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).

Energetics index: GCSE Notes on the basics of chemical energy changes – important to study and know before tackling any of the three Advanced Level Chemistry pages Parts 1–3 here * Part 1a–b ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes – reaction, formation, combustion : 1.2a & 1.2b(i)–(iii) Thermochemistry – Hess's Law and Enthalpy Calculations – reaction, combustion, formation etc. : 1.2b(iv) Bond Enthalpy Calculations  : 1.3a–b Experimental methods for determining enthalpy changes and treatment of results : 1.4 Some enthalpy data patterns : 1.4a The combustion of linear alkanes and linear aliphatic alcohols : 1.4b Some patterns in Bond Enthalpies and Bond Length : 1.4c Enthalpies of Neutralisation : 1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and evidence of aromatic ring structure in benzene : Extra Q page A set of practice enthalpy calculations with worked out answers ** Part 2 ΔH Enthalpies of ion hydration, solution, atomisation, lattice energy, electron affinity and the Born–Haber cycle : 2.1a–c What happens when a salt dissolves in water and why? : 2.1d–e Enthalpy cycles involving a salt dissolving : 2.2a–c The Born–Haber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1a–g Introduction to Entropy : 3.2 Examples of entropy values and comments * 3.3a ΔS, Entropy and change of state : 3.3b ΔS, Entropy changes and the feasibility of a chemical change : 3.4a–d More on ΔG, Free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants : 3.6 Kinetic stability versus thermodynamic feasibility * PLEASE note that delta H/S/G values vary slightly from source to source, so I apologise in advance for any inconsistencies that may arise as I've researched and developed each section.

 

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 a quick recap on the calculations so far.

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  ΔSθtot = ΔSθsys +  ΔSθsurr

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 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.

Example

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3.4b Calculating free energy changes for chemical reactions – applications

Using the free energy equation: ΔGθsys = ΔHθsys – TΔSθsys

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

OR in the case of cells, ΔGθsys =  –neEθF (see section 3.4c)

which gives the electrical energy available.

The free energy change of a feasible reaction, ΔGθsys, must be <0

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 (see end of section 3.3 Entropy).

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

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

Examples


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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


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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 e.g.

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.

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.


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