Doc Brown's A Level Chemistry Advanced Level Theoretical Physical Chemistry AS A2 Level Revision Notes Basic Thermodynamics GCE Thermodynamicsthermochemistry subindex 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 halfcells 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. 

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 ΔS_{tot} must be >0 and the overall free energy change ΔG_{sys} must be <0. If ΔG_{sys} 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: ΔS^{θ}_{sys} = = ΣS^{θ}_{products } ΣS^{θ}_{reactants} and ΔS^{θ}_{surr} = ΔH^{θ}_{sysreaction}/T giving the overall entropy change equation: ΔS^{θ}_{tot} = ΔS^{θ}_{sys} + ΔS^{θ}_{surr} This has been dealt with in section 3.3b ΔS, Entropy changes and the feasibility of a chemical change
3.4b Calculating free energy changes for chemical reactions applications Using the free energy equation: ΔG^{θ}_{system/reaction} = ΔH^{θ}_{system/reaction} TΔS^{θ}_{system/reaction}
In the case of cells the free energy equation is: ΔG^{θ}_{sys} = neE^{θ}F (see section 3.4c)
The free energy change of a feasible reaction, ΔG^{θ}_{sys}, must be <0, effectively the net cell p.d. must be positive. If ΔG^{θ}_{sys} 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 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 ΔG_{1} = ΔH_{1} TΔS_{1} and ΔG_{2} = ΔH_{2} TΔS_{2} ∴ ΔH_{1} TΔS_{1} = ΔH_{2} TΔS_{2} ∴ TΔS_{2} TΔS_{1} = ΔH_{2} ΔH_{1} ∴ T(ΔS_{2} TΔS_{1}) = ( ΔH_{2} ΔH_{1}) ∴ T = ( ΔH_{2} ΔH_{1}) / (ΔS_{2} ΔS_{1}) Introduction to the examples of free energy calculations and the linked concepts ΔG^{θ}_{system/reaction} = ΔH^{θ}_{system/reaction} TΔS^{θ}_{system/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.
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!) ΔH^{θ}_{f} = standard enthalpy of formation of a compound at 298K, 1 atm./101kPa pressure in kJmol^{1}
ΔG^{θ}_{f} = standard free energy of formation of a compound at 298K, 1 atm./101kPa pressure in kJmol^{1}
S^{θ}_{f} = standard entropy energy content at 298 K, 1 atm./101kPa pressure in JK^{}1mol^{1}
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.
Example (1) The thermal decomposition of lithium carbonate
To calculate the enthalpy change
To calculate the free energy change
To calculate the entropy change
Substituting into the Gibbs free energy change equation
To calculate the temperature at which the decomposition of lithium carbonate becomes feasible When ΔG^{θ}_{sys} = 0, you have reached the point of first feasibility Therefore, at this point: ΔH^{θ}_{sys} TΔS^{θ}_{sys} = 0 So T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} = 226 / 0.1628 = 1388 K (1115^{o}C) 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? ΔG^{θ}_{sys} = ΔH^{θ}_{sys} TΔS^{θ}_{sys} 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 = ΔG^{θ}_{sys}, gradient = m = ΔS^{θ}_{sys}, x = T(K), constant = c = ΔH^{θ}_{sys} So, by plotting a graph of ΔG^{θ}_{sys} versus T, you can derive ΔH^{θ}_{sys} and TΔS^{θ}_{sys} 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^{1}mol^{1} or +163JK^{1}mol^{1} From the constant c, ΔH = +226 kJmol^{1}
Example (2) The 'synthesis' of ethene from its elements
To calculate the enthalpy change
To calculate the free energy change
To calculate the entropy change
Substituting into the Gibbs free energy equation for other temperatures
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^{1}mol^{1} or 48.7JK^{1}mol^{1} From the constant c, ΔH = +52.3 kJmol^{1}
Example (3) The decomposition of hydrazoic acid (hydrogen azide, a colourless volatile explosive liquid) I could only find thermodynamic data on the gaseous state.
To calculate the enthalpy change
To calculate the free energy change
To calculate the entropy change
Substituting into the Gibbs free energy equation for other temperatures
Both terms are always negative, therefore ΔG^{θ}_{sys} 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^{1}mol^{1} or +117JK^{1}mol^{1} From the constant c, ΔH = +52.3 kJmol^{1}
Example (4) The synthesis of ammonia
To calculate the enthalpy change
To calculate the free energy change
To calculate the entropy change
Substituting into the Gibbs free energy equation for other temperatures
Which means it is not favoured by high temperatures, no matter how fast you want the reaction to take place!
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^{1}mol^{1} or 99.4JK^{1}mol^{1} From the constant c, ΔH = +46.2 kJmol^{1} I did find a set of values for the equilibrium constant K_{p}, for the equation N_{2}(g) + 3H_{2}(g) 2NH_{3}(g) K_{p} = p_{NH3}^{2}/p_{N2}p_{H2}^{3} 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.
Example (5) The thermal decomposition of ammonium chloride (a sublimation, solid ===> gases)
To calculate the enthalpy change
To calculate the free energy change
To calculate the entropy change
Substituting into the Gibbs free energy equation for other temperatures
To calculate the temperature at which the decomposition of ammonium chloride becomes feasible When ΔG^{θ}_{sys} = 0, you have reached the point of first feasibility Therefore, at this point: ΔH^{θ}_{sys} TΔS^{θ}_{sys} = 0 So T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} = 176.5 / 0.3294 = 539 K (263^{o}C) 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 kJmol1, 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.
Enthalpy change
Free energy change
Entropy change
Gibbs free energy change equation for other temperatures
To calculate the temperature at which this specific cracking reaction becomes feasible When ΔG^{θ}_{sys} = 0, you have reached the point of first feasibility Therefore, at this point: ΔH^{θ}_{sys} TΔS^{θ}_{sys} = 0 So, T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} = 229.6 / 0.259 = 886 K (613^{o}C)
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)
Enthalpy change
Free energy change
Entropy change
Gibbs free energy change equation for other temperatures
To calculate the temperature at which the decomposition of calcium carbonate becomes feasible When ΔG^{θ}_{sys} = 0, you have reached the point of first feasibility Therefore, at this point: ΔH^{θ}_{sys} TΔS^{θ}_{sys} = 0 So T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} = 178 / 0.0.1611 = 1105 K (263^{o}C)
so, the decomposition will occur at any temperature above 1105 K (832^{o}C), 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 E^{θ}_{cell(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 recalculate! The E_{batterycell} 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 E_{cell} rule of >0V for feasibility). ΔG^{θ}_{cell(sys)} = nE^{θ}F 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, E^{θ} is the Emf for the overall reaction in volts, F = the Faraday constant (96500 C mol^{1}). Ex 3.4c1 for the zinccopper Daniel cell producing +1.10V 2 electrons transferred: M^{2+}_{(aq)} + 2e^{} <=> M_{(s)} full cell redox reaction: Cu^{2+}_{(aq)} + Zn_{(s)} ==> Cu_{(s)} + Zn^{2+}_{(aq)} ΔG^{θ}_{cell} = 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 + Zn^{2+} ==>) 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.
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
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