Posts Tagged ‘enthalpy’

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In terms of simplicity, purely mechanical systems have an advantage over thermodynamic systems in that stability and instability can be defined solely in terms of potential energy. For example the center of mass of the tower at Pisa, in its present state, must be higher than in some infinitely near positions, so we can conclude that the structure is not in stable equilibrium. This will only be the case if the tower attains the condition of metastability by returning to a vertical position or absolute stability by exceeding the tipping point and falling over.

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Thermodynamic systems lack this simplicity, but in common with purely mechanical systems, thermodynamic equilibria are always metastable or stable, and never unstable. This is equivalent to saying that every spontaneous (observable) process proceeds towards an equilibrium state, never away from it.

If we restrict our attention to a thermodynamic system of unchanging composition and apply the further constraint that the system is closed, i.e. no mass transfer takes place between system and surroundings, the fundamental relation of thermodynamics can be written as an exact differential expression exhibiting two independent variables: dU = TdS–PdV. There are therefore 22–1 possible Legendre transformations, each representing a new function. We thus arrive at a set of four state functions

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where the independent variables, shown in parentheses, are the natural (canonical) variables for that function [S = entropy, V = volume, T = temperature, P = pressure].

If the above pairs of natural variables are held constant, the energy function is predictably a minimum at equilibrium. But what if the energy function and one of its natural variables are instead held constant? Which extremum will the other natural variable reach at equilibrium – minimum or maximum? There are eight permutations, giving a grand total of twelve sets of equilibrium conditions.

Of these twelve, only four have straightforward proofs based on the fundamental criteria [derived in my previous blogpost Reversible and Irreversible Change] that

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The proofs of the other eight involve a lengthier train of logical argument which is now rarely seen either in textbooks or on the internet. To redress this issue, CarnotCycle herewith presents all twelve proofs in their entirety.

– – – –

The Twelve Conditions of Equilibrium and Stability

Here are the four straightforward proofs

1. For given U and V that S is a maximum

For a closed system, the First Law may be written

cse05 (3)

Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the volume is also kept constant during the change, it follows from the First Law that dU = 0. The criteria for these conditions may be expressed as follows

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Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when U and V are held constant.

2. For given H and P that S is a maximum

For a closed system, the First Law may be written

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Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the pressure P of the system is kept constant during the change, and is made to differ only infinitesimally from that of the surroundings, equation (3) becomes

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By the definition of enthalpy, we find

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In this case therefore, H and P are constant, and the criteria for these conditions may be expressed as follows

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Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when H and P are held constant.

3. For given T and V that A is a minimum

For a closed system undergoing a change at constant temperature, we may write the differential expression

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But by the first law

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so that

cse11 (4)

Suppose the system is capable only of PV work. If the volume is kept constant, no work can be done and therefore

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Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of A:

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Since observable change always proceeds towards equilibrium, A will decrease towards a minimum at equilibrium when T and V are held constant.

4. For given T and P that G is a minimum

For a closed system undergoing a change at constant temperature, we may write the differential expression (cf. equation 4 above)

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Suppose the system is capable only of PV work. If the pressure P of the surroundings is kept constant during the change, and is made to differ only infinitesimally from that of the system, then

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Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of G:

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Since observable change always proceeds towards equilibrium, G will decrease towards a minimum at equilibrium when T and P are held constant.

– – – –

A word of introduction to the remaining eight not-so-straightforward proofs.

On page 2 of his masterwork On the Equilibrium of Heterogeneous Substances, under the heading Criteria of Equilibrium and Stability, J. Willard Gibbs constructs the foundations of chemical thermodynamics on the equivalence of two propositions relating to an isolated system – that for given U and V that S is a maximum, and for given S and V that U is a minimum.

Gibbs shows by neat argument that the truth (according to Clausius) of the first proposition necessitates the truth of the second proposition. In parallel fashion, the proof already given for the first proposition (condition 1 above) will be used to prove that for given S and V that U is a minimum. This sequitur argument repeats throughout, with condition 2 used to prove condition 6, condition 3 used to prove condition 7 etc.

The train of logical argument for these remaining proofs is somewhat labyrinthine, but the method is same in each case. It gets easier to follow as you go from one to the next.

– – – –

5. For given S and V that U is a minimum

Imagine a closed system in some state of equilibrium for which the values of the entropy, volume and internal energy are S1, V1 and U1. Then in any non-equilibrium state with the same values V1 and U1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (1) of observable change at constant U and V requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, volume and internal energy are S2, V1 and U2. Compared with the first equilibrium state the volume is unaltered but S2 < S1. Now U and S always change in the same direction since dU = TdS – PdV implies (∂U/∂S)V = T which is always positive. Hence in the second equilibrium state U2 < U1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and volume are the same but U2 < U1. Since observable change always proceeds towards equilibrium, U will decrease towards a minimum at equilibrium when S and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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6. For given S and P that H is a minimum

Imagine a closed system in some state of equilibrium for which the values of the entropy, pressure and enthalpy are S1, P1 and H1. Then in any non-equilibrium state with the same values P1 and H1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (2) of observable change at constant H and P requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, pressure and enthalpy are S2, P1 and H2. Compared with the first equilibrium state the pressure is unaltered but S2 < S1. Now H and S always change in the same direction since dH = TdS + VdP implies (∂H/∂S)P = T which is always positive. Hence in the second equilibrium state H2 < H1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and pressure are the same but H2 < H1. Since observable change always proceeds towards equilibrium, H will decrease towards a minimum at equilibrium when S and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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7. For given A and V that T is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, volume and temperature are A1, V1 and T1. Then in any non-equilibrium state with the same values V1 and T1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, volume and temperature are A2, V1 and T2. Compared with the first equilibrium state the volume is unaltered but A2 > A1. Now A and T always change in opposite directions since dA = –SdT – PdV implies (∂A/∂T)V = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and volume are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when A and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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8. For given G and P that T is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, pressure and temperature are G1, P1 and T1. Then in any non-equilibrium state with the same values P1 and T1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, pressure and temperature are G2, P1 and T2. Compared with the first equilibrium state the pressure is unaltered but G2 > G1. Now G and T always change in opposite directions since dG = VdP – SdT implies (∂G/∂T)P = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and pressure are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when G and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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9. For given U and S that V is a minimum

Imagine a closed system in some state of equilibrium for which the values of the internal energy, entropy and volume are U1, S1 and V1. Then in any non-equilibrium state with the same values S1 and V1, the internal energy U2 must be greater than U1 because observable change always proceeds towards equilibrium, and the condition (5) of observable change at constant S and V requires the internal energy to decrease.

Now imagine a second equilibrium state for which the values of the internal energy, entropy and volume are U2, S1 and V2. Compared with the first equilibrium state the entropy is unaltered but U2>U1. Now U and V always change in opposite directions since dU = TdS – PdV implies (∂U/∂V)P = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the internal energy and entropy are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when U and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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10. For given H and S that P is a maximum

Imagine a closed system in some state of equilibrium for which the values of the enthalpy, entropy and pressure are H1, S1 and P1. Then in any non-equilibrium state with the same values S1 and P1, the enthalpy H2 must be greater than H1 because observable change always proceeds towards equilibrium, and the condition (6) of observable change at constant S and P requires enthalpy to decrease.

Now imagine a second equilibrium state for which the values of the enthalpy, entropy and pressure are H2, S1 and P2. Compared with the first equilibrium state the entropy is unaltered but H2 > H1. Now H and P always change in the same direction since dH = TdS + VdP implies (∂H/∂P)S = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the enthalpy and entropy are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when H and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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11. For given A and T that V is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, temperature and volume are A1, T1 and V1. Then in any non-equilibrium state with the same values T1 and V1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, temperature and volume are A2, T1 and V2. Compared with the first equilibrium state the temperature is unaltered but A2 > A1. Now A and V always change in opposite directions since dA = –SdT – PdV implies (∂A/∂V)T = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and temperature are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when A and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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12. For given G and T that P is a maximum

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, temperature and pressure are G1, T1 and P1. Then in any non-equilibrium state with the same values T1 and P1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, temperature and pressure are G2, T1 and P2. Compared with the first equilibrium state the temperature is unaltered but G2 > G1. Now G and P always change in the same direction since dG = VdP – SdT implies (∂G/∂P)T = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and temperature are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when G and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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– – – –

P Mander April 2015

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Historical background*

It was the American physicist Josiah Willard Gibbs (1839-1903) pictured above who first introduced the thermodynamic potentials ψ, χ, ζ which we today call Helmholtz free energy (A), enthalpy (H) and Gibbs free energy (G).

In his milestone treatise On the Equilibrium of Heterogeneous Substances (1876-1878), Gibbs springs these functions on the reader with no indication of where he got them from. Using an esoteric lexicon of Greek symbols he simply states:

Let
ψ = ε – tη
χ = ε + pv
ζ = ε – tη + pv

As with much of Gibbs’ writings, the clues to his sudden pronouncements need to be sought on other pages or – as in this case – another publication.

In an earlier paper entitled A method of geometrical representation of the thermodynamic properties of substances by means of surfaces, Gibbs shows that the state of a body in terms of its volume, entropy and energy can be represented by a surface:

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Gibbs’ thermodynamic surface of 1873, realized by James Clerk Maxwell in 1874

It can be demonstrated from purely geometrical considerations that the tangent plane at any point on this surface represents the U-related function

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Now this is none other than Gibbs’ zeta (ζ ) function. The question is, did he recognize it for what it was – a Legendre transform? A key feature of On the Equilibrium of Heterogeneous Substances is the business of finding an extremum for a multivariable function subject to various kinds of constraint, and it is known that Gibbs was familiar with Lagrange’s method of multipliers – he mentions the technique by name on page 71, immediately after equation 41. The point here is that the Legendre transformation can be phrased in the same terms – for example, the multiplier expression for finding the stationary value of U when T and P are held constant yields the Legendre transform shown above.

But suggestive though this is, it actually gets us no closer to determining whether or not Gibbs was aware that ψ, χ, ζ  were Legendre transforms. Gibbs gave no indication in his writings either that he knew the transformation trick, or that he had discovered it for himself. We can only estimate likelihoods and have hunches.

*Text revised following input from Bas Mannaerts (see comments below)

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In the CarnotCycle thermodynamics library, the first textbook reference to Legendre transformation is by P.S. Epstein in 1937. Epstein was a Russian mathematical physicist who was recruited by Caltech in 1921. He was a renowned commentator on Gibbs’ work, especially in statistical mechanics.

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Thermodynamics and the Legendre transformation

The fundamental relation of thermodynamics dU = TdS–PdV is an exact differential expression

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where the coefficients Ci are functions of the independent variables Xi. By means of Legendre transformations (ℑ) the above expression generates three new state functions whose natural variables contain one or more Ci in place of the conjugate Xi

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The equation of the tangent plane to the thermodynamic surface generates ℑ3, with ℑ1 and ℑ2 following procedurally from

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How the Legendre transformation works

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defines a new Y-related function Z by transforming

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into

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Proof
dZ = dY – d(C1X1)
dZ = dY – C1dX1 – X1dC1
Substitute dY with the original differential expression
dZ = C1dX1 + C2dX2 – C1dX1 – X1dC1
The C1dX1 terms cancel, leaving
dZ = C2dX2 – X1dC1

The independent (natural) variables are transformed from Y(X1,X2) to Z(X2, C1)
The same procedural principle applies to ℑ2 and ℑ3.

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The Legendre Wheel

Since exact differential expressions in two independent (natural) variables can be written for the internal energy (U), the enthalpy (H), the Gibbs free energy (G) and the Helmholtz free energy (A), each of these state functions can generate the other three via the Legendre transformations ℑ1, ℑ2, ℑ3. This is neatly demonstrated by the Legendre Wheel, which executes the transformation functions

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from any of the four starting points:

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 [click on image to enlarge]

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Legendre transformations and the Gibbs-Helmholtz equations

For an exact differential expression

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the transforming function

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can be written in terms of the natural variables of Y

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This Legendre transformation is the means by which we obtain the Gibbs-Helmholtz equations. Taking Y=G(T,P) as an example, ℑ1 executes the clockwise transformation

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while the transforming function

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reverses the positions of the natural variables and executes the counterclockwise transformation

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Setting Y=G(T,P) generates six Gibbs-Helmholtz equations, in each of which one of the two natural variables is held constant. Since there are four state functions – U, H, G and A – the total number of Gibbs-Helmholtz equations generated by this procedure is twenty-four. To this can be added a parallel set of twenty-four equations where U, H, G and A are replaced by ΔU, ΔH, ΔG and ΔA.

These equations are particularly useful since they relate a state function’s dependence on either of its natural variables to an adjacent state function on the Legendre Wheel.

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Who was Legendre?

Adrien Legendre (1752-1833) was a French mathematician. He wrote a popular and influential geometry textbook Éléments de géométrie (1794) and contributed to the development of calculus and mechanics. The Legendre transformation and Legendre polynomials are named for him.

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P Mander September 2014

"What [the National Weather Service] has found, year after year, is that the levels of carbon pollution in our atmosphere have increased dramatically. That science, accumulated and reviewed over decades, tells us that our planet is changing in ways that will have profound impacts on all of humankind." Speech delivered by President Obama at Georgetown University, Washington D.C. on Tuesday 25th June 2013.

“What [the National Weather Service] has found, year after year, is that the levels of carbon pollution in our atmosphere have increased dramatically. That science, accumulated and reviewed over decades, tells us that our planet is changing in ways that will have profound impacts on all of humankind.”
Speech delivered by President Obama at Georgetown University, Washington D.C. on Tuesday 25th June 2013.

In the week (week 26, 2013 – Ed) that President Obama announced his bold plan for the US to reduce atmospheric carbon pollution and put the brakes on global warming, what better than a blog post that examines the thermochemical relationship between heat production and CO2 emissions in hydrocarbon fuels.

But before we begin…

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The camping gas cartridge pictured above contains a mixture of butane (C4H10) and propane (C3H8). Combustion of this mixture generates heat, along with the combustion products carbon dioxide and water.

Now here’s a question for you. If instead of the mixture, the cartridge contained only butane, would you expect combustion to produce more or less carbon dioxide in relation to the amount of heat generated?

Base your reasoning solely on the information given above. Jot down your answer, and we will come back to it later.

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Thermochemistry is the branch of thermodynamics which studies thermal processes associated with chemical reactions and other forms of chemical change. From its beginnings in the 18th century, thermochemistry played an important role in overcoming the misunderstandings about heat which arose from its confusion with the concept of temperature.

Laplace and Lavoisier's ice calorimeter of 1780. James Clerk Maxwell wrote of it: "Nothing can be more perfect than the theory and design of this apparatus. It is worthy of Laplace and Lavoisier, and in their hands it furnished good results."

Laplace and Lavoisier’s ice calorimeter of 1780. James Clerk Maxwell wrote of it: “Nothing can be more perfect than the theory and design of this apparatus. It is worthy of Laplace and Lavoisier, and in their hands it furnished good results.”

Through the development and use of the calorimeter, two important early discoveries were made. In France in 1780, those men of genius – Antoine Lavoisier and Pierre-Simon Laplace – discovered that the heat gained or lost in a chemical reaction is equal in magnitude and opposite in sign for the reverse reaction. And in Russia in 1840, the Swiss-born Germain Hess discovered that if a given set of reactants is transformed into a given set of products by more than one reaction path, the total amount of heat gained or lost is the same for each path.

The second of these two discoveries, now called Hess’s law, has for decades been peculiarly subject to factual distortion in science education. Hess measured heat changes at constant pressure (qp), but because of the later-discovered identity qp = ΔH, Hess’s law is commonly stated in terms of enthalpy change. Students are thus misled into thinking that Hess discovered an enthalpy summation relation, when in fact he had been dead for 25 years before the concept of enthalpy was first defined by Gibbs in 1875. This kind of timeline falsification, although motivated by an understandable desire to keep things simple in textbooks and internet resources, does the proper study of thermodynamics a disservice in my view.

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The development in the first half of the 20th century of the concept of the covalent bond led directly to a new class of thermochemical data called bond enthalpies. Tables of these averaged quantities allow the estimation of enthalpy of reaction using the rule:

ΔHrxn = ΣΔH(bonds broken) – ΣΔH(bonds formed)

Care needs to be taken with average values, since there can be considerable variations in certain circumstances and you need eyes in the back of your head for things like phase transitions and unusual oxidation states, but for applications involving uncomplicated single-phase reactions, bond enthalpies can be a very useful resource.

Here’s an example using an analytical technique of my own. Consider the combustion of alkanes in the vapor phase, which can be represented by the general equation:

CnH2n+2(g) + ½(3n+1)O2(g) → nCO2(g) + (n+1)H2O(g)

Say you are interested in the relationship between the heat generated by the reaction (ΔHrxn) and the number of moles (n) of carbon dioxide produced.

The procedure is as follows. First, expressions in n are found for bonds broken and formed:

bonds broken

total

avg. bond enthalpy

C–H

2n+2

a

O=O

½(3n+1)

b

C–C

n-1

c

bonds formed  

 

C=O

2n

d

O–H 2n+2

e


The enthalpy of reaction ΔHrxn can then be expressed as the sum of the bond enthalpies of the reactants, minus the sum of the bond enthalpies of the products, all bond enthalpies being regarded as positive quantities.

ΔHrxn =  [(2n+2)a + ½(3n+1)b + (n-1)c] – [(2n)d + (2n+2)e]    …(1)

The incremental value of ΔHrxn for an integer increase in n can be found by differentiation:

d(ΔHrxn)/dn = [2a + (3/2)b + c] – [2d + 2e]    …(2)

It is evident from inspection of (1) that (2) represents the limiting value of ΔHrxn per mole of CO2 produced as n becomes very large. This fixes the far end of the alkane scale; the other end is found by setting n=1 in (1):

ΔHrxn(n=1) = [4a + 2b] – [2d + 4e]    …(3)

By inserting the values of the average bond enthalpies into expressions (3) and (2), the range of ΔHrxn per mole of CO2 produced by the combustion of alkanes is determined. The absolute values may not be very accurate, but the trend as n increases will have sufficient accuracy to indicate which alkanes generate the most energy per mole of CO2 produced.

Note: The analytical technique described above is original work by the author of CarnotCycle. If you make use of it for your own purposes, please include a reference to this blog/blogpost. Many thanks.

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At a time when atmospheric CO2 levels have passed the 400 ppm mark for the first time since the Pliocene period, it seems an appropriate moment to take a look at the principal hydrocarbon fuels and see which of them generate most energy in relation to carbon dioxide produced.

Energy generated (kJ)

Reaction enthalpy

per gram fuel per mole O2

per mole CO2

Natural gas CH4 + 2O2=CO2 + 2H2O

810

51.6 405

810

Ethanol C2H5OH + 3O2 = 2CO2 + 3H2O

1257

27.3 419

628

Petroleum 2(-CH2-) +3O2 = 2CO2 + 2H2O

1220

43.6 407

610

Coal 4(-CH-) + 5O2 = 4CO2 + 2H2O

2046

39.3 409

512

Cellulose (-CHOH-) + O2 = CO2 + H2O

447

14.9 447

447


According to these figures widely quoted on the internet, natural gas gives the best quotient among the fossil fuels, delivering 42% more energy per mole of CO
2 produced than coal, which explains the “cleaner alternative” phrase favored by energy ministers when asked about all the sudden interest in shale gas.

In terms of transport fuels, there is little difference on this analysis between ethanol and petroleum. The value of bioethanol supplementation of gasoline would appear to be linked more to sustainability and supply security than energy/emission quotient, although for fair comparison one should take biomass origin and CO2 emissions during industrial processing into the overall reckoning.

Cellulose gives the lowest quotient, although again it must be acknowledged that the carbon it contains came from atmospheric CO2 in the first place. The only observation I would make about the popular “CO2-neutral” appellation is that while living plants generally take up CO2 at average atmospheric concentrations, power stations burning biomass pump out CO2 at vastly higher concentrations. Under continual operation, a steady-state gradient with a high local CO2 concentration is likely to be established. If there is water nearby (power stations are often placed next to large water masses for cooling purposes), CO2 will readily dissolve in it. So in overall terms, CO2-neutral could actually mean accelerated atmosphere-to-ocean transfer, which in view of the resultant ocean acidification may not be entirely neutral in its environmental impact.

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One last point to cover is the underlying reason for the differences in energy/emission quotient shown in the final column of the above table. The figures in the penultimate column for energy generated per mole of oxygen consumed show remarkably little variation between fuels, which indicates that the explanation of the difference lies in the other element in the CO2 molecule – carbon.

The covalently bonded carbon atom can vary in its oxidation state depending on the atoms other than carbon to which it is attached. Bonding to hydrogen decreases the oxidation state, while bonding to oxygen increases it. And this affects the amount of energy released from a hydrocarbon fuel on combustion – the lower the average oxidation number of the carbon, the higher the enthalpy of combustion, and vice versa. The table below demonstrates the correlation nicely.

kJ per mole of CO2

Avg. oxidation number of carbon in fuel

Natural gas CH4 + 2O2 = CO2+2H2O

810

–4

Ethanol C2H5OH + 3O2 = 2CO2 + 3H2O

628

–2

Petroleum 2(-CH2-) +3O2 = 2CO2 + 2H2O

610

–2

Coal 4(-CH-) + 5O2 = 4CO2 + 2H2O

512

–1

Cellulose (-CHOH-) + O2 = CO2 + H2O

447

0


We have reached the end, so let’s go back to that question posed at the beginning –

“Would you expect butane to produce more or less carbon dioxide in relation to the amount of heat generated than a butane-propane mixture?”

The answer is more carbon dioxide.

The average oxidation number of carbon in butane is –10/4, while that of propane is –8/3. Butane alone has a less negative (i.e. higher) average carbon oxidation number than any butane-propane mixture, and so should produce more carbon dioxide in relation to the amount of heat generated. Enthalpy of combustion determinations show this prediction to be correct.

– – – –

 

left: Julius Thomsen (1826-1909) right: Marcellin Berthelot (1827-1907)

left: Julius Thomsen (1826-1909) right: Marcellin Berthelot (1827-1907)

Historical Background

In the mid-19th century, chemistry was a predominantly experimental science. There were of course the gas laws, and a handful of other stand-alone laws. But what was missing from chemistry was a theoretical foundation upon which a framework of understanding could be built.

Then around 1850, with the law of energy conservation beginning to be understood thanks to the pioneering work of Robert Mayer and James Joule, progress seemed to be made. Chemists started thinking about the ‘driving force’ causing chemical change, and concluded that the heat produced in chemical reactions provided a good quantitative measure of this force.

According to the Thomsen-Berthelot principle, formulated in slightly different versions by two professors of chemistry – Julius Thomsen in Denmark in 1854 and Marcellin Berthelot in France in 1864 – all chemical changes were accompanied by heat production, and the actual process that occurred was the one which produced most heat.

Right from the start, the Thomsen-Berthelot principle attracted criticism, but more from physicists than chemists. By the late 1870s, most chemists still supported it, but physicists regarded it as ill-conceived because the theory rested solely on the first law of thermodynamics, from which no conclusions could be drawn as to the direction of chemical change. That was the province of the second law.

What finally overthrew the Thomsen-Berthelot principle was a landmark paper “Die Thermodynamik chemischer Vorgänge” (On the Thermodynamics of Chemical Processes), written in 1882 by the 61 year-old German physicist Hermann von Helmholtz. In it, Helmholtz gave physical chemistry a solid theoretical foundation by showing that the driving force of a chemical reaction was measured not by heat production but by the maximum work produced when the reaction was carried out reversibly. Helmholtz also provided a name for this reversible maximum work – he called it “free energy”, for reasons we shall see.

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Hermann von Helmholtz (1821-1894)

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Enthalpy change is a first law concept

Physicists criticized the Thomsen-Berthelot principle on theoretical grounds. They argued that the direction of chemical change must involve the second law of thermodynamics. But heat production during chemical change involves only the first law and so cannot in itself be a measure of the driving force.

The fact that enthalpy change is solely dependent on first law considerations can immediately be seen from the textbook derivation:

Consider a chemical change carried out at constant pressure rather than at constant volume. Under such conditions, the work w done by the system as a result of volume change is not zero.

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(U+PV), like U, is a state function, as U, P and V are all state functions. It is called enthalpy and is defined

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The enthalpy change of a system is equal to the heat produced at constant pressure, assuming that the system does only PV work.

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Helmholtz finds a new thermodynamic function

Physicists were unified in their view that the force driving chemical reactions must involve the second law of thermodynamics, which Rudolf Clausius had incorporated into his Mechanical Theory of Heat via the fundamental relation

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It was this equation for a reversible process which seems to have served as a starting point for Helmholtz to generate a new state function using a method functionally equivalent to a Legendre transformation (I have not been able to access Helmholtz’s original 1882 paper, so what follows is based on secondary sources and may not be reliable in all details).

Helmholtz substituted TdS in the above equation with

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giving

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and generating a new state function we now know as the Helmholtz free energy A

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The question was, what did this new function represent? Helmholtz recognised that for isothermal reversible processes the new function identified with the maximum work

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If during a reversible change the system does work, wrev will be negative. dA will also be negative and A will decrease. Helmholtz saw that under isothermal conditions A was the equivalent in chemical systems of potential energy in mechanical systems: it was a measure of the maximum work the system can do on its surroundings.

Since U was the total energy (gessamt Energie) and A was the maximum work, Helmholtz argued that U–A represented the energy that was unavailable for conversion into work. He called the TS term “bound energy” (gebundene Energie) and therefore by logical extension, A was the “free energy” (freie Energie) for isothermal processes.

And so just as potential energy was the driver of change in mechanical systems, free energy was the driver of change in chemical systems. Heat production, the quantity erroneously used as a measure of driving force of chemical change in classical thermochemistry, identified with the loss of bound energy.

Helmholtz went on to consider the new function A under constant volume conditions, and was the first to derive what is now generically called a “Gibbs-Helmholtz” equation

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The cornerstone of the Thomsen-Berthelot principle was that chemical reactions were only possible if they produced heat; endothermic (heat-absorbing) reactions were impossible. Helmholtz was now in a position to challenge this assertion on theoretical grounds: there was nothing in the latter equation to suggest that ΔA could not be less negative than T(∂ΔA/∂T)V, in which case ΔU would be positive corresponding to an absorption of heat from the surroundings. Endothermic reactions were thus compatible with the new concept of free energy as the driver of chemical change.

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Why is the Gibbs-Helmholtz equation named for Gibbs when he didn’t deduce it?

Good question. It is a matter of historical fact that, although Gibbs was first to state the relation

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he did not specifically state the Gibbs-Helmholtz equation

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nor did he explore its chemical significance.

As the American physical chemist Wilder Bancroft commented, “This is an equation which Helmholtz did deduce and which Gibbs could have and perhaps should have deduced, but did not.”

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Wilder Bancroft (1867-1953) Born in Middletown, Rhode Island, he received a BA from Harvard University in 1888 and in 1892 gained a PhD from Leipzig University where Wilhelm Ostwald was professor. Commenting on the apparently inappropriate inclusion of Gibbs’ name in the Gibbs-Helmholtz equation, he explained “The error goes back to Ostwald who was advertising Gibbs at the time.” [Ostwald was the first to translate Gibbs’ milestone monograph on chemical thermodynamics into German, and to promote it in Europe]