**History**

The Gibbs-Helmholtz equation was first deduced by the German physicist Hermann von Helmholtz in his groundbreaking 1882 paper *“Die Thermodynamik chemischer Vorgänge”* (On the Thermodynamics of Chemical Processes). In it, he introduced the concept of free energy (*freie Energie*) and used the equation to demonstrate that free energy – not heat production – was the driver of spontaneous change in isothermal chemical reactions, thereby overthrowing the famously incorrect * Thomsen-Berthelot principle*.

Although Gibbs was first to state the relations A = U – TS and G = U + PV – TS, he did not explicitly state the Gibbs-Helmholtz equation, nor did he explore its chemical significance. So the honors for this equation really belong more to Helmholtz than to Gibbs.

But from the larger historical perspective, both of these gentlemen can rightly be considered the founders of chemical thermodynamics – Gibbs for his hugely long and insanely difficult treatise *“On the Equilibrium of Heterogeneous Substances”* (1875-1878), and Helmholtz for his landmark paper referred to above. These works had a significant influence on the development of physical chemistry.

**– – – –**

**Derivation**

The confusing thing about the Gibbs-Helmholtz equation is that it comes in ** three different versions**, but most physical chemistry texts don’t say why. This is not helpful to students. Chemical thermodynamics is difficult enough already, so CarnotCycle will begin by giving the reason.

It just so happens that the form of calculus used in the Gibbs-Helmholtz equation has the following property:

For a thermodynamic state function (f) and its natural variables (x,y)

If we choose as our state function (f) the Gibbs Free Energy G and assign its natural variables, temperature T to (x) and pressure P to (y), we obtain:

Since (∂G/∂T)_{P} = –S, and G ≡ H – TS

This is the Gibbs-Helmholtz equation. If we apply this equation to the initial and final states of a process occurring at constant temperature and pressure, and take the difference, we obtain:

where ΔH is the enthalpy change of a process taking place in a closed system capable of PV work; the three equivalent versions of the equation are determined by the properties of calculus:

A further useful relation can be derived from (2) and (3) using the equation ΔG^{°} = –RTlnK_{p} for a gas reaction where each of the reactants and products is in the standard state of 1 atm pressure.

Substituting –RlnK_{p} for ΔG^{°}/T in (2) and (3) yields

These are equivalent forms of the van ‘t Hoff equation, named after the Dutch physical chemist and first winner of the Nobel Prize in Chemistry, J.H. van ‘t Hoff (1852-1911). Approximate integration yields

**– – – –**

**Applications of the Gibbs-Helmholtz equation**

**1. Calculate ΔH _{rxn} from ΔG and its variation with temperature at constant pressure**

This application of (1) is useful particularly in relation to reversible reactions in electrochemical cells, where ΔG identifies with the electrical work done –nFE. Scroll down to see the worked example GH1

**2. Calculate ΔG _{rxn} for a reaction at a temperature other than 298K**

ΔH usually varies slowly with temperature, and can with reasonable accuracy be regarded as constant. Integration of (2) or (3) enables you to compute ΔG

_{rxn}for a constant-pressure process at a temperature T

_{2}from a knowledge of ΔG and ΔH at temperature T

_{1}. Scroll down to see the worked example GH2

**3. Calculate the effect of a temperature change on the equilibrium constant K _{p}**

ΔH usually varies slowly with temperature, and can with reasonable accuracy be regarded as constant. The integrated van ‘t Hoff equation (6) allows the equilibrium constant K

_{p}at T

_{2}to be calculated with knowledge of K

_{p}and ΔH° at T

_{1}. Scroll down to see the worked example GH3

**– – – –**

*Insight: The Φ function and the meaning of –ΔG/T*

By 1897, Hermann von Helmholtz was dead and Max Plank was professor of theoretical physics at the University of Berlin where he published ** Treatise on Thermodynamics**, a popular textbook which ran to several editions. In it, Planck introduced the Φ function (originally deduced by François Massieu in 1869) as a measure of chemical stability:

S is the entropy of the system, and (U+pV)/T is the enthalpy H of the system divided by its temperature T. Since G ≡ H – TS, we can immediately identify Φ with –G/T.

Planck’s formula indicates that Φ tends to be large when S is large and H is small, i.e. when the energy levels are closely spaced and the ground level is low – the criteria for chemical stability.

The Φ function gets even more interesting when one considers the meaning of ΔΦ.

ΔΦ = –ΔG/T = –ΔH/T + ΔS = ΔS_{surroundings} + ΔS_{system} = ΔS_{universe}

ΔΦ and –ΔG/T equate to the increase in the entropy of the universe: a measure of the ultimate driving force behind chemical reactions. The larger the value, the more strongly the reaction will want to go.

A direct association between –ΔG/T and the equilibrium constant K is thus implied, and this can be confirmed by rearranging the relation ΔG° = –RTlnK_{p} to: –ΔG°/T = RlnK_{p}

**– – – –**

*Worked Example GH1*

An electrochemical cell has the following half reactions:

Anode (oxidation): Ag(s) + Cl^{–} → AgCl(s) + 1e^{–}

Cathode (reduction): ½Hg_{2}Cl_{2}(s) + 1e^{–} → Hg(l) + Cl^{–}

The EMF of the cell is +0.0455 volt at 298K and the temperature coefficient is +3.38 x 10^{-4} volt per kelvin. Calculate the enthalpy of the cell reaction, taking the faraday (F) as 96,500 coulombs.

Strategy

Use version 1 of the Gibbs-Helmholtz equation

Substitute for ΔG using the relation ΔG = –nFE

Calculation

[note that if the EMF is positive, the reaction proceeds spontaneously in the direction shown in the half reactions. If the EMF is negative, the reaction goes in the opposite direction]

The complete cell reaction for one 1 faraday is:

Ag(s) + ½Hg_{2}Cl_{2}(s) → AgCl(s) + Hg(l)

Each mole of silver transfers one mole of electrons (1e^{–}) to one mole of Cl^{–} ions. So n = 1.

F = 96,500 C

E = 0.0455 V

T = 298 K

(∂E/∂T)_{P} = 3.38 x 10^{-4} VK^{-1}

ΔH_{rxn} = –1 x 96,500 (0.0455 – 298 (3.38 x 10^{-4})) joules

Dimensions check: remember that V= J/C, so C x V = J

ΔH_{rxn} = 5329 J = 5.329 kJ

[note that this electrochemical cell makes use of a spontaneous endothermic reaction]

**– – – –**

*Insight: The effect of temperature on EMF*

From version 1 of the Gibbs-Helmholtz equation

and the relations ΔG = –nFE and (∂ΔG/∂T)_{P} = –ΔS, it can be seen that

For many redox reactions that are used to power electrochemical cells, ΔS_{rxn} is typically small (less than 50 JK^{-1}). As a result (∂E/∂T)_{P} is usually in the 10^{-4} to 10^{-5} range, and hence electrochemical cells are relatively insensitive to temperature.

**– – – –**

*Worked Example GH2*

The Haber Process for the production of ammonia is one of the most important industrial processes: N_{2}(g) + 3H_{2}(g) = 2NH_{3}(g)

ΔG°(298K) = –33.3 kJ

ΔH°(298K) = –92.4kJ

Calculate ΔG° at 500K

Strategy

Use version 3 of the Gibbs-Helmholtz equation

Making the assumption that ΔH° remains approximately constant, perform integration

Solve for ΔG°_{(T2)}

Calculation

ΔG°_{(T1)} = –33.3 kJ

ΔH° = –92.4 kJ

T_{2} = 500K

T_{1} = 298K

Inserting these values into the integrated equation yields the result ΔG°(500K) ≈ 6.76 kJ. Compared with the negative value of ΔG° at 298K, the small positive value of ΔG° at 500K shows that the reaction has just become unfeasible at this temperature, pressure remaining constant at 1 atm.

**– – – –**

*Insight: Reading the Haber process equation*

N_{2}(g) + 3H_{2}(g) = 2NH_{3}(g) ΔH°(298K) = –92.4 kJ

The Haber process is exothermic (negative ΔH) and results in a halving of volume (negative ΔS). Since ΔG = ΔH – TΔS, increasing the temperature will drive ΔG in a positive direction, leading to an upper temperature limit on reaction feasibility.

Le Châtelier’s principle shows that the Haber process is thermodynamically favored by low temperature and high pressure. In practice however a compromise has to be struck, since low temperature slows the rate at which equilibrium is achieved while high pressure increases the cost of equipment and maintenance.

**– – – –**

*Worked Example GH3*

The Haber Process for the production of ammonia is one of the most important industrial processes:

N_{2}(g) + 3H_{2}(g) = 2NH_{3}(g) ΔH°(298K) = –92.4 kJ

The equilibrium constant K_{P} at 298K is 6.73 x 10^{5}. Calculate K_{P} at 400K.

Strategy

Use the approximate integral of the van ‘t Hoff equation (6) to solve for K_{P}(T_{2})

Calculation

K_{P}(T_{1}) = 6.73 x 10^{5}, ln K_{P}(T_{1}) = 13.42

ΔH°(298K) = –92400 J (mol^{-1})

R = 8.314 J K^{-1} mol^{-1}

T_{2} = 400K

T_{1} = 298K

Inserting these values into the integrated equation yields the approximate result ln K_{P}(T_{2}) ≈ 3.91, therefore K_{P}(400K) ≈ 49.92. This is reasonably close to the measured value of K_{P}(400K) = 48.91.

Compared to the large value of K_{P} at 298K, the small positive value of K_{P} at 400K shows that the reaction is approaching the point where it will shift to become reactant-favorable rather than product favorable.

P Mander August 2015

Laid out nicely!

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