# Gibbs energy

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Physical size
Surname Free enthalpy,
Gibbs energy
Size type energy
Formula symbol ${\ displaystyle G}$
Size and
unit system
unit dimension
SI J  = kg · m 2 · s -2 L 2 · M · T −2

The Gibbs energy (also free enthalpy ), named after Josiah Willard Gibbs , is a thermodynamic potential , i.e. a state variable in thermodynamics . It is an extensive quantity with the dimension of energy . In the SI system of units , it is measured in the unit joule . Its symbol is and its natural variables are temperature , pressure and particle numbers . ${\ displaystyle G}$ ${\ displaystyle p}$${\ displaystyle \ {N_ {i} \}}$

The Gibbs energy of a system results from its enthalpy through a Legendre transformation with respect to the entropy, in that the enthalpy is reduced by the product of the absolute temperature and the entropy : ${\ displaystyle H}$ ${\ displaystyle T}$ ${\ displaystyle S}$

${\ displaystyle G (T, p, \ {N_ {i} \}) = H (S, p, \ {N_ {i} \}) - T \, S}$.

or starting from the internal energy : ${\ displaystyle U}$

${\ displaystyle G (T, p, \ {N_ {i} \}) = U (S, V, \ {N_ {i} \}) + p \, VT \, S}$.

The molar Gibbs energy (unit: J / mol ) is the Gibbs energy related to the amount of substance : ${\ displaystyle n}$

${\ displaystyle G_ {m} = {\ frac {G} {n}}}$.

The specific Gibbs energy (unit: J / kg) is the Gibbs energy related to the mass : ${\ displaystyle m}$

${\ displaystyle g = {\ frac {G} {m}}}$.

The molar and the specific Gibbs energy are intense quantities : If two identical subsystems have the same molar or specific Gibbs energy, then the overall system formed from them also has this molar or specific Gibbs energy.

## overview

The Gibbs energy is measured in joules, but it is not an independent form of energy , such as kinetic energy or the internal energy contained in the system or a converted heat of reaction . It must not be added up when calculating the total energy. It is just an abbreviation for the common expression . With their help, statements can be made about the behavior of the system, such as the direction of voluntary processes or the position of states of equilibrium. ${\ displaystyle U + p \, VT \, S}$

### Gibbs energy as equilibrium criterion

A process in a given thermodynamic system takes place voluntarily if and only if it is associated with an increase in the total entropy of the system and its surroundings. If one restricts oneself to processes that take place at constant temperature and constant pressure, then the processes that increase the entropy of the system and the environment are precisely those that reduce the Gibbs energy of the system . A closed system kept at constant temperature and under constant pressure therefore assumes that state as a state of equilibrium in which its Gibbs energy has the smallest possible value. If the system is not in equilibrium, it goes over voluntarily (provided there are no other inhibitions) into states of lower Gibbs energy until equilibrium is reached. Since numerous physical and chemical processes take place at constant temperature ( isothermal ) and constant pressure ( isobaric ), the Gibbs energy provides a frequently applicable criterion for the direction in which the process takes place voluntarily and for the position of equilibrium.

For example, for an isothermal and isobaric chemical reaction :

• If the Gibbs energy of the reaction products is less than the Gibbs energy of the starting materials ( ), then the reaction proceeds in the direction of the products.${\ displaystyle \ Delta G <0}$
• If the Gibbs energy of the reaction products is greater than the Gibbs energy of the starting materials ( ), the reaction proceeds in the opposite direction.${\ displaystyle \ Delta G> 0}$
• The Gibbs energy of the mixture of starting materials and products generally depends on the mixture and therefore changes in the course of a reaction. If it passes through a minimum, i.e. when a certain mixture is reached ( ), then in this state the forward and backward reactions run at the same speed, the reacting system has reached a chemical equilibrium .${\ displaystyle \ mathrm {d} G = 0}$

Two phases of a substance are in equilibrium with each other if the molar (or the specific) Gibbs energies of the substance are the same in both phases. If the molar Gibbs energies of the phases involved are known, one can immediately see whether there is equilibrium or not. Conversely, when there is equilibrium, the molar Gibbs energy of all phases is known as soon as it is known for a phase. For example, if liquid water is in equilibrium with its vapor, then the two phases have the same molar Gibbs energy. That of the steam can easily be calculated (approximately as an ideal gas), the numerical value found also applies to the liquid water.
If the phases consist of mixtures of several substances, the equilibrium criterion applies separately to each substance contained in the mixture.

### Chemical potential

In a one-component system, the chemical potential is identical to the molar Gibbs energy of the system. In a multi-component system, the chemical potentials are identical to the partial molar Gibbs energies of the system.

### Change in the Gibbs energy of a system

If you change the temperature of a system at constant pressure and constant amounts of substance, the Gibbs energy of the system changes proportionally to the temperature change, the proportionality constant is the negative of the entropy of the system; If the pressure is changed under the same conditions, the Gibbs energy changes proportionally to the change in pressure (proportionality constant: volume of the system).

If you change the amount of substance of one of the substances contained in the system at constant temperature, constant pressure and constant amounts of substance of the other substances, the Gibbs energy of the system changes proportionally to the change in the amount of substance, the proportionality constant is the chemical potential of the substance under the conditions prevailing in the system .

If a certain amount of reversible physical work (except work for volume change ) is performed on a closed system kept at constant temperature and under constant pressure , the Gibbs energy of the system increases by the relevant amount. In this way, the Gibbs energy of a system can be increased in a targeted manner by, for example, doing a known amount of lifting work or electrical work on the system. Conversely, if the system performs work under the conditions mentioned (except for volume change work), its Gibbs energy decreases by the relevant amount. ${\ displaystyle p \, \ mathrm {d} V}$

The equality of the amounts of converted work and change in Gibbs energy only applies in the case of reversible work. In the irreversible case, depending on the extent of the irreversibility, the work to be done on the system is greater or the work done by the system is less than the change in Gibbs energy.
In this context, the Gibbs energy provides a measure of the “driving force” of the process, such as the “affinity” of the reactants in a chemical reaction.

### As a fundamental equation

The entire thermodynamic information about the system can be derived from the Gibbs energy. The prerequisite, however, is that it is given as a function of the variables temperature , pressure and number of moles of the chemical components contained in the system. These are the "natural variables" of Gibbs energy. It can also be applied as a function of other variables, but then no longer contains the complete thermodynamic information. ${\ displaystyle T}$${\ displaystyle p}$ ${\ displaystyle n_ {i}}$

## Minimum principle of Gibbs energy

According to the second law of thermodynamics , a closed system among the attainable states takes as a state of equilibrium that which has the highest entropy for the given internal energy . From this maximum principle of entropy, other, equivalent extremal principles can also be derived, such as a minimum principle of internal energy: If the entropy is kept constant, a system assumes the state of equilibrium that has the lowest internal energy.

A similar minimum principle exists for the Gibbs energy: A system, the temperature and pressure of which are kept constant and which does no work except volume change work, takes the one in which the Gibbs energy of all achievable conditions with this temperature and this pressure is as a state of equilibrium has the smallest possible value.

To prove this, consider a closed system, the temperature and pressure of which are kept at a constant value. The temperature can be kept constant in that the system under consideration is in contact via a heat-permeable wall with a second system which invariably has the desired temperature (in thermodynamic expression: a heat reservoir ). In the event of a temperature difference, the system under consideration can exchange heat with the heat reservoir via a heat flow through the contact wall until it has adjusted its temperature to that of the reservoir again. The pressure can be kept constant in that the system is in contact via a heat-impermeable but flexible wall with a system that invariably has the desired pressure (a volume reservoir). By deforming the flexible wall, the system under consideration can “exchange volume” with the volume reservoir in the event of a pressure difference until it has adjusted its pressure to that of the volume reservoir again.

In the course of any process, the entropies of the system and the heat reservoir usually change (the volume reservoir does not exchange any heat or matter with the system under consideration, i.e. no entropy either). According to the second law of thermodynamics , the entropy of the closed overall system formed from the system under consideration and the heat reservoir increases or, at best, remains the same:

${\ displaystyle \ mathrm {d} (S _ {\ mathrm {Sys}} + S _ {\ mathrm {Res}}) = \ mathrm {d} S _ {\ mathrm {Sys}} + \ mathrm {d} S _ {\ mathrm {Res}} \ geq 0}$,

or

${\ displaystyle \ mathrm {d} S _ {\ mathrm {Sys}} \ geq - \ mathrm {d} S _ {\ mathrm {Res}}}$.

The “larger” symbol applies to processes that increase the entropy of the overall system and therefore run on their own initiative. These processes run by themselves (if there is no other inhibition ) in the direction of the equilibrium state. The equals sign applies when the overall system has assumed the greatest possible entropy under the given conditions and has thus reached the state of thermal equilibrium.

The change in entropy of the reservoir is by definition related to the heat flowing into the reservoir and the temperature of the reservoir${\ displaystyle \ mathrm {d} S _ {\ mathrm {Res}}}$${\ displaystyle \ mathrm {d} Q _ {\ mathrm {Res}}}$${\ displaystyle T _ {\ mathrm {Res}}}$

${\ displaystyle \ mathrm {d} S _ {\ mathrm {Res}} = \ mathrm {d} Q _ {\ mathrm {Res}} / T _ {\ mathrm {Res}}}$.

Because the reservoir and the system under consideration exchange heat exclusively with one another , and since the system and the reservoir have the same temperature according to the assumption, is . Hence it follows from the above inequality ${\ displaystyle \ mathrm {d} Q _ {\ mathrm {Sys}} = - \ mathrm {d} Q _ {\ mathrm {Res}}}$${\ displaystyle T _ {\ mathrm {Sys}} = T _ {\ mathrm {Res}}}$

${\ displaystyle \ mathrm {d} S _ {\ mathrm {Sys}} \ geq \ mathrm {d} Q _ {\ mathrm {Sys}} / T _ {\ mathrm {Sys}}}$.

It was thus possible to formulate the entropy criterion, which considers the entropies of the system and the reservoir, exclusively using quantities of the system under consideration, which greatly simplifies the application. Since a distinction is no longer necessary, the indices on the quantities of the system are now omitted and the inequality reads

${\ displaystyle \ mathrm {d} S \ geq \ mathrm {d} Q / T}$( Clausius' inequality ).

It is also assumed that the system is kept under constant pressure ( isobaric process , ) and that the system is also designed in such a way that it can not do any other type of work other than volume change work. Then the amount of heat added or removed is numerically equal to the change in enthalpy of the system ( , see → Enthalpy ), and it follows from Clausius' inequality ${\ displaystyle \ mathrm {d} p = 0}$${\ displaystyle \ mathrm {d} H = \ mathrm {d} Q}$

${\ displaystyle \ mathrm {d} S \ geq \ mathrm {d} H / T}$

or moved

${\ displaystyle \ mathrm {d} HT \, \ mathrm {d} S \ leq 0 \ quad (*)}$.

On the other hand, the change in the Gibbs energy of the system is according to its definition

${\ displaystyle \ mathrm {d} G = \ mathrm {d} (HT \, S) = \ mathrm {d} U + p \ mathrm {d} V + V \ mathrm {d} pT \ mathrm {d} SS \ mathrm {d} T}$,

which in the present case is due to the assumed constancy of temperature ( ) and pressure ( ) ${\ displaystyle \ mathrm {d} T = 0}$${\ displaystyle \ mathrm {d} p = 0}$

${\ displaystyle \ mathrm {d} G = \ mathrm {d} HT \, \ mathrm {d} S \ quad (*)}$

simplified. A comparison of the marked equations finally yields the following statement: ${\ displaystyle (*)}$

${\ displaystyle \ mathrm {d} G \ leq 0}$.

The "smaller" symbol applies to processes that run voluntarily. The equals sign applies as soon as the system has reached equilibrium.

The maximum principle for the entropy of the overall system thus leads to the Gibbs energy of the system under consideration assuming a minimum on the subset of the states with constant temperature and constant pressure. If the system is not yet in equilibrium, it will move (if isothermal and isobaric conditions are present and the system is not doing non-volume work) into states of lower Gibbs energy. The equilibrium is reached when the Gibbs energy has the smallest possible value under the given conditions.

If one wanted to determine the state of equilibrium directly with the help of the (general and always valid) entropy criterion, the maximum of the total entropy would have to be determined, i.e. the sum of the entropies of the system under investigation and its surroundings. Therefore, not only the change in the system entropy in the event of a change of state would have to be considered, but also the entropy change that the system generates by reacting to the environment there. The Gibbs energy criterion is a reformulation of the entropy criterion, which only includes properties of the system under consideration and which automatically takes into account the effect on the environment (under isothermal and isobaric conditions) by the term , because is under the given conditions . When using the Gibbs energy criterion, the determination of the (isothermal and isobaric) equilibrium state can be limited to the consideration of the system, which makes the investigations noticeably easier. ${\ displaystyle -T \, S}$${\ displaystyle \ mathrm {d} (-T \, S) = \ mathrm {d} (T _ {\ mathrm {Res}} S _ {\ mathrm {Res}})}$

For a real physical or chemical process, the atmosphere can often serve as a heat and volume reservoir. Because of their large volume, their temperature and pressure do not change significantly when a system transfers heat or volume to them. The prerequisites for the applicability of the minimum principle of Gibbs energy are particularly fulfilled when a system is exposed to the free atmosphere and its processes are therefore isothermal and isobaric. However, they are also fulfilled, for example, if a subsystem is considered within a larger system which, due to its size, represents a heat and volume reservoir for the subsystem (e.g. for a grape in a fermentation vat).

Compare the minimum principle of Gibbs energy under isothermal and isobaric conditions with the extremal principles of other thermodynamic potentials, which represent the equilibrium conditions in closed systems under different boundary conditions:

Sizes kept constant Equilibrium condition
Inner energy and volume${\ displaystyle U}$${\ displaystyle V}$ Maximum of entropy ${\ displaystyle S (U, V, n_ {1}, \ dots, n_ {r})}$
Entropy and volume${\ displaystyle S}$${\ displaystyle V}$ Minimum of internal energy ${\ displaystyle U (S, V, n_ {1}, \ dots, n_ {r})}$
Entropy and pressure${\ displaystyle S}$${\ displaystyle p}$ Minimum of enthalpy ${\ displaystyle H (S, p, n_ {1}, \ dots, n_ {r})}$
Temperature and volume${\ displaystyle T}$${\ displaystyle V}$ Minimum of free energy ${\ displaystyle F (T, V, n_ {1}, \ dots, n_ {r})}$
Temperature and pressure${\ displaystyle T}$${\ displaystyle p}$ Minimum Gibbs energy ${\ displaystyle G (T, p, n_ {1}, \ dots, n_ {r})}$

This section assumes that the system is doing nothing other than volume change work. The following section covers systems that do other forms of work as well.

## Gibbs energy and maximum non-volume work

In the course of a process, a system usually exchanges heat and work with its environment. As will be shown in the following, in an isothermal, isobaric and reversible process, the decrease in the Gibbs energy of the system is numerically equal to the work done by the system on the environment, if the work that may have been done to change the volume is not taken into account. ${\ displaystyle p \, \ mathrm {d} V}$

### Maximum non-volume work

If the enthalpy of a system changes in the course of a process, a differential enthalpy change generally applies ${\ displaystyle H = U + p \, V}$

{\ displaystyle {\ begin {alignedat} {3} \ mathrm {d} H & = \ mathrm {d} (U + pV) && \\ & = \ mathrm {d} U && \, + \, p \, \ mathrm {d} V + V \, \ mathrm {d} p \\ & = \ mathrm {d} Q + \ mathrm {d} W && \, + \, p \, \ mathrm {d} V + V \, \ mathrm {d} p \\ & = \ mathrm {d} Q + \ mathrm {d} W_ {V} + \ mathrm {d} W_ {N} && \, + \, p \, \ mathrm {d} V + V \, \ mathrm {d} p \ end {alignedat}}}.

According to the first law of thermodynamics, the change in internal energy can be split up into the exchanged heat and the exchanged work . The work, in turn, can be divided into volume change work and other forms of work (e.g. mechanical, electrical, chemical work), which are summarized as non-volume work. ${\ displaystyle \ mathrm {d} U}$${\ displaystyle \ mathrm {d} Q}$${\ displaystyle \ mathrm {d} W}$ ${\ displaystyle \ mathrm {d} W_ {V}}$${\ displaystyle \ mathrm {d} W_ {N}}$

If the process is reversible,

• can be according to the Second Law , the exchanged heat through the entropy exchanged express: ,${\ displaystyle \ mathrm {d} Q = T \ mathrm {d} S}$
• the volume change work performed on the system is given by ,${\ displaystyle \ mathrm {d} W_ {V} = - p \, \ mathrm {d} V}$
• the maximum possible work is done on the system for the current change in state and it is .${\ displaystyle \ mathrm {d} W_ {N} = \ mathrm {d} W_ {N _ {\ mathrm {max}}}}$

In this reversible case, the above equation can be written as

{\ displaystyle {\ begin {alignedat} {5} \ mathrm {d} H & = T \ mathrm {d} S && \, - \, p \, \ mathrm {d} V && \, + \, \ mathrm {d} W_ {N _ {\ mathrm {max}}} && \, + \, p \, \ mathrm {d} V && \, + \, V \, \ mathrm {d} p \\ & = T \ mathrm {d} S &&&& \, + \, \ mathrm {d} W_ {N _ {\ mathrm {max}}} &&&& \, + \, V \, \ mathrm {d} p \ end {alignedat}}}.

Finally, for the change in the Gibbs energy of the system associated with the process, according to its definition and using the expression just derived for : ${\ displaystyle \ mathrm {d} H}$

{\ displaystyle {\ begin {alignedat} {6} \ mathrm {d} G & = \ mathrm {d} (H-TS) &&&&& \\ & = \ mathrm {d} H &&& \, - \, & T \ mathrm {d } S & \, - \, S \ mathrm {d} T \\ & = (T \ mathrm {d} S & \, + \, & \ mathrm {d} W_ {N _ {\ mathrm {max}}} + V \ mathrm {d} p) & \, - \, & T \ mathrm {d} S & \, - \, S \ mathrm {d} T \\ & = && \ mathrm {d} W_ {N _ {\ mathrm {max }}} + V \ mathrm {d} p &&& \, - \, S \ mathrm {d} T \ end {alignedat}}}

If the process takes place under isothermal ( ) and isobaric ( ) conditions, this is simplified to ${\ displaystyle \ mathrm {d} T = 0}$${\ displaystyle \ mathrm {d} p = 0}$

${\ displaystyle \ mathrm {d} G = \ mathrm {d} W_ {N _ {\ mathrm {max}}}}$.

If you want to count the work done by the system and made available to the environment positively, you have to reverse its sign

${\ displaystyle \ mathrm {d} W '_ {N _ {\ mathrm {max}}} = - \ mathrm {d} W_ {N _ {\ mathrm {max}}}}$

and it's the work released

${\ displaystyle \ mathrm {d} W '_ {N _ {\ mathrm {max}}} = - \ mathrm {d} G}$.

If the Gibbs energy of a system changes in the course of an isothermal, isobaric and reversible process, then the decrease in the Gibbs energy of the system is equal to the non-volume work given off by the system to the environment during the process. Conversely, if non-volume work is done on the system, its Gibbs energy increases by the corresponding amount.

This relationship can be used, for example, to calculate the electrical work that can be obtained from electrochemical cells or from fuel cells . Conversely, the measurement of the electrical work of a reaction can be used to determine its Gibbs reaction energy , if an electrochemical cell can be operated with this reaction. By applying a suitably selected counter voltage, the cell can be operated close to the equilibrium state, so that the electrochemical reaction takes place practically reversibly and delivers the maximum possible electrical work. ${\ displaystyle \ Delta _ {r} G}$

With a similar argument as above, it can be shown that the same relationship between non-volume work and Gibbs energy also applies to processes that temporarily assume other temperatures and pressures that deviate from the initial and final state. This is the case, for example, with combustion reactions. If a state is selected as the final state in which the system has again assumed the initial temperature - for example 20 ° C - after combustion, then the overall process including heating and cooling is described by the Gibbs energies of the starting materials and products at 20 ° C. However, free combustion is a reaction that takes place far from equilibrium and is therefore irreversible, so that it cannot deliver the maximum possible work performance. If a process in which the system temporarily assumes other temperatures than the temperature of the initial and final state and other pressures than the pressure of the initial and final state is to be reversible, it must be ensured in particular that, despite the variability of temperature and pressure the heat exchange with the heat reservoir takes place at temperature and the volume exchange with the pressure reservoir takes place at pressure . ${\ displaystyle T}$${\ displaystyle p}$${\ displaystyle T}$${\ displaystyle p}$

If an isothermal, isobaric and non- reversible process takes place, then not all of the decrease in Gibbs energy is converted into non-volume work; some of it is given off as heat. In this case, the non-volume work yield is less: ${\ displaystyle \ mathrm {d} W '_ {N _ {\ mathrm {irrev}}}}$

${\ displaystyle - \ mathrm {d} G = \ mathrm {d} W '_ {N _ {\ mathrm {max}}}> \ mathrm {d} W' _ {N _ {\ mathrm {irrev}}}}$.

### First example

As an example of the maximum non-volume work of a chemical reaction, let us determine what non-volume work an organism can gain from the combustion of one mole of glucose at a body temperature of 37 ° C and atmospheric pressure in order to carry out its functions (such as the electrical nerves - and muscle activity). The enthalpy of combustion for glucose is taken from the relevant tables

${\ displaystyle \ Delta _ {c} H = -2808 \ \ mathrm {\ tfrac {kJ} {mol}}}$

(the index c stands for combustion ) and the entropy change associated with the combustion

${\ displaystyle \ Delta _ {c} S = + 182 {,} 4 \ \ mathrm {\ tfrac {J} {K \ mol}}}$.

Strictly speaking, the numbers apply to 25 ° C and are used here approximately for 37 ° C. Since enthalpy and entropy are state variables and therefore only depend on the initial and final state, these numerical values ​​apply to the oxidation of glucose regardless of whether it takes place in open combustion or - as here - in an enzymatically catalyzed reaction. The Gibbs enthalpy of reaction is

${\ displaystyle \ Delta _ {c} G = \ Delta _ {c} HT \ \ Delta _ {c} S = -2808 \ \ mathrm {\ tfrac {kJ} {mol}} - (310 \ \ mathrm {K }) \ (182 {,} 4 \ \ mathrm {\ tfrac {J} {K \ mol}}) = - 2865 \ \ mathrm {\ tfrac {kJ} {mol}}}$.

A maximum of non-volume work can therefore be obtained from the combustion of one mole of glucose . If the combustion process is non-reversible, the work that can be gained is less depending on the degree of irreversibility. ${\ displaystyle 2865 \ \ mathrm {kJ}}$

### Second example

In the derivation of the barometric altitude formula discussed below, there is also an altitude-dependent term in addition to a pressure-dependent term: The Gibbs energy of the volume element under consideration increases by the amount of lifting work that is performed when lifting the volume element in the gravitational field.

### Processes without non-volume work

If an isothermal, isobaric and reversible process takes place, in the course of which the system does no non-volume work (i.e. no work, apart from any work on volume change), then the free energy of the system does not change during this process:

${\ displaystyle - \ mathrm {d} G = \ mathrm {d} W '_ {N _ {\ mathrm {max}}} = 0}$.

If, for example, part of a given amount of water is converted into a vapor phase by adding latent heat, which is in equilibrium with the rest of the water at the same temperature and pressure, then this process is isothermal, isobaric and reversible (because of the expansion of the water during evaporation ) associated with volume change work, but does not do any non-volume work. Consequently, the steam has the same molar Gibbs energy as the water with which it is in equilibrium. The equality of the molar Gibbs energy in different phases of a substance that are in equilibrium will be discussed in more detail later.

An isothermal, isobaric and irreversible process takes place during which the system does not do any non-volume work

${\ displaystyle \ mathrm {d} W '_ {N _ {\ mathrm {irrev}}} = 0}$,

then the inequality applies (see above)

${\ displaystyle - \ mathrm {d} G = \ mathrm {d} W '_ {N _ {\ mathrm {max}}}> \ mathrm {d} W' _ {N _ {\ mathrm {irrev}}} = 0 }$,

or

${\ displaystyle \ mathrm {d} G <0}$.

Spontaneous processes are always irreversible. If an isothermal and isobaric process, during which the system does not do any non-volume work, runs spontaneously, then it is associated with a decrease in the Gibbs energy. This result is already known from the previous section.

## Derivatives of Gibbs energy

Since the change in Gibbs free energy is an important aspect in their application to thermodynamic processes, is examined in this section as of their variables , and dependent. In preparation, the derivatives of internal energy and some related definitions are considered first. ${\ displaystyle T}$${\ displaystyle p}$${\ displaystyle n_ {1}, \ dots, n_ {r}}$

### The derivatives of internal energy

If you start from the internal energy as a function of its natural variables and form its total differential, you get: ${\ displaystyle U}$${\ displaystyle S, V, n_ {1}, \ dots, n_ {r}}$

${\ displaystyle \ mathrm {d} U (S, V, n_ {1}, \ dots, n_ {r}) = \ left ({\ frac {\ partial U} {\ partial S}} \ right) _ { V, n_ {1}, \ dots, n_ {r}} \ mathrm {d} S + \ left ({\ frac {\ partial U} {\ partial V}} \ right) _ {S, n_ {1}, \ dots, n_ {r}} \ mathrm {d} V \, + \, \ sum _ {i = 1} ^ {r} \ left ({\ frac {\ partial U} {\ partial n_ {i}} } \ right) _ {S, V, n_ {j \ neq i}} \ mathrm {d} n_ {i}}$.

The partial derivatives that occur here are interpreted in thermodynamics as the definitions of temperature , pressure and chemical potential of the i-th substance : ${\ displaystyle T}$${\ displaystyle p}$${\ displaystyle \ mu _ {i}}$

${\ displaystyle T: = \ left ({\ frac {\ partial U (S, V, n_ {1}, \ dots, n_ {r})} {\ partial S}} \ right) _ {V, n_ { 1}, \ dots, n_ {r}}}$
${\ displaystyle p: = - \ left ({\ frac {\ partial U (S, V, n_ {1}, \ dots, n_ {r})} {\ partial V}} \ right) _ {S, n_ {1}, \ dots, n_ {r}}}$
${\ displaystyle \ mu _ {i}: = \ left ({\ frac {\ partial U (S, V, n_ {1}, \ dots, n_ {r})} {\ partial n_ {i}}} \ right) _ {S, V, n_ {j \ neq i}}}$.

With these definitions the internal energy differential can also be written as

${\ displaystyle \ mathrm {d} U (S, V, n_ {1}, \ dots, n_ {r}) = T \, \ mathrm {d} Sp \, \ mathrm {d} V \, + \, \ sum _ {i = 1} ^ {r} \ mu _ {i} \, \ mathrm {d} n_ {i}}$.

### The derivatives of Gibbs energy

On the one hand, the total differential of the Gibbs energy as a function of its natural variables is formal ${\ displaystyle G}$${\ displaystyle T, p, n_ {1}, \ dots, n_ {r}}$

${\ displaystyle \ mathrm {d} G (T, p, n_ {1}, \ dots, n_ {r}) = \ left ({\ frac {\ partial G} {\ partial T}} \ right) _ { p, n_ {1}, \ dots, n_ {r}} \ mathrm {d} T + \ left ({\ frac {\ partial G} {\ partial p}} \ right) _ {T, n_ {1}, \ dots, n_ {r}} \ mathrm {d} p \, + \, \ sum _ {i = 1} ^ {r} \ left ({\ frac {\ partial G} {\ partial n_ {i}} } \ right) _ {T, p, n_ {j \ neq i}} \ mathrm {d} n_ {i} \ quad (*)}$.

and on the other hand, using their definition and using the expression just derived for : ${\ displaystyle \ mathrm {d} U}$

{\ displaystyle {\ begin {alignedat} {13} \ mathrm {d} G & \, = \ mathrm {d} (U &&&&& \, + \, & pV &&& \, - \, & TS) && \\ & \, = \ mathrm {d} U &&&&& \, + \, & p \, \ mathrm {d} V & \, + \, & V \, \ mathrm {d} p & \, - \, & T \, \ mathrm {d} S & \, - \ , & S \ mathrm {d} T \\ & \, = T \, \ mathrm {d} S & \, - \, & p \, \ mathrm {d} V & \, + \, & \ sum \ mu _ {i } \, \ mathrm {d} n_ {i} & \, + \, & p \, \ mathrm {d} V & \, + \, & V \, \ mathrm {d} p & \, - \, & T \, \ mathrm {d} S & \, - \, & S \ mathrm {d} T \\ & \, = &&&& \ sum \ mu _ {i} \, \ mathrm {d} n_ {i} &&& \, + \, & V \, \ mathrm {d} p &&& \, - \, & S \ mathrm {d} T \\\ end {alignedat}}}

so changed

${\ displaystyle \ mathrm {d} G = -S \, \ mathrm {d} T + V \, \ mathrm {d} p \, + \, \ sum \ mu _ {i} \, \ mathrm {d} n_ {i} \ quad (*)}$

so that it follows from the comparison of the coefficients in the marked equations${\ displaystyle (*)}$

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial T}} \ right) _ {p, n_ {1}, \ dots, n_ {r}} = - S}$,

such as

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial p}} \ right) _ {T, n_ {1}, \ dots, n_ {r}, \ dots} = V}$

and

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial n_ {i}}} \ right) _ {T, p, n_ {j \ neq i}} = \ mu _ {i}}$.

These simple relationships are discussed in more detail in the following sections.

The derivation shows at the same time how the addition of the terms and the list of the independent variables changes from in by removing the from and dependent terms in the total differential and adding dependent terms instead. ${\ displaystyle -T \, S}$${\ displaystyle p \, V}$${\ displaystyle (S, V, \ dots)}$${\ displaystyle (T, p, \ dots)}$${\ displaystyle \ mathrm {d} S}$${\ displaystyle \ mathrm {d} V}$${\ displaystyle \ mathrm {d} T}$${\ displaystyle \ mathrm {d} p}$

The second of the marked equations is a "differential fundamental function", namely the frequently required differential Gibbs energy as a function of its natural variables:

${\ displaystyle \ mathrm {d} G (T, p, n_ {1}, \ dots, n_ {r}) = - S \, \ mathrm {d} T + V \ mathrm {d} p \, + \ , \ sum _ {i = 1} ^ {r} \ mu _ {i} \, \ mathrm {d} n_ {i}}$.

## Gibbs energy and chemical potential

Consider a homogeneous phase with the temperature and the pressure , which consists of a mixture of substances, the th substance in each case being present in the quantity . From the equation derived above ${\ displaystyle T}$${\ displaystyle p}$${\ displaystyle r}$${\ displaystyle i}$${\ displaystyle n_ {i}}$

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial n_ {i}}} \ right) _ {T, p, n_ {j \ neq i}} = \ mu _ {i}}$

a clear interpretation of the chemical potential follows: ${\ displaystyle \ mu _ {i}}$

The chemical potential of the -th component of the phase indicates the (infinitesimal) amount by which the Gibbs energy of the phase changes if the amount of the -th component in the phase changes by the (infinitesimal) amount , where the temperature , the pressure and the amounts of the other components are kept constant. The amount of the -th component can change, for example, in the course of a chemical reaction taking place in the phase, or because it is added to the phase from the environment (this can also be another phase of the system). ${\ displaystyle \ mu _ {i}}$${\ displaystyle i}$${\ displaystyle \ mathrm {d} G}$${\ displaystyle i}$${\ displaystyle \ mathrm {d} n_ {i}}$${\ displaystyle T}$${\ displaystyle p}$${\ displaystyle n_ {j}}$${\ displaystyle i}$

The transition from the infinitesimal size to the total Gibbs energy of the system seems to require integration at first. This would also be difficult because it can depend in a complicated way on the amounts of substance present , which in turn can change in a complicated way in the course of the process. It turns out, however, that the relationship between (or the ) and is surprisingly simple. ${\ displaystyle \ mathrm {d} G}$${\ displaystyle G}$${\ displaystyle \ mathrm {d} G}$${\ displaystyle n_ {1}, \ dots, n_ {r}}$${\ displaystyle \ mathrm {d} G}$${\ displaystyle \ mu _ {i}}$${\ displaystyle G}$

The starting point for the considerations is the generally applicable differential fundamental function derived above

${\ displaystyle \ mathrm {d} G = -S \, \ mathrm {d} T + V \, \ mathrm {d} p + \ mu _ {1} \, \ mathrm {d} n_ {1} + \ dots + \ mu _ {r} \, \ mathrm {d} n_ {r}}$.

Now imagine the size of the phase multiplied. During this process, the temperature and the pressure remain unchanged because they are intense quantities. In this case, and and the differential is simplified to ${\ displaystyle \ mathrm {d} T = 0}$${\ displaystyle \ mathrm {d} p = 0}$

${\ displaystyle \ mathrm {d} G = \ mu _ {1} \, \ mathrm {d} n_ {1} + \ dots + \ mu _ {r} \, \ mathrm {d} n_ {r}}$.

Since the chemical potentials are also intense quantities, they also remain unchanged. The direct integration of the differential therefore provides ${\ displaystyle \ mu _ {1}, \ dots, \ mu _ {r}}$

{\ displaystyle {\ begin {aligned} \ Delta G = \ int \ mathrm {d} G & = \ int \ mu _ {1} \, \ mathrm {d} n_ {1} + \ dots + \ int \ mu _ {r} \, \ mathrm {d} n_ {r} \\ & = \ mu _ {1} \ int \ mathrm {d} n_ {1} + \ dots + \ mu _ {r} \ int \ mathrm { d} n_ {r} \\ & = \ mu _ {1} \ Delta n_ {1} + \ dots + \ mu _ {r} \ Delta n_ {r}, \ end {aligned}}}

because they can be drawn in front of the respective integrals as constant quantities. ${\ displaystyle \ mu _ {1}, \ dots, \ mu _ {r}}$

If the phase is increased to -fold, then the numerical values ​​of the extensive variables , ... increase to -fold and it applies ${\ displaystyle k}$${\ displaystyle G}$${\ displaystyle n_ {1}}$${\ displaystyle n_ {r}}$${\ displaystyle k}$

${\ displaystyle \ Delta G = k \, GG = (k-1) \, G}$
${\ displaystyle \ Delta n_ {i} = k \, n_ {i} -n_ {i} = (k-1) \, n_ {i}}$

Substituting into the previous equation leads to

${\ displaystyle (k-1) G = \ mu _ {1} (k-1) n_ {1} + \ dots + \ mu _ {r} (k-1) n_ {r}}$

and thus

${\ displaystyle G = n_ {1} \ mu _ {1} + \ dots + n_ {r} \ mu _ {r}}$.

So there is a simple connection between the Gibbs energy of the phase and the chemical potentials of the substances contained in the phase:

If each chemical potential is multiplied by the amount of the substance in question and the sum of all substances is formed, the result is equal to the Gibbs energy of the phase.${\ displaystyle \ mu _ {i}}$${\ displaystyle n_ {i}}$${\ displaystyle i}$

Both and those are only determined up to one constant. ${\ displaystyle G}$${\ displaystyle \ mu _ {i}}$

Division by the total amount of substance in the phase provides the often used molar Gibbs energy ${\ displaystyle n}$

${\ displaystyle {\ frac {G} {n}} = {\ frac {n_ {1}} {n}} \ mu _ {1} + \ dots + {\ frac {n_ {r}} {n}} \ mu _ {r} = x_ {1} \ mu _ {1} + \ dots + x_ {r} \ mu _ {r}}$

with the amount of substance . ${\ displaystyle x_ {i} = n_ {i} / n}$

For a system that consists only of a single substance ( ) is particular ${\ displaystyle r = 1}$

${\ displaystyle {\ frac {G} {n}} = \ mu}$.

In a multicomponent system, the chemical potentials are identical to the partial molar Gibbs energies of the system. In a one-component system, the chemical potential is identical to the molar Gibbs energy of the system.

## Chemical potential in phases that are in equilibrium

Thermodynamic potentials usually take on different values ​​in the different phases of a system. For example, the molar enthalpies of water and water vapor, which are in equilibrium, differ by the amount of the enthalpy of vaporization of the water. The chemical potential, on the other hand, assumes the same value in all phases of a system that is in equilibrium.

To prove this, consider a one-component system consisting of several phases and in equilibrium and initially assume that the chemical potential has different values or values at different locations and in the system . If one thinks that a quantity of substance is transported from place to place , the Gibbs energy changes at the place by the amount and at the place by the amount , so the total Gibbs energy of the system changes by . If , then the change in Gibbs energy is negative and the transport takes place spontaneously, contrary to the assumption that the system is already in equilibrium. The system can only be in equilibrium when is. ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle \ mu _ {A}}$${\ displaystyle \ mu _ {B}}$${\ displaystyle \ mathrm {d} n}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A}$${\ displaystyle - \ mu _ {A} \ mathrm {d} n}$${\ displaystyle B}$${\ displaystyle + \ mu _ {B} \ mathrm {d} n}$${\ displaystyle \ mathrm {d} G = (\ mu _ {B} - \ mu _ {A}) \ mathrm {d} n}$${\ displaystyle \ mu _ {B} <\ mu _ {A}}$${\ displaystyle \ mu _ {A} = \ mu _ {B}}$

Since the two compared locations can be in different phases but also in the same phase, the equality of the chemical potential follows in all phases, but also in all locations within each phase.

Since in the case under consideration of a single substance the chemical potential is identical to the molar Gibbs energy, the constancy of the molar Gibbs energy also follows in all phases.

## Temperature dependence of the Gibbs energy

### Temperature dependence

From the equation derived above

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial T}} \ right) _ {p, n_ {1}, \ dots, n_ {r}} = - S}$

it follows immediately that the temperature dependence of the Gibbs energy at constant pressure and constant amounts of substance is given by the negative of the entropy of the system. Since the entropy is always positive according to the third law of thermodynamics , the Gibbs energy always decreases under these conditions when the temperature increases. In gases with their high entropy, the temperature dependence of the Gibbs energy is greater than in liquids or solids.

Division by the amount of substance in the phase gives the corresponding equation for the molar sizes of the phase:

${\ displaystyle \ left ({\ frac {\ partial G_ {m}} {\ partial T}} \ right) _ {p, x_ {1}, \ dots, x_ {r-1}} = - S_ {m }}$

### example

As an example of the temperature dependence, consider ice and liquid water, which are in equilibrium. The temperature common to the two phases is therefore the melting temperature corresponding to the common pressure (for example 0 ° C. if the pressure is atmospheric pressure). Since equilibrium is assumed, both phases have the same molar Gibbs energy:

${\ displaystyle G_ {m, \ \ mathrm {ice}} (T) = G_ {m, \ \ mathrm {water}} (T)}$.

If the temperature is increased while the pressure is kept constant, the molar Gibbs energies change proportionally to the negative of their respective molar entropies:

{\ displaystyle {\ begin {alignedat} {7} & G_ {m, \ \ mathrm {ice cream}} (T ') && \, = \, G_ {m, \ \ mathrm {ice cream}} (T) && - S_ {m, \ \ mathrm {ice}} && \ mathrm {d} T \\ & G_ {m, \ \ mathrm {water}} (T ') && \, = \, G_ {m, \ \ mathrm {water} } (T) && - S_ {m, \ \ mathrm {water}} && \ mathrm {d} T. \ end {alignedat}}}

Since the liquid water has the larger molar entropy compared to the ice with its lattice structure, its molar Gibbs energy decreases faster with increasing temperature, and it finally has a lower molar Gibbs energy than the ice at the increased final temperature. The equilibrium between the two phases is thus disturbed. It could be restored either by melting the ice and accepting the lower Gibbs molar energy of water, or by freezing the water and accepting the higher Gibbs molar energy of ice. Since the Gibbs energy of the entire system decreases during the melting process (at constant pressure and a constant new temperature), it is the melting process that the system selects and runs through spontaneously until a new equilibrium has been established.

Conversely, if the temperature is reduced, the molar Gibbs energy of the water increases more than that of the ice and the disturbed equilibrium is restored spontaneously, as the water freezes and assimilates the lower molar Gibbs energy of the ice.

## Pressure dependence of the Gibbs energy

### Pressure dependency

From the equation derived above

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial p}} \ right) _ {T, n_ {1}, \ dots, n_ {r}} = V}$

it follows immediately that the pressure dependence of the Gibbs energy at constant temperature and constant amounts of substance is given by the volume of the system. Since the volume is always positive, the Gibbs energy always increases under these conditions when the pressure increases.

Division by the amount of substance in the phase gives the corresponding equation for the molar sizes of the phase:

${\ displaystyle \ left ({\ frac {\ partial G_ {m}} {\ partial p}} \ right) _ {T, x_ {1}, \ dots, x_ {r-1}} = V_ {m} }$.

### example

As an example of the pressure dependence, consider again ice and liquid water, which are in equilibrium. Since equilibrium is assumed, both phases have the same molar Gibbs energy:

${\ displaystyle G_ {m, \ \ mathrm {ice}} (p) = G_ {m, \ \ mathrm {water}} (p)}$

If the pressure is increased while the temperature is kept constant, the molar Gibbs energies change proportionally to the respective molar volume:

{\ displaystyle {\ begin {alignedat} {7} & G_ {m, \ \ mathrm {ice cream}} (p ') && \, = \, G_ {m, \ \ mathrm {ice cream}} (p) && + V_ {m, \ \ mathrm {ice}} && \ mathrm {d} p \\ & G_ {m, \ \ mathrm {water}} (p ') && \, = \, G_ {m, \ \ mathrm {water} } (p) && + V_ {m, \ \ mathrm {water}} && \ mathrm {d} p. \ end {alignedat}}}

Since ice has a larger molar volume than liquid water, its molar Gibbs energy increases faster when the pressure increases, and at the increased final pressure it finally has a higher molar Gibbs energy than water. The balance between the two phases is disturbed. It is restored as the ice melts and takes on the lower molar Gibbs energy of water. Since this melting process (at constant temperature and constant new pressure) reduces the total Gibbs energy of the system, melting occurs spontaneously.

Despite the temperature being kept constant, the system, which is under increased pressure, is now completely liquid. The temperature would have to be reduced in order to restore equilibrium between liquid water and ice at the new pressure: the increase in pressure lowered the melting temperature. The properties of the state variable Gibbs energy allowed this conclusion to be drawn solely from the knowledge of the molar volumes of ice and liquid water.

Conversely, if the pressure is reduced, the molar Gibbs energy of the ice decreases faster than that of the water and the disturbed equilibrium is restored spontaneously, as the water freezes and assimilates the lower molar Gibbs energy of the ice.

### Inferences

If the Gibbs energy at a reference pressure , for example the standard pressure, is known, it can be determined for any other pressure by integration: ${\ displaystyle p ^ {\ circ}}$${\ displaystyle p}$

${\ displaystyle G (p) = G (p ^ {\ circ}) + \ int _ {p ^ {\ circ}} ^ {p} V \, \ mathrm {d} p}$.

The following applies accordingly to the molar Gibbs energy:

${\ displaystyle G_ {m} (p) = G_ {m} (p ^ {\ circ}) + \ int _ {p ^ {\ circ}} ^ {p} V_ {m} \, \ mathrm {d} p}$.

In the case of liquids and solids, the molar volume is only slightly variable with the pressure. can then be regarded as approximately constant and drawn in front of the integral: ${\ displaystyle V_ {m}}$${\ displaystyle V_ {m}}$

${\ displaystyle G_ {m} (p) = G_ {m} (p ^ {\ circ}) + (pp ^ {\ circ}) \, V_ {m}}$.

If the pressure increases are not too extreme, the following is usually negligible and the free energy of liquids and solids is practically independent of pressure: ${\ displaystyle (pp ^ {\ circ}) \, V_ {m}}$

${\ displaystyle G_ {m} (p) = G_ {m} (p ^ {\ circ})}$

In the case of gases, on the other hand, the molar volume is too variable to be set constant. Here the integral has to be evaluated.

## Gibbs energy of the ideal gas

With ideal gases, the integral can be calculated immediately, since the molar volume is as

${\ displaystyle V_ {m} = {\ frac {R \, T} {p}}}$

can be expressed and inserted into the integral:

{\ displaystyle {\ begin {aligned} G_ {m} (p) & = G_ {m} (p ^ {\ circ}) + R \, T \, \ int _ {p ^ {\ circ}} ^ { p} {\ frac {1} {p}} \ mathrm {d} p \\ & = G_ {m} (p ^ {\ circ}) + R \, T \, \ ln \ left ({\ frac { p} {p ^ {\ circ}}} \ right). \ end {aligned}}}

For a system that only consists of a single substance, the molar Gibbs energy is identical to the chemical potential of the substance (see above) and it applies ${\ displaystyle G_ {m}}$${\ displaystyle \ mu}$

${\ displaystyle \ mu = \ mu ^ {\ circ} + R \, T \, \ ln \ left ({\ frac {p} {p ^ {\ circ}}} \ right)}$,

where the molar Gibbs energy of the substance in the standard state ,, now called the chemical potential of the substance in the standard state , is denoted. ${\ displaystyle G_ {m} (p ^ {\ circ})}$${\ displaystyle \ mu ^ {\ circ}}$

The chemical potential of an ideal gas can therefore easily be calculated as a function of the pressure. This is used in a variety of ways, because in this way not only the chemical potential of approximately ideal gases can be determined. For example, if one looks at a liquid that is in equilibrium with its vapor, the molar Gibbs energy - and thus also the chemical potential - is identical in both phases (see above), and the chemical potential of the liquid is therefore known, provided that the Steam may be treated as an ideal gas in a sufficiently good approximation.

## Gibbs energy of a solvent

As an example, consider a solution , i.e. a mixture of a solvent and a substance dissolved in it. According to Raoult's law , the vapor pressure of the solvent is proportional to the amount of substance with which it is represented in the solution: ${\ displaystyle p}$${\ displaystyle x}$

${\ displaystyle p \, = \, c \, x}$.

If you change the mole fraction of the solvent from (it is assumed the validity of the solution under consideration below In most cases, the Raoult's law only in the limit applies very dilute solutions.) To behave the associated vapor pressures and : so how the mole fractions ${\ displaystyle x_ {0}}$${\ displaystyle x}$${\ displaystyle p_ {0}}$${\ displaystyle p}$

${\ displaystyle {\ frac {p} {p_ {0}}} \, = \, {\ frac {x} {x_ {0}}}}$.

If one also assumes that the solvent vapor can be treated as an ideal gas in a sufficient approximation, then the formula from the last section is applicable and applies to the chemical potentials and the vapor over the two solvent components considered: ${\ displaystyle \ mu _ {0}}$${\ displaystyle \ mu}$

${\ displaystyle \ mu = \ mu _ {0} + R \, T \, \ ln \ left ({\ frac {p} {p_ {0}}} \ right) = \ mu _ {0} + R \ , T \, \ ln \ left ({\ frac {x} {x_ {0}}} \ right)}$.

If you choose the pure solvent (marked by an asterisk ) as the starting point , then the chemical potential of the solvent vapor is above the solvent with a proportion${\ displaystyle ^ {*}}$${\ displaystyle x_ {0} = x ^ {*} = 1}$${\ displaystyle x}$

${\ displaystyle \ mu = \ mu ^ {*} + R \, T \, \ ln (x)}$.

Because the solvent is in equilibrium with its vapor, it has the same chemical potential. Since the mole fraction of the solvent is less than one, its logarithm has a negative sign. The presence of a dissolved substance therefore reduces the chemical potential of the solvent under the conditions mentioned. This is the cause of phenomena such as osmosis, depression of the freezing point in solutions and the like. ${\ displaystyle x}$

If the mole fraction of the dissolved substance is referred to, then becomes ${\ displaystyle x_ {2}: = 1-x}$

{\ displaystyle {\ begin {aligned} \ mu & = \ mu ^ {*} + R \, T \, \ ln (x) \\ & = \ mu ^ {*} + R \, T \, \ ln (1-x_ {2}) \\ & \ approx \ mu ^ {*} - R \, T \, x_ {2}, \ end {aligned}}}

the chemical potential of the solvent thus decreases proportionally to the mole fraction of the dissolved substance. It should be noted that the change in the chemical potential of the solvent does not depend on the nature of the dissolved substance, but only on its mole fraction, see → colligative property .

## Gibbs energy of mixing of ideal gases

Consider a container with two chambers containing the pure gases and respectively . Your amounts of substance are respectively (with ). Their temperatures are the same, as are their pressures . Since the Gibbs energy is an extensive quantity, the Gibbs energy of the entire system is calculated as the sum of the Gibbs energies of both subsystems. Using the molar Gibbs energies, which in turn are identical to the chemical potentials, results ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle n _ {\ mathrm {A}}}$${\ displaystyle n _ {\ mathrm {B}}}$${\ displaystyle n _ {\ mathrm {A}} + n _ {\ mathrm {B}} = n}$${\ displaystyle T}$${\ displaystyle p}$

${\ displaystyle G _ {\ mathrm {beginning}} = n _ {\ mathrm {A}} \ G _ {\ mathrm {m, A}} + n _ {\ mathrm {B}} \ G _ {\ mathrm {m, B} } = n _ {\ mathrm {A}} \ \ mu _ {\ mathrm {A}} + n _ {\ mathrm {B}} \ \ mu _ {\ mathrm {B}}}$.

In the case of ideal gases in particular, the Gibbs energy using the formula derived in the previous section for the chemical potential of an ideal gas is:

${\ displaystyle G _ {\ mathrm {beginning}} = n _ {\ mathrm {A}} \ left (\ mu _ {\ mathrm {A}} ^ {\ circ} + R \, T \, \ ln \ left ( {\ frac {p} {p ^ {\ circ}}} \ right) \ right) + n _ {\ mathrm {B}} \ left (\ mu _ {\ mathrm {B}} ^ {\ circ} + R \, T \, \ ln \ left ({\ frac {p} {p ^ {\ circ}}} \ right) \ right)}$.

If the partition between the chambers is removed, the gases mix. In the mixed state they have the partial pressures or (with ). With these pressures the Gibbs energy of the system becomes in the final mixed state ${\ displaystyle p _ {\ mathrm {A}}}$${\ displaystyle p _ {\ mathrm {B}}}$${\ displaystyle p _ {\ mathrm {A}} + p _ {\ mathrm {B}} = p}$

${\ displaystyle G _ {\ mathrm {Ende}} = n _ {\ mathrm {A}} \ left (\ mu _ {\ mathrm {A}} ^ {\ circ} + R \, T \, \ ln \ left ( {\ frac {p _ {\ mathrm {A}}} {p ^ {\ circ}}} \ right) \ right) + n _ {\ mathrm {B}} \ left (\ mu _ {\ mathrm {B}} ^ {\ circ} + R \, T \, \ ln \ left ({\ frac {p _ {\ mathrm {B}}} {p ^ {\ circ}}} \ right) \ right)}$.

The difference between the Gibbs energies in the final and initial state is the Gibbs mixing energy : ${\ displaystyle \ Delta _ {\ mathrm {mix}} G}$

${\ displaystyle \ Delta _ {\ mathrm {mix}} G = G _ {\ mathrm {end}} -G _ {\ mathrm {beginning}} = n _ {\ mathrm {A}} \ R \, T \, \ ln \ left ({\ frac {p _ {\ mathrm {A}}} {p}} \ right) + n _ {\ mathrm {B}} \ R \, T \, \ ln \ left ({\ frac {p_ { \ mathrm {B}}} {p}} \ right)}$.

According to Dalton's law , the partial pressure can be expressed as the product of the amount of substance and the total pressure : ${\ displaystyle p _ {\ mathrm {A}}}$${\ displaystyle x _ {\ mathrm {A}}}$${\ displaystyle p}$

${\ displaystyle p _ {\ mathrm {A}} = x _ {\ mathrm {A}} \ p}$.

In addition, it is due to the definition of the mole fraction

${\ displaystyle n _ {\ mathrm {A}} = x _ {\ mathrm {A}} \ n}$,

and corresponding expressions also apply to gas . Thus the Gibbs mixing energy of ideal gases becomes: ${\ displaystyle B}$

${\ displaystyle \ Delta _ {\ mathrm {mix}} G = n \ R \, T \, (x _ {\ mathrm {A}} \ ln x _ {\ mathrm {A}} + x _ {\ mathrm {B} } \ ln x _ {\ mathrm {B}})}$.

Since, according to their definition, the molar fractions are always less than one in the case under consideration, the logarithms are always negative, and Gibbs' mixing energy is always negative. The Gibbs energy of the overall system of ideal gases always decreases when they are mixed and the mixing of ideal gases is therefore a voluntary process, which also corresponds to experience with real but approximately ideal gases.

If one mixes strongly non-ideal substances, other formulas for the Gibbs mixing energy result. It can then even assume positive values ​​under certain circumstances. In such a case a mixture separates itself voluntarily, the components involved are immiscible. If the Gibbs mixing energy assumes positive values ​​only in a certain range of compositions of the mixture, the mixture is only unstable for these compositions - there is a miscibility gap.

In the case under consideration of non-reacting ideal gases, the amounts of both components remained constant during the mixing process. If the mixed substances can chemically react with one another, the amounts of substance are variable. They then voluntarily adjust themselves in such a way that the Gibbs mixing energy assumes the smallest value compatible with the conditions of the reaction. The state that arises in this way is the state of chemical equilibrium . The position of this state of equilibrium can therefore be calculated in advance if the Gibbs mixing energy of the system is known.

## application

### osmosis

The effect of osmosis becomes apparent, for example, when a solvent that contains dissolved substances is separated from the pure solvent by a membrane that is permeable to the solvent but impermeable to the dissolved substances. Pure solvent then flows spontaneously through the membrane into the solution, even without there being a pressure difference.

To illustrate this, consider a phase that is a mixture of a solvent and solutes. It is in contact with a second phase, which consists of the pure solvent, via a semipermeable membrane that is only permeable to the solvent. An example would be a phase with sugar water and a phase with pure water, which are separated from each other by a cellophane film . The temperature and pressure are identical on both sides.

In the mixture, the solvent has a lower chemical potential than in the pure state (compare the above discussion of the free energy of a solvent ). The different chemical potentials of the solvent in the two phases set a balancing solvent flow in motion, which transports the solvent from the side with the higher chemical potential to the side with the lower chemical potential (namely to the side with the mixture).

If you want to prevent osmotic transport, you have to ensure that the solvent has the same chemical potential in both the pure phase and in the mixed phase. This can be done by abandoning the requirement for the same pressure on both sides and increasing the pressure in the mixed phase (while the temperatures remain identical). As explained above , an increase in pressure increases the chemical potential. The additional pressure that has to be applied to create equilibrium is called osmotic pressure. If the pressure is in the pure solvent and the pressure in the mixture, then it is in equilibrium ${\ displaystyle \ Pi}$${\ displaystyle p}$${\ displaystyle p ^ {\ prime}}$

${\ displaystyle \ Pi = p'-p}$.

The chemical potential of the pure solvent is denoted by; the star marks the pure substance and the index the pressure under which the substance is. ${\ displaystyle \ mu _ {p} ^ {*}}$${\ displaystyle p}$

For the sake of simpler treatment, it is assumed for the mixed phase that it is an ideal mixture. Then the chemical potential of the solvent (mole fraction ) in the mixed phase under the pressure is given by ${\ displaystyle x}$${\ displaystyle p '}$

${\ displaystyle \ mu _ {p '} ^ {*} + R \, T \, \ ln x}$.

In osmotic equilibrium, the two chemical potentials are the same:

${\ displaystyle \ mu _ {p} ^ {*} = \ mu _ {p '} ^ {*} + R \, T \, \ ln x}$,

or changed and taking into account the pressure dependence of the chemical potential of the pure solvent:

{\ displaystyle {\ begin {aligned} R \, T \, \ ln x & = \ mu _ {p} ^ {*} - \ mu _ {p '} ^ {*} \\ & = \ int _ {p '} ^ {p} V_ {m} \ \ mathrm {d} p, \ end {aligned}}}

where is the molar volume of the pure solvent. With as the mean value of the molar volume over the pressure interval , this becomes: ${\ displaystyle V _ {\ mathrm {m}}}$${\ displaystyle {\ overline {V}} _ {\ mathrm {m}}}$${\ displaystyle p-p '}$

${\ displaystyle R \, T \, \ ln x = {\ overline {V}} _ {\ mathrm {m}} \, (p-p ') = - {\ overline {V}} _ {\ mathrm { m}} \ \ Pi}$.

The following applies to the osmotic pressure:

${\ displaystyle \ Pi = - {\ frac {R \, T} {{\ overline {V}} _ {\ mathrm {m}}}} \, \ ln x}$.

With as the sum of the mole fractions of all dissolved substances ${\ displaystyle \ textstyle \ sum x _ {\ mathrm {i}}}$

${\ displaystyle \ Pi = - {\ frac {R \, T} {{\ overline {V}} _ {\ mathrm {m}}}} \, \ ln (1- \ sum x _ {\ mathrm {i} })}$,

what little too ${\ displaystyle \ textstyle \ sum x _ {\ mathrm {i}}}$

${\ displaystyle \ Pi \ approx {\ frac {R \, T} {{\ overline {V}} _ {\ mathrm {m}}}} \, \ sum x _ {\ mathrm {i}}}$

simplified, the van-'t-Hoff law . The osmotic pressure of sufficiently dilute solutions is therefore proportional to the sum of the molar proportions of the dissolved substances.

### Clausius-Clapeyron equation

In the above discussion of the temperature and pressure dependence of the Gibbs energy it was shown that an existing equilibrium of two phases of a substance is disturbed if either the temperature of the system is changed while the pressure is kept constant or the pressure of the system is changed while the temperature is kept constant. However, it is possible to change the pressure and temperature together so that the phases remain in equilibrium if the changes and are appropriately adjusted to each other. ${\ displaystyle \ mathrm {d} T}$${\ displaystyle \ mathrm {d} p}$

If the considered system of two phases and a substance is initially in equilibrium, the specific Gibbs energies of the two phases are the same. After the changes and the phases should also be in equilibrium, their specific Gibbs energies must therefore be the same again - usually with changed numerical values. The specific Gibbs energies of both phases must have changed by the same amount: ${\ displaystyle 1}$${\ displaystyle 2}$${\ displaystyle \ mathrm {d} T}$${\ displaystyle \ mathrm {d} p}$

${\ displaystyle \ mathrm {d} g_ {1} = \ mathrm {d} g_ {2}}$.

Since the system only consists of a single substance, the differential Gibbs function (see above) is reduced to the expression , and from ${\ displaystyle \ mathrm {d} g = -s \, \ mathrm {d} T + v \, \ mathrm {d} p}$

${\ displaystyle v_ {1} \, \ mathrm {d} p-s_ {1} \, \ mathrm {d} T = v_ {2} \, \ mathrm {d} p-s_ {2} \, \ mathrm {d} T}$

follows the Clapeyron equation

${\ displaystyle {\ frac {\ mathrm {d} p} {\ mathrm {d} T}} = {\ frac {s_ {2} -s_ {1}} {v_ {2} -v_ {1}}} }$.

The difference is the difference in the specific entropies of the two phases. It is identical to the specific latent enthalpy that must be added to a unit of mass of the substance at the given temperature and pressure in order to convert it reversibly from phase to phase , divided by the present temperature : ${\ displaystyle s_ {2} -s_ {1}}$${\ displaystyle \ Delta h}$${\ displaystyle 1}$${\ displaystyle 2}$${\ displaystyle T}$

${\ displaystyle s_ {2} -s_ {1} = {\ frac {\ Delta h} {T}}}$.

Substituting this expression gives the Clausius-Clapeyron equation:

${\ displaystyle {\ frac {\ mathrm {d} p} {\ mathrm {d} T}} \, = \, {\ frac {\ Delta h} {T \, \ Delta v}}}$.

### Vapor pressure over drops

The saturation vapor pressure of a liquid at a given temperature is the pressure at which the liquid is in equilibrium with its vapor. It is usually assumed that the liquid surface is flat. The saturation vapor pressure takes on different values ​​over curved surfaces: it is higher over convex surfaces (e.g. over drops) and lower over concave surfaces (e.g. over the meniscus in a partially filled capillary) than over a flat surface .

The reason for this is the changed pressure under which the liquid is when the surface is curved. With a flat surface (since equilibrium is assumed) the pressure in the liquid phase is equal to the pressure in the vapor phase. The liquid in a drop of the radius , however, is under a higher pressure because the surface tension creates an additional capillary pressure${\ displaystyle r}$${\ displaystyle \ gamma}$

${\ displaystyle p _ {\ mathrm {k}} = 2 \ gamma / r}$

generated in the drop. The total pressure in the drop is the sum of the capillary pressure and the saturation vapor pressure exerted on the drop by the vapor phase. We are looking for the new saturation vapor pressure that has to be established over the curved surface under these changed pressure conditions in order to maintain equilibrium.

The pressure dependency of the saturation vapor pressure can be determined by considering the pressure dependence of the chemical potentials of the liquid and the vapor (the index l stands for liquid , the index g for gas ). In any equilibrium of liquid and vapor, the chemical potentials of the two phases are the same: ${\ displaystyle \ mu _ {\ mathrm {l}}}$${\ displaystyle \ mu _ {\ mathrm {g}}}$

${\ displaystyle \ mu _ {\ mathrm {g}} = \ mu _ {\ mathrm {l}}}$.

If there is a change that leads to a new equilibrium, in general and , but they remain the same, both must change in the same way: ${\ displaystyle \ mu _ {\ mathrm {g}}}$${\ displaystyle \ mu _ {\ mathrm {l}}}$

${\ displaystyle \ mathrm {d} \ mu _ {\ mathrm {g}} = \ mathrm {d} \ mu _ {\ mathrm {l}}}$.

If the change consists in a change in pressure in the liquid, then the chemical potential of the liquid changes (see pressure dependence of the Gibbs energy ) ${\ displaystyle \ mathrm {d} p _ {\ mathrm {l}}}$

${\ displaystyle \ mathrm {d} \ mu _ {\ mathrm {l}} = V _ {\ mathrm {m, l}} \, \ mathrm {d} p _ {\ mathrm {l}}}$,

where is the molar volume of the liquid. A similar expression applies to the vapor phase and it follows ${\ displaystyle V _ {\ mathrm {m, l}}}$

${\ displaystyle V _ {\ mathrm {m, g}} \, \ mathrm {d} p _ {\ mathrm {g}} = V _ {\ mathrm {m, l}} \, \ mathrm {d} p _ {\ mathrm {l}}}$.

This equation can be used to determine what pressure change in the vapor phase is necessary to re-establish the equilibrium if the pressure in the liquid phase changes by. ${\ displaystyle \ mathrm {d} p _ {\ mathrm {g}}}$${\ displaystyle \ mathrm {d} p _ {\ mathrm {l}}}$

For the sake of simplicity, it is assumed that the steam behaves like an ideal gas. Its molar volume is then given by and it applies ${\ displaystyle V _ {\ mathrm {m, g}} = R \, T / p _ {\ mathrm {g}}}$

${\ displaystyle {\ frac {\ mathrm {d} p _ {\ mathrm {g}}} {p _ {\ mathrm {g}}}} = {\ frac {V _ {\ mathrm {m, l}} \, \ mathrm {d} p _ {\ mathrm {l}}} {R \, T}}}$

This formula is now integrated, from the initial state without additional pressure to the final state in which there is additional pressure in the liquid . In the initial state, the pressures in the vapor and in the liquid are equal to the normal saturated vapor pressure . In the final state, the pressure in the liquid is increased by the amount , in the vapor the pressure to be determined prevails : ${\ displaystyle \ Delta p}$${\ displaystyle p_ {0}}$${\ displaystyle \ Delta p}$${\ displaystyle p_ {0} + \ Delta p}$${\ displaystyle p}$

${\ displaystyle \ int _ {p_ {0}} ^ {p} {\ frac {\ mathrm {d} p _ {\ mathrm {g}}} {p _ {\ mathrm {g}}}} = {\ frac { 1} {R \, T}} \ int _ {p_ {0}} ^ {p_ {0} + \ Delta p} V _ {\ mathrm {m, l}} \, \ mathrm {d} p _ {\ mathrm {l}}}$.

If, as a further simplification, it is assumed that the molar volume of the liquid is constant in the pressure range under consideration (i.e. the liquid is incompressible), it follows

{\ displaystyle {\ begin {aligned} \ ln \ left ({\ frac {p} {p_ {0}}} \ right) & = {\ frac {V _ {\ mathrm {m, l}}} {R \ , T}} \, (p_ {0} + \ Delta p-p_ {0}) \\ & = {\ frac {V _ {\ mathrm {m, l}}} {R \, T}} \, \ Delta p \ end {aligned}}}

or reshaped

${\ displaystyle p = p_ {0} \, e ^ {\ frac {V _ {\ mathrm {m, l}} \, \ Delta p} {R \, T}}}$.

This formula describes how the saturation vapor pressure to increase as an additional pressure on the liquid is applied, regardless of the manner in which the increase in pressure generated in the liquid. ${\ displaystyle p_ {0}}$${\ displaystyle p}$${\ displaystyle \ Delta p}$${\ displaystyle \ Delta p}$

An open water surface, for example, is exposed to atmospheric pressure rather than just its own saturated vapor pressure; the saturation vapor pressure in moist atmospheric air is therefore slightly higher than in a pure water vapor atmosphere at the same temperature ( Poynting effect ). The saturation vapor pressure of pure water in equilibrium with an atmosphere consisting only of water vapor at the triple point is 612 Pa. If air is added to the vapor phase until the total atmospheric pressure of 101325 Pa is reached, the equilibrium partial pressure of the water vapor in the mixture of steam and air (in so-called "humid air") is around 0 because of the 10 5 Pa pressure increase .5 Pa higher than in the pure water vapor atmosphere.

If the pressure increase is due to capillary forces, because instead of the liquid with a flat surface a drop of the radius is considered, then is and applies to the saturation vapor pressure above the drop ${\ displaystyle r}$${\ displaystyle \ Delta p = 2 \ gamma / r}$

${\ displaystyle p _ {\ mathrm {Tr}} = p_ {0} \, e ^ {\ frac {2 \ gamma \, V _ {\ mathrm {m, l}}} {R \, T \, r}} }$

This is the Kelving equation . For example, above a water droplet with a radius of 0.001 mm, the saturation vapor pressure at 25 ° C is greater by a factor of 1.001 than above a level water surface. Over a concave meniscus with the same radius, the saturation vapor pressure is lower by the same factor.

### Barometric altitude formula

This short alternative derivation of the barometric altitude formula demonstrates the change in the Gibbs energy of a volume element in a fluid when the applied hydrostatic pressure changes and when non-volume work is done on the element.

A vertical column of a fluid (for example water or air) in equilibrium is given in the homogeneous gravity field of the field strength . The Gibbs energy of a volume element is examined as a function of its height in the fluid column. For this purpose it is assumed that the volume element has been transported from one point to a point at another height. The change in its specific (i.e. related to the mass) Gibbs energy during this process consists of two contributions: ${\ displaystyle {\ overline {g}}}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle g}$

The change in hydrostatic pressure along the way changes the specific Gibbs energy ${\ displaystyle P}$

${\ displaystyle \ Delta g_ {P} = \ int _ {A} ^ {B} v \, \ mathrm {d} P}$,

where the specific volume of the fluid is under the respective pressure. ${\ displaystyle v}$

The specific lifting work performed (assumed to be reversible) on the volume element during the change in height is non-volume work, i.e. it increases the specific Gibbs energy of the volume element by ${\ displaystyle \ Delta h}$

${\ displaystyle \ Delta g_ {W} = \ mathrm {d} w_ {N _ {\ mathrm {max}}} = \ int _ {A} ^ {B} {\ overline {g}} \, \ mathrm {d } H}$.

Overall, the specific Gibbs energy changes with the change in altitude

${\ displaystyle \ Delta g = \ Delta g_ {P} + \ Delta g_ {W} = \ int _ {A} ^ {B} v \, \ mathrm {d} P + \ int _ {A} ^ {B} {\ overline {g}} \, \ mathrm {d} h}$.

Since it is assumed that the fluid column is in equilibrium, the specific Gibbs energy must have the same value at all heights, so it is what follows ${\ displaystyle \ Delta g = 0}$

${\ displaystyle v \, \ mathrm {d} P + {\ overline {g}} \, \ mathrm {d} h = 0}$.

Adjusting and setting the density supplies ${\ displaystyle \ textstyle \ rho = {\ frac {1} {v}}}$

${\ displaystyle {\ frac {\ mathrm {d} P} {\ mathrm {d} h}} = - {\ frac {\ overline {g}} {v}} = - {\ overline {g}} \, \ rho}$.

This is the well-known relationship between the change in altitude and the change in hydrostatic pressure in a fluid. Integration of this equation provides the barometric altitude formula.

### Chemical reaction equilibrium of ideal gases

#### Introductory example

As an introductory example, consider a simple equilibrium chemical reaction of the type

${\ displaystyle \ mathrm {A \ rightleftharpoons B}}$

considered, which takes place at constant temperature and constant pressure. The system initially consists of pure raw material , and the product is increasingly formed from it . Since it is an equilibrium reaction according to the prerequisites, a reverse conversion from to takes place at the same time (for example, it could be an isomerization that can take place in both directions). The reverse conversion rate is initially zero because none exist yet. However, the more that has arisen through the forward reaction , the greater the reverse conversion rate. At a certain composition of the chemical system, back and forth reactions take place at the same conversion rate, and the composition of the system no longer changes - the equilibrium composition is reached. Since experience shows that such a system goes into this state voluntarily and the process takes place under isothermal and isobaric conditions, the equilibrium composition is characterized by the fact that it is the composition with the lowest Gibbs energy. This fact makes it possible to calculate the expected equilibrium composition. ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle B}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle B}$${\ displaystyle A \ to B}$

The progress of the reaction during the course of the reaction is measured by the conversion variable . In the initial state has the value zero. If one mole has been converted in the reaction, i.e. the amount of mole has decreased by one mole and the amount of mole has increased by one mole, then the value has reached 1 mole, and so on. ${\ displaystyle \ xi}$${\ displaystyle \ xi}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle \ xi}$

According to the assumption, the reaction takes place at constant temperature and constant pressure, in the differential fundamental function (see above) ${\ displaystyle \ mathrm {d} G}$

${\ displaystyle \ mathrm {d} G = -S \ \ mathrm {d} T + V \ \ mathrm {d} p + \ mu _ {A} \ \ mathrm {d} n_ {A} + \ mu _ {B } \ \ mathrm {d} n_ {B}}$

So drop by and dependent terms continue and for the change of stay under these conditions only the terms ${\ displaystyle \ mathrm {d} T}$${\ displaystyle \ mathrm {d} p}$${\ displaystyle \ mathrm {d} G}$${\ displaystyle G}$

{\ displaystyle {\ begin {aligned} (\ mathrm {d} G) _ {T, p} & = \ mu _ {A} \ \ mathrm {d} n_ {A} + \ mu _ {B} \ \ mathrm {d} n_ {B} \\ & = - \ mu _ {A} \ \ mathrm {d} \ xi + \ mu _ {B} \ \ mathrm {d} \ xi \ end {aligned}}}

Shifting supplies the equation

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial \ xi}} \ right) _ {T, p} = \ mu _ {B} - \ mu _ {A}}$.

For a given mixture of and, this equation provides information about the reaction direction in which the Gibbs energy decreases, i.e. in which direction the system runs voluntarily starting from the present mixture. The decisive factors for this are the chemical potentials and . In the case , the right side is negative, decreases in the positive reaction direction, the reaction therefore runs in the direction ; in the case she runs towards . In the course of the reaction, the chemical potentials depending on the mixture also change, and as soon as becomes, is also . The Gibbs energy has reached the minimum and the system has reached its equilibrium in which its composition no longer changes. The general thermodynamic equilibrium condition for such a chemical system is simple ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle \ mu _ {A}}$${\ displaystyle \ mu _ {B}}$${\ displaystyle \ mu _ {B} <\ mu _ {A}}$${\ displaystyle G}$${\ displaystyle B}$${\ displaystyle \ mu _ {B}> \ mu _ {A}}$${\ displaystyle A}$${\ displaystyle \ mu _ {A} = \ mu _ {B}}$${\ displaystyle \ mathrm {d} G = 0}$

${\ displaystyle 0 = \ mu _ {B} - \ mu _ {A}}$

or

${\ displaystyle \ mu _ {B} = \ mu _ {A}}$.

This is where the reason for the name “chemical potential” becomes apparent: The chemical system strives for “equipotential bonding” and has reached equilibrium when the two chemical potentials have become equal.

If the reacting substances are especially ideal gases, the expressions derived above for their chemical potentials depending on the partial pressures and can be used: ${\ displaystyle p_ {A}}$${\ displaystyle p_ {B}}$

{\ displaystyle {\ begin {aligned} \ left ({\ frac {\ partial G} {\ partial \ xi}} \ right) _ {T, p} & = \ mu _ {B} - \ mu _ {A } \\ & = \ mu _ {B} ^ {\ circ} + RT \ \ ln \ left ({\ frac {p_ {B}} {p ^ {\ circ}}} \ right) - \ mu _ { A} ^ {\ circ} -RT \ \ ln \ left ({\ frac {p_ {A}} {p ^ {\ circ}}} \ right) \\ & = \ mu _ {B} ^ {\ circ } - \ mu _ {A} ^ {\ circ} + RT \ \ ln \ left ({\ frac {p_ {B}} {p_ {A}}} \ right) \ end {aligned}}}.

The derivation on the left describes the change in Gibbs energy per mole of formula conversion with the present composition of the reacting mixture. It is also known for short as Gibbs reaction energy. ${\ displaystyle \ left ({\ frac {\ partial G} {\ partial \ xi}} \ right) _ {T, p}}$${\ displaystyle \ Delta _ {r} G}$

The chemical potential relates to the standard state and therefore to the pure substance . In this case, however, as described above, the chemical potential is identical to the molar Gibbs energy. The difference is therefore equal to the difference in the molar Gibbs energies of the pure substances and in the standard state . This difference is also known as Gibbs standard molar reaction energy . ${\ displaystyle \ mu _ {A} ^ {\ circ}}$${\ displaystyle A}$${\ displaystyle \ mu _ {B} ^ {\ circ} - \ mu _ {A} ^ {\ circ}}$${\ displaystyle G_ {m, B} ^ {\ circ} -G_ {m, A} ^ {\ circ}}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle \ Delta _ {r} G ^ {\ circ}}$

With these definitions, the equation can be written shorter than

${\ displaystyle \ Delta _ {r} G = \ Delta _ {r} G ^ {\ circ} + RT \ \ ln \ left ({\ frac {p_ {B}} {p_ {A}}} \ right) }$.

Is in balance , so ${\ displaystyle \ Delta _ {r} G = 0}$

${\ displaystyle - {\ frac {\ Delta _ {r} G ^ {\ circ}} {RT}} = \ ln \ left ({\ frac {p_ {B}} {p_ {A}}} \ right) }$

or

${\ displaystyle {\ frac {p_ {B}} {p_ {A}}} = e ^ {- {\ frac {\ Delta _ {r} G ^ {\ circ}} {RT}}} =: K}$

While the individual values and are of known initial quantities influenced and may take various different scheduled for experiments in equilibrium values, is their ratio in the steady state solely by and determines the temperature. For given substances and as well as a given temperature, it is a constant, the equilibrium constant . Equilibrium is reached as soon as the ratio of the partial pressures has assumed this numerical value. ${\ displaystyle p_ {A}}$${\ displaystyle p_ {B}}$${\ displaystyle \ Delta _ {r} G ^ {\ circ}}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle K}$

#### General case

Change in Gibbs energy during an equilibrium reaction. Complete conversion of the starting materials is not achieved since there is always a reverse reaction of the products.

In the general case of a reaction with more complex stoichiometry caused by

${\ displaystyle a \ A + b \ B + \ dots \, = \, p \ P + q \ Q + \ dots}$

is symbolized, the reaction equation is rearranged as when transforming a mathematical equation so that all summands are on the right-hand side and a reaction equation of the form is obtained

${\ displaystyle 0 = \ sum _ {J} \ nu _ {J} \ J}$.

The stoichiometric number belonging to the substance has a negative sign for the starting materials on the left side of the original equation and a positive sign for the products on the right side. ${\ displaystyle J}$ ${\ displaystyle \ nu _ {J}}$

Below the turnover variable to continue, then the molar amount of the substance changes according to their stoichiometric number to and is the change in the Gibbs energy in the course of the isothermal and isobaric reaction ${\ displaystyle \ xi}$${\ displaystyle \ mathrm {d} \ xi}$${\ displaystyle J}$${\ displaystyle \ mathrm {d} n_ {J} = \ nu _ {J} \ \ mathrm {d} \ xi}$

${\ displaystyle (\ mathrm {d} G) _ {T, p} = - S \ \ mathrm {d} T + V \ \ mathrm {d} p + \ sum _ {J} \ mu _ {J} \ \ mathrm {d} n_ {J} = \ sum _ {J} \ mu _ {J} \ \ nu _ {J} \ \ mathrm {d} \ xi}$.

Rearranging results

${\ displaystyle \ left ({\ frac {\ partial G} {\ partial \ xi}} \ right) _ {T, p} = \ sum _ {J} \ nu _ {J} \ \ mu _ {J} }$.

In the equilibrium state, the left side is zero and the general thermodynamic equilibrium condition for a reaction with this stoichiometry is

${\ displaystyle 0 = \ sum _ {J} \ nu _ {J} \ \ mu _ {J}}$.

Here the system aims to balance the chemical potentials weighted with the (partly positive, partly negative) stoichiometric numbers.

If the reacting substances are especially ideal gases, the known expressions for their chemical potentials can again be used. Using the definition for the Gibbs reaction energy as well as the definition of the molar Gibbs standard reaction energy extended to several summands results : ${\ displaystyle \ left ({\ frac {\ partial G} {\ partial \ xi}} \ right) _ {T, p} =: \ Delta _ {r} G}$${\ displaystyle \ sum _ {J} \ nu _ {J} \ \ mu _ {J} ^ {\ circ} =: \ Delta _ {r} G ^ {\ circ}}$

{\ displaystyle {\ begin {aligned} \ Delta _ {r} G & = \ sum _ {J} \ nu _ {J} \ \ mu _ {J} \\ & = \ sum _ {J} \ nu _ { J} \ left (\ mu _ {J} ^ {\ circ} + RT \ \ ln \ left ({\ frac {p_ {J}} {p ^ {\ circ}}} \ right) \ right) \\ & = \ left (\ sum _ {J} \ nu _ {J} \ \ mu _ {J} ^ {\ circ} \ right) + RT \ \ sum _ {J} \ nu _ {J} \ \ ln \ left ({\ frac {p_ {J}} {p ^ {\ circ}}} \ right) \\ & = \ Delta _ {r} G ^ {\ circ} + RT \ \ sum _ {J} \ ln \ left ({\ frac {p_ {J}} {p ^ {\ circ}}} \ right) ^ {\ nu _ {J}} \\ & = \ Delta _ {r} G ^ {\ circ} + RT \ \ ln \ prod _ {J} \ left ({\ frac {p_ {J}} {p ^ {\ circ}}} \ right) ^ {\ nu _ {J}} \ end {aligned}} }

In a state of equilibrium, that is ${\ displaystyle \ Delta _ {r} G = 0}$

${\ displaystyle - {\ frac {\ Delta _ {r} G ^ {\ circ}} {RT}} = \ ln \ prod _ {J} \ left ({\ frac {p_ {J}} {p ^ { \ circ}}} \ right) ^ {\ nu _ {J}}}$

or

${\ displaystyle \ prod _ {J} \ left ({\ frac {p_ {J}} {p ^ {\ circ}}} \ right) ^ {\ nu _ {J}} = e ^ {- {\ frac {\ Delta _ {r} G ^ {\ circ}} {RT}}} =: K}$

For example, is the reaction

${\ displaystyle 2 \ A + 3 \ B \ rightleftharpoons C + 2 \ D}$

before, then the corresponding stoichiometric numbers , , and . ${\ displaystyle \ nu _ {A} = - 2}$${\ displaystyle \ nu _ {B} = - 3}$${\ displaystyle \ nu _ {C} = 1}$${\ displaystyle \ nu _ {D} = 2}$

The general thermodynamic equilibrium condition for any substances , , and is ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle C}$${\ displaystyle D}$

${\ displaystyle 0 = -2 \ \ mu _ {A} -3 \ \ mu _ {B} + \ mu _ {C} +2 \ \ mu _ {D}}$.

In the case of an ideal gas mixture in particular, the use of the formulas for the chemical potentials of ideal gases yields the standard molar Gibbs reaction energy

${\ displaystyle \ Delta _ {r} G_ {m} ^ {\ circ} = - 2 \ mu _ {A} ^ {\ circ} -3 \ \ mu _ {B} ^ {\ circ} + \ mu _ {C} ^ {\ circ} +2 \ \ mu _ {D} ^ {\ circ}}$

and the equilibrium constant

${\ displaystyle K = e ^ {- {\ frac {\ Delta _ {r} G ^ {\ circ}} {RT}}}}$.

The equilibrium condition follows for the partial pressures

${\ displaystyle {\ frac {\ left ({\ frac {p_ {C}} {p ^ {\ circ}}} \ right) \ left ({\ frac {p_ {D}} {p ^ {\ circ} }} \ right) ^ {2}} {\ left ({\ frac {p_ {A}} {p ^ {\ circ}}} \ right) ^ {2} \ left ({\ frac {p_ {B} } {p ^ {\ circ}}} \ right) ^ {3}}} = K}$.

Here, too, the numerical values ​​that arise depend, among other things, on the initial quantities used, but the expression mentioned for the ratio of the partial pressures is determined solely by and the temperature, i.e. it is a constant for the substances and the given temperature in a reaction of the given stoichiometry , the equilibrium constant . The reaction has reached equilibrium as soon as the ratio of the partial pressures fulfills the above equilibrium condition. ${\ displaystyle p_ {A}, \ dots, p_ {D}}$${\ displaystyle \ Delta _ {r} G_ {m} ^ {\ circ}}$${\ displaystyle A, \ dots, D}$${\ displaystyle K}$

To determine the molar Gibbs standard reaction energy, only the molar Gibbs energies of the pure substances involved in the standard state are required, which can be found in the relevant tables. ${\ displaystyle \ Delta _ {r} G_ {m} ^ {\ circ}}$

### Electrochemistry

In electrochemistry (see electrochemical voltage series ), the useful work performed in a voluntary conversion of chemical substances (e.g. a fuel cell ) can be determined using the following relationship:

${\ displaystyle \ Delta _ {\ mathrm {r}} G = -z \ cdot F \ cdot E}$

With

• ${\ displaystyle \ Delta _ {\ mathrm {r}} G}$ - molar free enthalpy of reaction
• ${\ displaystyle z}$ - Number of electrons transferred in the reaction under consideration
• ${\ displaystyle F}$- Faraday constant : 96485.3399 (24) A s / mol
• ${\ displaystyle E}$ - equilibrium cell voltage

## Gibbs energy as a fundamental function

If one considers a system whose properties are given by the state variables entropy , volume and molar numbers of the chemical components, then the internal energy of the system, expressed as a function of the stated state variables (namely all extensive variables of the system), is ${\ displaystyle S}$${\ displaystyle V}$${\ displaystyle n_ {1}, \ dots, n_ {r}}$${\ displaystyle r}$${\ displaystyle U}$

${\ displaystyle U = U (S, V, n_ {1}, \ dots, n_ {r})}$

a fundamental function of the system. It describes the system completely; all thermodynamic properties of the system can be derived from it.

However, these variables are often unfavorable for practical work and it would be preferable to have the temperature or the pressure in the list of variables. In contrast to the usual procedure, however, a variable change in the present case must not be made by a simple substitution, otherwise information will be lost. If, for example, the entropy is to be replaced by the temperature , the functions and could be eliminated in order to obtain a function of the form . However, since the temperature is defined thermodynamically as the partial derivative of the internal energy according to the entropy ${\ displaystyle T}$${\ displaystyle S}$${\ displaystyle U (S, V, n_ {i})}$${\ displaystyle T (S, V, n_ {i})}$${\ displaystyle U (T, V, n_ {i})}$

${\ displaystyle T = \ left ({\ frac {\ partial U} {\ partial S}} \ right) _ {V, n_ {1}, \ dots, n_ {r}}}$

this formulation would be synonymous with a partial differential equation for , which would only define functions up to indefinite. This would still be a description of the system under consideration, but it would no longer be a complete description and thus no longer a fundamental function. ${\ displaystyle U}$${\ displaystyle U}$${\ displaystyle U}$

A Legendre transformation must be carried out to change variables while maintaining the complete information . For example, if you want to go to the list of variables , the transformation is: ${\ displaystyle (T, p, n_ {1}, \ dots, n_ {r})}$

{\ displaystyle {\ begin {aligned} G (T, p, n_ {1}, \ dots, n_ {r}) &: = U (S, V, n_ {1}, \ dots, n_ {r}) - \ left ({\ frac {\ partial U} {\ partial S}} \ right) _ {V, n_ {1}, \ dots, n_ {r}} \ cdot S- \ left ({\ frac {\ partial U} {\ partial V}} \ right) _ {S, n_ {1}, \ dots, n_ {r}} \ cdot V \\ & = UT \, S + p \, V \\ & = HT \, S \ end {aligned}}}

The Legendre transform is called Gibbs energy. It is in turn a fundamental function if it is given as a function of the variables - these are the natural variables of the Gibbs energy. It can also be expressed as a function of other variables, but is then no longer a fundamental function. ${\ displaystyle G = HT \ S}$${\ displaystyle (T, p, n_ {1}, \ dots, n_ {r})}$

The origin of the Gibbs energy from a Legendre transformation explains the additive terms and : They compensate for the loss of information that would otherwise be associated with the variable change. ${\ displaystyle -T \, S}$${\ displaystyle p \, V}$

The following table shows some examples of how other thermodynamic quantities can be derived from the fundamental function "Gibbs energy":

• State equations:
${\ displaystyle s (T, p) = - \ left ({\ frac {\ partial g} {\ partial T}} \ right) _ {p}}$
${\ displaystyle v (T, p) = \ left ({\ frac {\ partial g} {\ partial p}} \ right) _ {T}}$
${\ displaystyle h (T, p) = gT \ left ({\ frac {\ partial g} {\ partial T}} \ right) _ {p}}$
• Derivation of the caloric equations of state:
${\ displaystyle c_ {p} (T, p): = \ left ({\ frac {\ partial h} {\ partial T}} \ right) _ {p} = - T \ left ({\ frac {\ partial ^ {2} g} {\ partial T ^ {2}}} \ right) _ {p}}$
${\ displaystyle \ left ({\ frac {\ partial h} {\ partial p}} \ right) _ {T} = vT \ left ({\ frac {\ partial v} {\ partial T}} \ right) _ {p}}$
• Derivatives of entropy:
${\ displaystyle \ left ({\ frac {\ partial s} {\ partial T}} \ right) _ {p} = {\ frac {c_ {p} (T, p)} {T}}}$
${\ displaystyle \ left ({\ frac {\ partial s} {\ partial p}} \ right) _ {T} = - \ left ({\ frac {\ partial v} {\ partial T}} \ right) _ {p}}$

## literature

• Ulrich Nickel: Textbook of Thermodynamics. A clear introduction. 3rd, revised edition. PhysChem, Erlangen 2019, ISBN 978-3-937744-07-0 .
• Hans Rudolf Christen: Fundamentals of general and inorganic chemistry. 5th edition. Verlag Sauerländer, 1977, pp. 291-313.
• Handbook of Chemistry and Physics. CRC-Press, Florida 1981.

## Remarks

1. The following argumentation assumes that the system exchanges heat only with the heat reservoir and volume (and thus volume change work) only with the volume reservoir. If matter were to flow across the system boundary, it could transport additional heat and do work, which must therefore be ruled out.
2. If the whole done by the system work are examined, ie including the change in volume of work, the use of which provides free energy instead of the Gibbs energy, because in an isothermal reversible process, the decrease of the free energy is numerically equal to the total delivered work: .${\ displaystyle \ mathrm {d} W '_ {\ mathrm {total}} = - \ mathrm {d} F}$
3. The specific Gibbs energy in a gravitational field is nothing other than the potential of the gravitational field.
If a particle of mass moves in a gravitational field, then the work done by the particle is ${\ displaystyle m}$
${\ displaystyle \ mathrm {d} W_ {N} = {\ vec {K}} \ cdot {\ vec {\ mathrm {d} l}} \, = \, K \, \ cos (\ vartheta) \, \ mathrm {d} l}$
an example of mechanical non-volume work. This is the amount of the force vector acting on the particle, the length of the path covered by the particle in the force field and the angle between the force vector and path. For the sake of simplicity, let the temperature and pressure be constant and the work performed reversible. The following then applies to the change in the Gibbs energy of the particle ${\ displaystyle K}$${\ displaystyle \ mathrm {d} l}$${\ displaystyle \ vartheta}$
${\ displaystyle \ mathrm {d} G = - \ mathrm {d} W_ {N} \, = \, - K \, \ cos (\ vartheta) \, \ mathrm {d} l}$.
For the specific (i.e. mass related) Gibbs energy is
${\ displaystyle \ mathrm {d} g = - \ mathrm {d} w_ {N} \, = \, - k \, \ cos (\ vartheta) \, \ mathrm {d} l}$
and the change of the specific Gibbs energy at a movement of the particle from point to point is ${\ displaystyle A}$${\ displaystyle B}$
${\ displaystyle \ Delta g \, = \, - \ int _ {A} ^ {B} k \, \ cos (\ vartheta) \, \ mathrm {d} l}$.
The negative integral over the work done by a unit mass is the potential of the force field.
Similar relationships can also be established between the Gibbs energy and other ( conservative ) force fields.
4. a b In the choice of the set of variables there is implicitly the prerequisite that the system is the only form of work to only do volume change work, because with this set of variables there is only a "work coordinate" available that can describe the course of the work. If the system is to perform other forms of work as well , the derivation of the expression for the extended differential fundamental function shown in the section on maximum non-volume work follows${\ displaystyle S, V, n_ {1}, \ dots, n_ {r}}$${\ displaystyle \ mathrm {d} V}$${\ displaystyle W_ {N}}$${\ displaystyle \ mathrm {d} G}$
${\ displaystyle \ mathrm {d} G (T, p, {\ color {NavyBlue} \ dots}, n_ {1}, \ dots, n_ {r}) = - S \, \ mathrm {d} T + V \ mathrm {d} p \, + \, \ sum _ {i = 1} ^ {r} \ mu _ {i} \, \ mathrm {d} n_ {i} + {\ color {NavyBlue} \ mathrm { d} W_ {N}}}$,
whereby the variable list is to be expanded by suitable working coordinates for describing . The other conclusions in this section remain unaffected.${\ displaystyle W_ {N}}$
5. ↑ In addition to the Gibbs energy, the chemical potential can also be derived from other thermodynamic potentials. It then results as a function of the natural variables of the relevant thermodynamic potential. The following applies to the internal energy , the enthalpy , the free energy and the Gibbs energy : ${\ displaystyle U}$ ${\ displaystyle H}$ ${\ displaystyle F}$${\ displaystyle G}$
${\ displaystyle \ left ({\ frac {\ partial U (S, V, n_ {1}, \ dots, n_ {r})} {\ partial n_ {i}}} \ right) _ {S, V, n_ {j \ neq i}} = \ mu _ {i} (S, V, n_ {1}, \ dots, n_ {r})}$
${\ displaystyle \ left ({\ frac {\ partial H (S, p, n_ {1}, \ dots, n_ {r})} {\ partial n_ {i}}} \ right) _ {S, p, n_ {j \ neq i}} = \ mu _ {i} (S, p, n_ {1}, \ dots, n_ {r})}$
${\ displaystyle \ left ({\ frac {\ partial F (T, V, n_ {1}, \ dots, n_ {r})} {\ partial n_ {i}}} \ right) _ {T, V, n_ {j \ neq i}} = \ mu _ {i} (T, V, n_ {1}, \ dots, n_ {r})}$
${\ displaystyle \ left ({\ frac {\ partial G (T, p, n_ {1}, \ dots, n_ {r})} {\ partial n_ {i}}} \ right) _ {T, p, n_ {j \ neq i}} = \ mu _ {i} (T, p, n_ {1}, \ dots, n_ {r})}$
6. It must also be assumed that the dissolved substance has such a low vapor pressure that it is not essentially present in the vapor space above the solution, as otherwise the vapor above the solution is a mixture and thus - contrary to the assumption - is no longer a one-component ideal gas .
7. In a mixture, the total Gibbs energy is generally not the sum of the molar Gibbs energies of the components in the pure state (weighted with the amounts of substance), but the sum of the partial molar Gibbs energies (weighted with the amounts of substance) , depending on the extent the interactions of the components with one another can be very complicated functions of temperature, pressure and the amount of substance present. However, by definition, the ideal gases considered in the present case do not interact with one another, so that the formula derived in the previous section for the chemical potential of a pure ideal gas can also be used for the ideal gases in a mixture.
8. Similar to “ideal gases” being idealized gases with particularly simple behavior, “ideal mixtures” are idealized mixtures with particularly simple behavior. Ideal gases do not necessarily have to be defined by the fact that they obey the ideal gas law. Alternatively, they can also be defined by their chemical potential by the formula
${\ displaystyle \ mu = \ mu ^ {\ circ} + R \, T \, \ ln \ left ({\ frac {p} {p ^ {\ circ}}} \ right)}$
is described - they then automatically obey the ideal gas law. Correspondingly, ideal mixtures can be defined by the chemical potentials of all their components by formulas of the shape
${\ displaystyle \ mu _ {i} = \ mu _ {i} ^ {*} + R \, T \, \ ln x_ {i}}$
- they then automatically have the simple properties corresponding to ideality. is the chemical potential of the -th substance in its pure state. is the amount of substance with which the -th substance is represented in the mixture. Note that the mole fractions are always less than one, their logarithm and thus the entire term therefore assume negative values. As expected, these formulas therefore describe that the chemical potential of a substance in an (ideal) mixture is smaller than in the pure state.${\ displaystyle \ mu _ {i} ^ {*}}$${\ displaystyle i}$${\ displaystyle x_ {i}}$${\ displaystyle i}$${\ displaystyle R \, T \, \ ln x_ {i}}$
9. The term "osmotic pressure" in the broader sense denotes hydrostatic pressure differences that arise due to osmotic effects. In a narrower sense, however, this means the special numerical value for which, according to van-'t-Hoff's law, characterizes the osmotic properties of a solution. It does not matter whether and to what extent in the present situation the osmotic effects actually lead to a hydrostatic pressure difference between the phases. If the solvent passing through the membrane as a result of the osmosis can flow off unhindered, no hydrostatic pressure difference occurs despite the presence of osmosis. If the solvent flows into a closed volume, the hydrostatic pressure increases there until it has reached the numerical value of the “osmotic pressure ” parameter that characterizes the solution . Depending on the inflow and outflow conditions, hydrostatic pressures can also result between these extremes. The parameter "osmotic pressure" can be used to characterize the solution instead of the amount of substance in the dissolved substances, just as the dew point temperature of the air can serve to characterize the air humidity instead of the water vapor partial pressure (and is usually not identical to the air temperature) .${\ displaystyle \ Pi}$${\ displaystyle p}$${\ displaystyle \ Pi}$
10. The derivation can just as well be carried out with the molar quantities, the molar entropies, latent enthalpies and volumes then appear in the resulting equations.
11. In addition, it is to be assumed that no non-volume work is done in the course of the process, since otherwise an additional term would appear in the differential Gibbs function .${\ displaystyle \ mathrm {d} w_ {N}}$
12. Strictly speaking, the pressure in the liquid changes not only by the pressure change imposed by the external cause , but also by the resulting increase in vapor pressure, which has an effect on the liquid. The contribution of the increase in vapor pressure is, however, usually small and neglected.${\ displaystyle \ Delta p}$${\ displaystyle \ Delta p}$
13. Because air dissolves in the originally pure water at the same time, the saturation vapor pressure is reduced by 0.014 Pa due to Raoult's law .
14. The overline only serves to distinguish the magnitude of the gravitational field strength from the specific Gibbs energy .${\ displaystyle {\ overline {g}}}$${\ displaystyle g}$
15. There must be a composition of the mixture for which , since according to the prerequisite, it is a reaction that strives for a state of equilibrium and finally reaches it.${\ displaystyle \ mu _ {A} = \ mu _ {B}}$
16. In the introductory example was and , so .${\ displaystyle \ nu _ {A} = - 1}$${\ displaystyle \ nu _ {B} = 1}$${\ displaystyle \ mathrm {d} \ xi = - \ mathrm {d} n_ {A} = \ mathrm {d} n_ {B}}$

## Individual evidence

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2. Entry on Gibbs energy (function) . In: IUPAC Compendium of Chemical Terminology (the “Gold Book”) . doi : 10.1351 / goldbook.G02629 Version: 2.0.2.
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5. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 107.
6. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 114.
7. a b E. Keszei: Chemical Thermodynamics. Springer, Berlin / Heidelberg 2012, ISBN 978-3-642-19863-2 .
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10. K. Denbigh: The Principles of Chemical Equilibrium. 4th ed., Cambridge University Press, Cambridge 1981, ISBN 0-521-28150-4 .
11. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 119 (the 2864 kJ there correctly rounded to 2865 kJ).
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15. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 131.
16. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 132.
17. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 132.
18. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 171.
19. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 176.
20. JC Kotz, PM Treichel, JR Townsend: Chemistry & Chemical Reactivity. Thomson Brooks / Cole, Belmont, CA, 2009, ISBN 978-0-495-38703-9 , p. 635 limited preview in the Google book search
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22. ^ JE McDonald: Intermolecular Attractions and Saturation Vapor Pressure. Journal of the Atmospheric Sciences, Volume 20 (March 1963), No. 2, doi : 10.1175 / 1520-0469 (1963) 020 <0178: IAASVP> 2.0.CO; 2
23. ,${\ displaystyle \ gamma = 0 {,} 073 \ \ mathrm {N \, m ^ {- 1}}}$${\ displaystyle V _ {\ mathrm {m, l}} = 18 \ \ mathrm {cm ^ {3} \, mol ^ {- 1}} = 1 {,} 8 \ cdot 10 ^ {- 5} \ \ mathrm {m ^ {3} \, mol ^ {- 1}}}$
24. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 161.
25. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 220 ff.
26. ^ PW Atkins: Physical Chemistry. 2nd reprint d. 1st edition. VCH, Weinheim 1990, ISBN 3-527-25913-9 , p. 223 f.
27. ^ HD Baehr: Thermodynamics. 12th edition. Springer, Berlin / Heidelberg / New York 2005, ISBN 3-540-23870-0 , p. 140.