS = –0.0.0487 kJK–1mol–1  or –48.7JK–1mol–1

From the constant c, ΔH = +52.3 kJmol–1


 

(3) ΔH positive (endothermic), ΔS positive (entropy increases)

ΔG   = ΔH – TΔS
  negative, exothermic ΔS positive, entropy increases

Since both ΔH and –TΔS terms are always negative, ΔG is always < 0, so the reaction is ALWAYS feasible.

Example (3) The decomposition of hydrazoic acid (hydrogen azide, a colourless volatile explosive liquid)

I could only find thermodynamic data on the gaseous state.

Thermodynamic data HN3(s) ====> 1/2H2(g) + 3/2N2(g)
ΔHθformation (kJmol–1) +294   0   0
ΔGθformation (kJmol–1) +328   0   0
Sθspecies (JK–1mol–1) 237   131/2   3/2192

To calculate the enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = (0 + 0) – (+294) = –294 kJmol–1

To calculate the free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = (0 + 0) – (+328) = +328 kJmol–1

To calculate the entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = (131/2 + 3/2192) – (237) = +116.5 JK–1mol–1   (+0.1165 kJK–1mol–1)

Substituting into the Gibbs free energy equation for other temperatures

ΔGθsys = ΔHθsys – TΔSθsys

ΔGθsys = –294 – 0.1165T

Both terms are always negative, therefore ΔGθsys is always negative and the reaction feasible at any temperature!

Suppose you given a series of free energies at various temperatures for a given reaction?

How do you calculate the enthalpy change and entropy change?

The arguments are given in detail in Example (1) above, so I'm just starting from the given graph of free energy change versus temperature.

Using Excel software and applying the y = mx + c logic:

From the gradient: m = = –ΔS = –0.117, ΔS = 0.117 kJK–1mol–1  or +117JK–1mol–1

From the constant c, ΔH = +52.3 kJmol–1


 

(4) ΔH positive (endothermic), ΔS positive (entropy increases)
ΔG   = ΔH – TΔS
  negative, exothermic ΔS negative, entropy decreases

Since – – = +, at low temperatures the –TΔS term is not too positive to outweigh the negative ΔH and the reaction is feasible, BUT at high temperatures the –TΔS term (– – = +) becomes to positive (– – = +) and outweighs the negative ΔH, so ΔG becomes positive and the reaction becomes less and less feasible.

Example (4) The synthesis of ammonia

Thermodynamic data 1/2N2(s) + 3/2H2(g) ====> NH3(g)
ΔHθformation (kJmol–1) 0   –0   –46.2
ΔGθformation (kJmol–1) 0   0   –16.6
Sθspecies (JK–1mol–1) 1/2192   3/2131   193

To calculate the enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = (–46.2) – (0 + 0) = –46.2 kJmol–1

To calculate the free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = (–16.6) – (0 + 0) = –16.6 kJmol–1

To calculate the entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = (193) – (192/2 + /2131) = –99.5 JK–1mol–1   (0.0995 kJK–1mol–1)

Substituting into the Gibbs free energy equation for other temperatures

ΔGθsys = ΔHθsys – TΔSθsys

ΔGθsys = –46.2 – (–0.0995T) = –46.2 + 0.0995T

Which means it is not favoured by high temperatures, no matter how fast you want the reaction to take place!

For further discussion see

Applying Le Chatelier's Principle to Industrial Processes

The Haber Synthesis of ammonia

Suppose you given a series of free energies at various temperatures for a given reaction?

How do you calculate the enthalpy change and entropy change?

The arguments are given in detail in Example (1) above, so I'm just starting from the given graph of free energy change versus temperature.

You can see from the graph, the point of reaction feasibility, when ΔG is =<0, is around 460K or lower temperatures.

Using Excel software and applying the y = mx + c logic:

From the gradient: m = = –ΔS = 0.0994, ΔS = –0.994 kJK–1mol–1  or –99.4JK–1mol–1

From the constant c, ΔH = +46.2 kJmol–1

I did find a set of values for the equilibrium constant Kp, for the equation N2(g) + 3H2(g) (c) doc b 2NH3(g)

Kp = pNH32/pN2pH23

Although I've done the thermodynamic calculations based on the formation of 1 mole of ammonia, you can plainly see the dramatic decrease in the equilibrium constant.

In reality, a fast low yield of ammonia is acceptable at a moderate temperature, with the help of high pressure and a catalyst, see ....

Applying Le Chatelier's Principle to Industrial Processes

The Haber Synthesis of ammonia


Example (5) The thermal decomposition of ammonium chloride (a sublimation, solid ===> gases)

Thermodynamic data NH4Cl(s) ====> NH3(g) + HCl(g)
ΔHθformation (kJmol–1) –315   –46.2   –92.3
ΔGθformation (kJmol–1) –204   –16.6   –95.3
Sθspecies (JK–1mol–1) 94.6   39.2   187

To calculate the enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = (–46.2 –92.3) – (–315) = +176.5 kJmol–1

To calculate the free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = (–16.6 –95.3) – (–204) = +92.1 kJmol–1

To calculate the entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = (39.2 + 187) – 94.6 = +131.6 JK–1mol–1  (0.3294 kJK–1mol–1)

Substituting into the Gibbs free energy equation for other temperatures

ΔGθsys = ΔHθsys – TΔSθsys

ΔGθsys = 176.5 – 0.329T

To calculate the temperature at which the decomposition of ammonium chloride becomes feasible

When ΔGθsys = 0, you have reached the point of first feasibility

Therefore, at this point: ΔHθsys – TΔSθsys = 0

So T = ΔHθsys / ΔSθsys = 176.5 / 0.3294 = 539 K (263oC)

so, the decomposition will occur at any temperature above 539 K, below 539 it becomes less and less feasible.

You can see from the graph, the point of reaction feasibility, when ΔG is =<0, is around 540K or higher temperatures.

You see this when heat ammonium chloride, it readily decomposes on heating it in a test tube with a bunsen burner, but when the gases reach the upper cooler part of the test tube, the solid crystal reforms.

Although this is a very endothermic reaction and the lattice enthalpy is 620 kJmol–1, indicating a strong ionic bonding, it is the formation of gases and the accompanying large increase in entropy that drives the reaction forward at higher temperatures.

Its more feasible at lower temperatures than lithium carbonate and calcium carbonate because of the bigger entropy change - two moles of gas formed compared to one. Numerically the negative -TΔS term is about twice as big.


Example (6) Cracking an alkane hydrocarbon to produce ethene

Suppose you wanted to 'ideally' convert butane into (i) two molecules of ethene for polymers and (iii) one molecule of hydrogen for fuel or hydrogenation of vegetable oil.

Calculate whether or not the reaction is feasible and what sort of minimum temperature is theoretically needed.

Thermodynamic data C4H10(g) ====> 2C2H4(g) + H2(g)
ΔHθformation (kJmol–1) –125   2 x +52.3   0
ΔGθformation (kJmol–1) –15.7   2 x 68.1   0
Sθspecies (JK–1mol–1) 310   2 x 219   131

Enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = {(2 x 52.3) + 0} – (–125) = +229.6 kJmol–1

Free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = {(2 x 68.1) + 0} – (–15.7) = +151.9 kJmol–1

Entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = {(2 x 219) + 131} – (310) = +259 JK–1mol–1  (0.259 kJK–1mol–1)

Gibbs free energy change equation for other temperatures

ΔGθsys = +229.6  – 0.0259T

To calculate the temperature at which this specific cracking reaction becomes feasible

When ΔGθsys = 0, you have reached the point of first feasibility

Therefore, at this point: ΔHθsys – TΔSθsys = 0

So, T = ΔHθsys / ΔSθsys = 229.6 / 0.259 = 886 K (613oC)

therefore, the decomposition will occur at any temperature above 886 K, below 886 it becomes less and less feasible.

However, you might need a very special catalyst to give these useful products and nothing else!

You can see from the graph, the point of reaction feasibility, when ΔG is =<0, is around 900K or higher temperatures.


(7) The thermal decomposition of calcium carbonate (limestone ==> lime)

Thermodynamic data CaCO3(s) ====> CaO(s) + CO2(g)
ΔHθformation (kJmol–1) –1207   –635   –394
ΔGθformation (kJmol–1) –1129   –604   –395
Sθspecies (JK–1mol–1) 92.9   40   214

Enthalpy change

ΔHθsys = ΣΔHθf(products) – ΣΔHθf(reactants)

ΔHθsys = (–635 –394) – (–1207) = +178 kJmol–1

Free energy change

ΔGθsys = ΣΔGθf(products – ΣΔGθf(reactants)

ΔGθsys = (–604 –395) – (–1129) = +130 kJmol–1

Entropy change

ΔSθsys = ΣSθproducts – ΣSθreactants

ΔSθsys = (40 + 214) – 92.9 = +161.1 JK–1mol–1  (0.1611 kJK–1mol–1)

Gibbs free energy change equation for other temperatures

ΔGθsys = ΔHθsys – TΔSθsys

ΔGθsys = 178 – 0.1611T

(its numerically similar to the free energy equation for the thermal decomposition of lithium carbonate)

To calculate the temperature at which the decomposition of calcium carbonate becomes feasible

When ΔGθsys = 0, you have reached the point of first feasibility

Therefore, at this point: ΔHθsys – TΔSθsys = 0

So T = ΔHθsys / ΔSθsys = 178 / 0.0.1611 = 1105 K (263oC)

so, the decomposition will occur at any temperature above 1105 K (832oC), below 1105K it becomes less and less feasible.

You can see from the graph, the point of reaction feasibility, when ΔG is =<0, is around 1100K or higher temperatures.

Although this is a very endothermic reaction, and strong ionic bonding in calcium carbonate, it is the formation of carbon dioxide and the accompanying large increase in entropy that drives the reaction forward at higher temperatures.

Note: I've done the same calculation on the entropy calculations page, which gave a value of 1112K.

The graph for calcium carbonate would be similar to that for the thermal decomposition of lithium carbonate in example (1).


3.4c Electrochemical Cell Emf's, Feasibility and Gibb's Free Energy Change

Relating Eθcell(sys) to the direction of overall chemical change and feasibility of reaction. If you calculate a –ve cell voltage for a given reaction, that is not the way cell reaction goes! please reverse the equation and re–calculate!

The Ebattery–cell must be >0.00V for the cell, and any other redox reaction, to be theoretically feasible.

The free energy change must be negative <0 for the cell reaction to be feasible (matching the Ecell rule of >0V for feasibility).

ΔGθcell(sys) = –nEθF J (ΔG not needed by all A2 advanced level syllabuses)

n = number of moles of electrons transferred in the reaction per mol of reactants involved,

Eθ is the Emf for the overall reaction in volts,

F = the Faraday constant (96500 C mol–1).

Ex 3.4c1 for the zinc–copper Daniel cell producing +1.10V

2 electrons transferred: M2+(aq) + 2e– <=> M(s)

full cell redox reaction: Cu2+(aq) + Zn(s) ==> Cu(s) + Zn2+(aq)

ΔGθcell = –2 x 1.10 x 96500 = –212300 Jmol–1 or –212.3 kJmol–1

A very negative free energy value so very feasible!

If work the calculation for the reverse reaction (Cu + Zn2+ ==>) you get a very positive free energy value, so definitely NOT a feasible chemical change.

For calculating an equilibrium constant for a redox reaction see section 3.5b


3.4d Free Energy Changes and the Extraction of Metals

In order for a metal extraction to be thermodynamically feasible from an oxide the decrease in Gibbs free energy on oxide formation must be numerically less than the free energy released by the formation of carbon monoxide/carbon dioxide if carbon/carbon monoxide is used in its extraction.

This is conveniently explored via an Ellingham diagram like the one shown below.

An Ellingham diagram shows the variation of the free energy change with temperature for particular reactions reaction. The more negative the free energy change the more feasible it is.

A metal can be theoretically extracted using carbon, if the free energy change in forming carbon monoxide or carbon dioxide, is more negative than the free energy change in forming the oxide.

 

You can theoretically extract magnesium (Mg) from magnesium oxide (MgO) using carbon at a temperature of over 2300K, but technologically not very convenient since most furnace materials would melt at this temperature!

Aluminium would require an even higher temperature of well over 2500K, so not surprisingly, these highly reactive metals are extracted by electrolysis.

However, iron can be extracted from iron oxides using carbon at temperatures above 1300K which of course is what happens in a blast furnace. Zinc can be extracted above 1100K.

Note that many metal mineral ores are sulfides and carbonates which are often roasted to give the oxide which is then reduced by some means.

For more on extraction of metals and Ellingham free energy diagrams see section 3.4d


A summary of the maths of free energy, entropy and feasibility

  • In order for a reaction to feasible the overall entropy change ...

    • ΔSθtot must be >=0 and the overall free energy change ΔGθ must be <=0.

  • For a reaction the overall entropy change is given by:

  • Total entropy change = entropy change of system + entropy change of surroundings

    • ΔSθtot = ΔSθsys +  ΔSθsurr

    • ΔSθsys= ΣSθproducts – ΣSθreactants

    • and ΔSθsurr = –ΔHθsys–reaction/T(K)

      • Note the connection between ΔS and ΔG.

      • From above and substituting ...

      • ΔSθtot = ΔSθsys –ΔHθsys/T

      • multiplying through by T gives

      • TΔSθtot = TΔSθsys –ΔHθsys

      • changing signs throughout gives

      • –TΔSθtot = –TΔSθsys + ΔHθsys

      • –TΔSθtot is called the Gibbs Free Energy change symbol G,

      • so giving the familiar free energy equation ...

      • ΔGθsys = ΔHθsys – TΔSθsys 

      • The free energy can be thought of as heat energy that is available to do work.

        • OR in the case of cells, where ΔGθsys =  –neEθF,

        • this gives the electrical energy available to do useful work.

        • PLEASE NOTE you do not have to derive the Gibbs Free Energy equation but some syllabuses require you to be able to use when supplied with enthalpy and entropy data.

    • The free energy change of the reaction, ΔGθ must be <=0 to be feasible.

    • If ΔSθtot or ΔGθsys is close to zero, then you are likely to be dealing with an equilibrium situation.

 


Energetics–Thermochemistry–Thermodynamics Notes INDEX


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