# Enthalpy

Physical size
Surname Enthalpy
Formula symbol ${\ displaystyle H}$
Size and
unit system
unit dimension
SI J  = kg · m 2 · s -2 L 2 · M · T −2
The enthalpy of reaction when burning alcohol in air is negative. It is therefore an exothermic reaction in which heat is given off to the environment.
The melting enthalpy is the amount of energy that must be applied to melt the ice at constant pressure. It is withdrawn from the environment and cools the drink in the process.

The enthalpy ( oldgr. Ἐν en 'in' and θάλπειν thálpein ' to warm'), formerly also heat content , of a thermodynamic system is the sum of the internal energy of the system and the product of the pressure and volume of the system: ${\ displaystyle H}$ ${\ displaystyle U}$ ${\ displaystyle p}$ ${\ displaystyle V}$

${\ displaystyle H = U + pV}$.

It has the dimension of energy and is measured in the unit joule .

The enthalpy is an extensive quantity : the enthalpy of an overall system is the sum of the enthalpies of the subsystems.

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

${\ displaystyle H _ {\ mathrm {m}} = {\ frac {H} {n}}}$.

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

${\ displaystyle h = {\ frac {H} {m}}}$.

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

The enthalpy is just like , and a quantity of state ; it is uniquely determined by the current state of the system and is independent of the previous history of the system. ${\ displaystyle H}$${\ displaystyle U}$${\ displaystyle p}$${\ displaystyle V}$

The practical usefulness of enthalpy is based on the fact that a process changes the enthalpy of a system

${\ displaystyle \ mathrm {d} H = \ mathrm {d} (U + p \ V) = \ mathrm {d} U + p \ \ mathrm {d} V + V \ \ mathrm {d} p}$

by the simpler expression

${\ displaystyle \ mathrm {d} H = \ mathrm {d} U + p \ \ mathrm {d} V}$

will be described, when the process at constant pressure ( isobaric , ) runs out. This expression can, however, be interpreted as the “gross energy” that has to be added to the system if its internal energy is to be increased by the amount and the system uses part of the energy supplied for the volume change work to be performed during the process . In the case of an isobaric process, the “gross energy” to be expended can therefore be identified with the enthalpy supplied and the advantages offered by calculating with state variables. If the system does not do any other form of work besides the volume change work, the enthalpy conversion of the process is equal to the heat conversion. ${\ displaystyle \ mathrm {d} p = 0}$${\ displaystyle \ mathrm {d} U}$${\ displaystyle p \, \ mathrm {d} V}$${\ displaystyle \ mathrm {d} U + p \ \ mathrm {d} V}$${\ displaystyle \ mathrm {d} H}$

Numerous physical and chemical processes take place at constant pressure. This is often the case, for example, with phase transitions or chemical reactions, especially (but not only) when they take place under atmospheric pressure. The enthalpy is then a suitable quantity to describe the heat conversion of these processes.

In theoretical thermodynamics, the enthalpy is a fundamental function , from which the entire thermodynamic information about the system can be derived. The prerequisite, however, is that it is given as a function of the variables entropy , pressure and the number of moles of the chemical components contained in the system. These are the “natural variables” of enthalpy. It can also be applied as a function of other variables, but then no longer contains the complete thermodynamic information. ${\ displaystyle S}$${\ displaystyle p}$ ${\ displaystyle N_ {i}}$

The enthalpy is a Legendre transform of the internal energy. The internal energy is also a fundamental feature when used as a function of their natural variables , , is given. The transition to other sets of variables requires the application of a Legendre transformation if it is to occur without loss of information. The transformation of the internal energy is a fundamental function of natural variables , , generated returns the expression , so the enthalpy. The additive term resulting from the Legendre transformation compensates for the loss of information that would otherwise be associated with the variable change. ${\ displaystyle S}$${\ displaystyle V}$${\ displaystyle N_ {i}}$${\ displaystyle S}$${\ displaystyle p}$${\ displaystyle N_ {i}}$${\ displaystyle U + p \ V}$${\ displaystyle p \ V}$

A distinction must be made between the free enthalpy or Gibbs energy , the Legendre transformation of the enthalpy according to the entropy .

## introduction

The enthalpy is a mathematically defined abstract parameter for the description of thermodynamic systems (see → Thermodynamic properties of enthalpy ). It cannot be directly interpreted as a descriptive energy quantity of a system. Under certain conditions, however, energy quantities occur in a system which in terms of a formula correspond to the enthalpy or to enthalpy changes of the system. They can then be identified with the enthalpy or its change, which - because the enthalpy is an extensive state variable - offers mathematical advantages and with terms such as "enthalpy content" or "enthalpy supply" a compact and clear way of speaking when describing the system and its processes allowed. The enthalpy is often used to describe isobaric processes and stationary flowing fluids.

### Enthalpy in isobaric processes

The enthalpy has a particularly clear meaning in the case of a process that takes place at constant pressure ( i.e. isobaric ). Since numerous technical, chemical and physical processes take place under ambient pressure and thus isobaric, this situation is often encountered.

If one wants to increase the internal energy of a system, one has to supply energy from the outside ( first law of thermodynamics ). If the system has no way of releasing the supplied energy or part of it again in the form of heat or mechanical (or chemical, electrical, magnetic ...) work , the entire supplied energy contributes to increasing the internal energy. For example, if the energy is supplied in the form of a quantity of heat , the energy balance for such a system is simple . ${\ displaystyle U}$${\ displaystyle \ mathrm {d} Q}$${\ displaystyle \ mathrm {d} U = \ mathrm {d} Q}$

As a rule, however, the increase in internal energy is associated with an increase in volume, for example as a result of thermal expansion when the temperature rises, in the course of a phase transition or in the case of a chemical reaction with gas evolution. If the increase in volume is not prevented (this would be the case, for example, with rigidly clamped systems or with chemical reactions in a rigid container), the system expands by the volume against the ambient pressure and does the volume change work that no longer increases internal energy is available: ${\ displaystyle \ mathrm {d} V}$${\ displaystyle p}$ ${\ displaystyle p \ \ mathrm {d} V}$

${\ displaystyle \ mathrm {d} U = \ mathrm {d} Qp \ \ mathrm {d} V}$

or moved

${\ displaystyle \ mathrm {d} Q = \ mathrm {d} U + p \ \ mathrm {d} V \ quad (*)}$

The change in the enthalpy of the system is on the other hand, according to its definition and using the product rule

${\ displaystyle \ mathrm {d} H = \ mathrm {d} (U + p \ V) = \ mathrm {d} U + p \ \ mathrm {d} V + V \ \ mathrm {d} p}$,

which in the special case of constant pressure ( ) is reduced to ${\ displaystyle \ mathrm {d} p = 0}$

${\ displaystyle \ mathrm {d} H = \ mathrm {d} U + p \ \ mathrm {d} V \ quad (*)}$.

Comparison of the marked expressions provides

${\ displaystyle \ mathrm {d} H = \ mathrm {d} Q}$.

So if a system changes from an initial to an end state at constant pressure and there is no other form of work than work of volume change, then the change in the enthalpy of the system is numerically equal to the amount of heat supplied to the system.

In the older literature, the enthalpy was therefore also referred to as the “heat content” of the system. In today's language, this is no longer common, because one cannot see from the resulting energy content of the system whether the energy was supplied as heat or as work. More on this follows in the next section.

The aforementioned restriction to systems that do no work other than volume change work is just as important as the restriction to constant pressure. A common example of systems that can do another form of work are galvanic cells . These can do electrical work, and the heat they convert is not identical to the change in their enthalpy.

The use of enthalpy is not limited to isobaric processes. In the case of non-isobaric processes the enthalpy change is complicated, however, when heat: . ${\ displaystyle \ mathrm {d} H = \ mathrm {d} Q + V \ \ mathrm {d} p}$

### Enthalpy as a state variable

#### State and process variables

A state variable is clearly defined by the current state of the system. In particular, it is independent of the previous history of the system, i.e. of the process through which it came to the present state. Examples are temperature, pressure, internal energy and enthalpy.

A process variable describes the process that transfers the system from one state to another. If there are different process controls that lead from a given initial state to a given final state, the respective process variables can be different despite fixed initial and final states. Examples are the heat flow that the system exchanges with its environment during the process, or the mechanical work that it exchanges with the environment.

#### Heat content and enthalpy content

If a system is supplied with a flow of heat, one is intuitively tempted to imagine a "quantity of heat" that flows into the system, and the accumulated total quantity of which represents a supposed "state quantity of heat content" of the system. However, this idea is not tenable. Mechanical work can also be added to the system, the sum of which would then have to be seen as “work content”. Since the energy in the system can no longer be seen as to whether it was supplied as heat or work, it does not make sense to divide it conceptually into “heat content” and “work content”. In addition, there are usually different ways in which the new state can be achieved, and the total energy supplied to the system can be divided differently between heat and work depending on the process management. In the new state, the system could then have different “heat content” and “work content” depending on the process path followed, so that these could not be state variables either. By " heat " therefore refers in the parlance of thermodynamics only the process variable " heat flow ", the term "heat content" is no longer used.

In contrast to the “heat content”, the internal energy and the enthalpy are state variables, and since they are also extensive (i.e. quantity-like) variables, one can clearly speak of the content of internal energy and the enthalpy content of the system. Both change when an amount of heat is supplied, namely with isochoric process control (i.e. volume kept constant) the content of internal energy increases by exactly the amount of heat supplied , and with isobaric process control (pressure kept constant) the enthalpy content changes exactly by the amount supplied amount of heat . These two state variables, or respectively, do exactly what one would intuitively use the term “heat content” for the respective process type, and replace it. ${\ displaystyle \ mathrm {d} U = \ mathrm {d} Q}$${\ displaystyle \ mathrm {d} H = \ mathrm {d} Q}$${\ displaystyle U}$${\ displaystyle H}$

The state variables also have the advantage that the difference in internal energy or enthalpy between two states of a system can be determined solely from the knowledge of the two states and does not depend on the way in which the state change occurred. The determination of process variables, for example how the energy exchanged during the change of state is divided into heat and work, usually requires additional knowledge of details of the change process.

#### Enthalpy content and enthalpy supply

If a system is transferred from a state to a state by a suitable (isochoric) process , it has the enthalpy in the initial state and the enthalpy in the final state . What does the enthalpy difference mean, expressed in terms of measurable physical quantities? ${\ displaystyle 1}$${\ displaystyle 2}$${\ displaystyle H_ {1}}$${\ displaystyle H_ {2}}$

For an infinitesimally small enthalpy difference, the following applies (see above)

${\ displaystyle \ mathrm {d} H = \ mathrm {d} Q + V \ \ mathrm {d} p}$,

so that for a given current volume a suitable combination of heat supply and pressure change changes the enthalpy content of the system by the amount . It is obvious to call the enthalpy supply and the temporal supply rate the enthalpy flow. A finitely large difference in enthalpy is the integral over . ${\ displaystyle V}$${\ displaystyle \ mathrm {d} Q}$${\ displaystyle \ mathrm {d} p}$${\ displaystyle H}$${\ displaystyle \ mathrm {d} H}$${\ displaystyle \ mathrm {d} H}$${\ displaystyle {\ frac {\ mathrm {d} H} {\ mathrm {d} t}}}$${\ displaystyle \ Delta H = H_ {2} -H_ {1}}$${\ displaystyle \ mathrm {d} H}$

If the enthalpy is unknown and if it is to be determined by measurement, the quantities , and can be measured during the process and added up starting from. In the isobaric case, it is sufficient to measure and add up the heat supply , for example in a suitable calorimeter . The relationship between the isobaric enthalpy supply and the associated temperature change is described by the heat capacity at constant pressure (see section enthalpy and heat capacity ). If an isobaric system is known, the enthalpy difference corresponding to a given temperature difference is also known (see example 2 ). ${\ displaystyle H_ {2}}$${\ displaystyle \ mathrm {d} Q}$${\ displaystyle V}$${\ displaystyle \ mathrm {d} p}$${\ displaystyle H_ {1}}$${\ displaystyle \ mathrm {d} p = 0}$${\ displaystyle \ mathrm {d} Q}$${\ displaystyle c_ {p}}$${\ displaystyle c_ {p}}$

If the opposite is true and known, the enthalpy difference provides information on possible process management and, in the isobaric case, describes what heat conversion is to be expected. For numerous systems, the enthalpies associated with the various states can be taken from the relevant tables. ${\ displaystyle H_ {1}}$${\ displaystyle H_ {2}}$

If the enthalpy of the final state is less than that of the initial state, the enthalpy corresponding to the difference must be removed during the process - the process is exothermic . In the case of an isobaric process, the amount of heat to be dissipated is numerically equal to the determined enthalpy difference. ${\ displaystyle \ Delta H}$${\ displaystyle \ Delta Q}$

If the enthalpy of the final state is greater than that of the initial state, the enthalpy corresponding to the difference must be added - the process is endothermic . In the case of an isobaric process, the amount of heat to be supplied is numerically equal to the determined enthalpy difference.

Since the initial and final enthalpy only depend on the initial and final state, but not on the process in between, when calculating an unknown enthalpy difference, the real process can be replaced by a process that is easier to treat or even by a process chain that uses auxiliary states with known enthalpies, as long as it is this connects the same two states. Practical examples follow below.

In most cases the absolute enthalpy content of a system is not important and only the difference in enthalpy caused by the process is of interest. In this case one has the freedom to choose the zero point for the measurement of the enthalpy as desired.

Note that part of the energy supplied is dissipated into the environment as displacement work and only the rest remains in the system itself, increasing its energy content. The enthalpy supplied, however, is thought to be fully stored in the system . This way of speaking is permissible because the enthalpy is a state variable of the system. If the process is reversed, both the entire energy supplied to the system and the entire enthalpy supplied to the system are returned. The system recovers the energy released into the environment as negative displacement work; the enthalpy is thought to come from the enthalpy content of the system. It turns out again that the enthalpy is not a certain concrete "type of energy", but an abstract quantity, which however allows a very useful way of speaking.

In general, no conservation law applies to enthalpy. As an example, consider a thermally insulated container of constant volume in which a chemical reaction takes place. The change in enthalpy

${\ displaystyle H_ {2} -H_ {1} = U_ {2} -U_ {1} + p_ {2} \ V_ {2} -p_ {1} \ V_ {1}}$

is reduced because of (no energy exchange with the environment; internal energy is subject to energy conservation) and (constant volume of the container) ${\ displaystyle U_ {2} = U_ {1}}$${\ displaystyle V_ {2} = V_ {1}}$

${\ displaystyle H_ {2} -H_ {1} = V \ (p_ {2} -p_ {1})}$.

Although the system inside the container does not exchange energy or matter with the environment, its enthalpy changes when the pressure in the container changes in the course of the chemical reaction. The term “enthalpy supply” would in this case, if one wanted to use it at all, only to be understood as a way of speaking for “enthalpy increase”. However, no enthalpy is extracted from the environment and transported into the container.

In the isobaric case, in which the change in enthalpy is numerically identical to the amount of heat converted, the law of conservation of energy applies to the enthalpy to which the heat energy is subject.

### Examples

#### Example 1: Enthalpy change in a phase transition

Let liquid water be given in equilibrium with its vapor. The temperature of both phases is 10 ° C, the pressure in both phases is the saturation vapor pressure at the given temperature (approx. 1228 Pa). A water vapor table shows that at this temperature and this pressure

• the specific enthalpy of liquid water and${\ displaystyle h_ {1} = 42 \ \ mathrm {\ tfrac {kJ} {kg}}}$
• is the specific enthalpy of water vapor .${\ displaystyle h_ {2} = 2519 \ \ mathrm {\ tfrac {kJ} {kg}}}$

In order to evaporate water at the given temperature and constant pressure, the specific enthalpy has to be increased from to, so it has to be ${\ displaystyle h_ {1}}$${\ displaystyle h_ {2}}$

${\ displaystyle \ Delta h = 2519-42 = 2477 \ \ mathrm {\ tfrac {kJ} {kg}}}$

of specific enthalpy. Since it is an isobaric process, the increase in enthalpy can be achieved by supplying the specific amount of heat . This amount of energy is the so-called specific heat of vaporization or specific enthalpy of vaporization of water at 10 ° C. ${\ displaystyle 2477 \ \ mathrm {\ tfrac {kJ} {kg}}}$

Note also the manner of speaking used here: The enthalpy is actually defined as a state variable and thus represents a property of the system. The 2477 kJ / kg determined in the example are an enthalpy difference between two system states. But it is also called an enthalpy, which again reflects the idea that you are dealing with a quantity-like quantity and have to add a certain "amount of enthalpy" in order to increase the "enthalpy content" of the system by the relevant amount.

The water vapor table also shows that at this temperature and pressure

• the specific volume of the liquid water and${\ displaystyle v_ {1} = 1 {,} 0003 \ cdot 10 ^ {- 3} \ \ mathrm {\ tfrac {m ^ {3}} {kg}}}$
• is the specific volume of the water vapor .${\ displaystyle v_ {2} = 106 {,} 42 \ \ mathrm {\ tfrac {m ^ {3}} {kg}}}$

So the increase in specific volume is

${\ displaystyle \ Delta v \ approx 106 {,} 42 \ \ mathrm {\ tfrac {m ^ {3}} {kg}}}$

and the specific volume change work performed by this volume increase is

${\ displaystyle p \ \ Delta v = 1228 \, \ mathrm {Pa} \ cdot 106 {,} 42 \ \ mathrm {\ tfrac {m ^ {3}} {kg}} \ approx 131 \, \ mathrm {\ tfrac {kJ} {kg}}}$.

The heat supplied as heat is released back into the environment and the rest increase the internal energy of the system. ${\ displaystyle 2477 \ \ mathrm {\ tfrac {kJ} {kg}}}$${\ displaystyle 131 \ \ mathrm {\ tfrac {kJ} {kg}}}$${\ displaystyle 2346 \ \ mathrm {\ tfrac {kJ} {kg}}}$

#### Example 2: Advantage of enthalpy as a state function

This example shows the advantage that can be drawn from the fact that enthalpy is a state function (as opposed to “heat content”). Consider an oxyhydrogen gas reaction in which a water molecule is formed from two atoms of hydrogen and one atom of oxygen :

${\ displaystyle \ mathrm {H_ {2} + {\ tfrac {1} {2}} O_ {2} \ longrightarrow \ H_ {2} O}}$

A relevant set of tables states that at a temperature of 25 ° C and a pressure of one atmosphere, the enthalpy of one mole of the resulting water is 285.84 kJ lower than the enthalpy of the starting materials on the left:

${\ displaystyle \ Delta H _ {\ mathrm {m}} (\ mathrm {25 ^ {\ circ} C}) = - 285 {,} 84 \ \ mathrm {\ tfrac {kJ} {mol}}}$

What is the molar enthalpy difference on both sides if the pressure remains at one atmosphere but the temperature has been increased to 100 ° C? As additional information, the following molar heat capacities at constant pressure are known from experiments (see section enthalpy and heat capacity ), each as an average over the temperature range from 25 ° C to 100 ° C: ${\ displaystyle c_ {p}}$

 ${\ displaystyle \ mathrm {H_ {2}}:}$ ${\ displaystyle c_ {p} = 28 {,} 9 \ \ mathrm {\ tfrac {J} {mol \ K}}}$ ${\ displaystyle \ mathrm {O_ {2}}:}$ ${\ displaystyle c_ {p} = 29 {,} 4 \ \ mathrm {\ tfrac {J} {mol \ K}}}$ ${\ displaystyle \ mathrm {H_ {2} O}:}$ ${\ displaystyle c_ {p} = 75 {,} 5 \ \ mathrm {\ tfrac {J} {mol \ K}}}$

Instead of the original process to be examined, which leads from the initial state directly to the final state , a chain of substitute processes with known properties is computationally considered, which connects the same two states with one another. This is allowed because the enthalpy difference between the initial and final state is only determined by these states themselves and not by the special processes that transform the two into one another. ${\ displaystyle \ mathrm {H_ {2} (100 ^ {\ circ} C) + {\ tfrac {1} {2}} O_ {2} (100 ^ {\ circ} C)}}$${\ displaystyle \ mathrm {H_ {2} O (100 ^ {\ circ} C)}}$

 ${\ displaystyle \ mathrm {H_ {2} (100 ^ {\ circ} C) + {\ tfrac {1} {2}} O_ {2} (100 ^ {\ circ} C)}}$ ${\ displaystyle {\ color {Orange} \ longrightarrow \ Delta H _ {\ mathrm {m}} (\ mathrm {100 ^ {\ circ} C}) =? \ longrightarrow}}$ ${\ displaystyle \ mathrm {H_ {2} O (100 ^ {\ circ} C)}}$ ${\ displaystyle {\ color {RoyalBlue} \ downarrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ uparrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ Delta H _ {\ mathrm {m}, 1}}}$ ${\ displaystyle {\ color {RoyalBlue} \ uparrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ downarrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ uparrow}}$ ${\ displaystyle \ mathrm {H_ {2} (25 ^ {\ circ} C) + {\ tfrac {1} {2}} O_ {2} (100 ^ {\ circ} C)}}$ ${\ displaystyle {\ color {RoyalBlue} \ Delta H _ {\ mathrm {m}, 4}}}$ ${\ displaystyle {\ color {RoyalBlue} \ downarrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ uparrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ Delta H _ {\ mathrm {m}, 2}}}$ ${\ displaystyle {\ color {RoyalBlue} \ uparrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ downarrow}}$ ${\ displaystyle {\ color {RoyalBlue} \ uparrow}}$ ${\ displaystyle \ mathrm {H_ {2} (25 ^ {\ circ} C) + {\ tfrac {1} {2}} O_ {2} (25 ^ {\ circ} C)}}$ ${\ displaystyle {\ color {RoyalBlue} \ longrightarrow \ Delta H _ {\ mathrm {m}, 3} = \ Delta H _ {\ mathrm {m}} (\ mathrm {25 ^ {\ circ} C}) \ longrightarrow} }$ ${\ displaystyle \ mathrm {H_ {2} O (25 ^ {\ circ} C)}}$

In a first computational step, the hydrogen is cooled from 100 ° C to 25 ° C. In doing so, its molar enthalpy changes

${\ displaystyle \ Delta H _ {\ mathrm {m}, 1} = 28 {,} 9 \ \ mathrm {\ tfrac {J} {mol \ K}} \ cdot -75 \ \ mathrm {K} = -2 { ,} 17 \ \ mathrm {\ tfrac {kJ} {mol}}}$.

Then half a mole of oxygen is cooled to the same temperature, its enthalpy changes

${\ displaystyle \ Delta H _ {\ mathrm {m}, 2} = {\ tfrac {1} {2}} \ 29 {,} 4 \ \ mathrm {\ tfrac {J} {mol \ K}} \ cdot - 75 \ \ mathrm {K} = -1 {,} 10 \ mathrm {\ tfrac {kJ} {mol \ H_ {2} O}}}$.

The molar enthalpy of reaction at 25 ° C is known:

${\ displaystyle \ Delta H _ {\ mathrm {m}, 3} = - 285 {,} 84 \ \ mathrm {\ tfrac {kJ} {mol}}}$.

The heating of the resulting water to 100 ° C requires the supply of

${\ displaystyle \ Delta H _ {\ mathrm {m}, 4} = 75 {,} 5 \ \ mathrm {\ tfrac {J} {mol \ K}} \ cdot 75 \ \ mathrm {K} = 5 {,} 66 \ \ mathrm {\ tfrac {kJ} {mol}}}$.

The sum of the molar enthalpy changes during the replacement process

${\ displaystyle \ color {RoyalBlue} \ Delta H _ {\ mathrm {m}, 1} + \ Delta H _ {\ mathrm {m}, 2} + \ Delta H _ {\ mathrm {m}, 3} + \ Delta H_ {\ mathrm {m}, 4} = - 283 {,} 45 \ \ mathrm {\ tfrac {kJ} {mol}}}$

is identical to the molar enthalpy change on the direct process path, so is

${\ displaystyle \ color {Orange} \ Delta H _ {\ mathrm {m}} (\ mathrm {100 ^ {\ circ} C}) = - 283 {,} 45 \ \ mathrm {\ tfrac {kJ} {mol} }}$.

### Stationary flowing fluids

In addition to isobaric processes, the enthalpy is also a useful quantity when describing systems in steady state, such as heat engines or throttles, where it can be used to clearly describe the flow of energy.

Consider a heat engine of a stationary flow of a working fluid is flowed. In the supply pipe with the cross-sectional area , the pressure prevailing in the inflowing fluid forced the volume into the machine in the period under consideration . The flowing fluid has exerted the force on the volume and shifted it by the distance . The mechanical work done on the volume is therefore . If you take into account the internal energy of the test volume and neglect its kinetic energy, then the machine became the energy ${\ displaystyle A_ {1}}$${\ displaystyle p_ {1}}$${\ displaystyle V_ {1}}$${\ displaystyle F = p_ {1} \ A_ {1}}$${\ displaystyle \ Delta x = V_ {1} / A_ {1}}$${\ displaystyle F \ cdot \ Delta x = p_ {1} \ A_ {1} \ cdot V_ {1} / A_ {1} = p_ {1} \ cdot V_ {1}}$${\ displaystyle U_ {1}}$

${\ displaystyle U_ {1} + p_ {1} \ cdot V_ {1} = H_ {1}}$

fed. Corresponding considerations apply to a volume flowing under the pressure through the cross section of the drain pipe during the same period . If, for the sake of generality, one adds an amount of heat supplied to the machine in the period under consideration and a useful work performed by the machine, the energy balance is simple ${\ displaystyle p_ {2}}$${\ displaystyle A_ {2}}$${\ displaystyle V_ {2}}$${\ displaystyle Q_ {1}}$${\ displaystyle W_ {2}}$

${\ displaystyle H_ {1} + Q_ {1} = H_ {2} + W_ {2}}$.

In such a stationary flowing working fluid, the term can be clearly interpreted as the displacement work performed continuously by the fluid (and thus, so to speak, “transported”) . ${\ displaystyle pV}$

An important special case is a throttle , i.e. a pipe in which a constriction or a porous plug reduces the pressure of the flowing fluid from to . If the pipe wall is adiabatic ( ) and no mechanical work is removed from the system ( ), then the energy balance is the throttle ${\ displaystyle p_ {1}}$${\ displaystyle p_ {2}}$${\ displaystyle Q_ {1} = 0}$${\ displaystyle W_ {2} = 0}$

${\ displaystyle H_ {1} = H_ {2}}$,

the throttling is therefore an isenthalpic process . The sizes , and the test volume usually have different values ​​after throttling than before throttling, but the changes are related in such a way that the size combination remains unchanged. ${\ displaystyle U}$${\ displaystyle p}$${\ displaystyle V}$${\ displaystyle U + p \ V}$

The equality of the enthalpies of the test volumes can only be determined for one fluid state before the throttling and one after the throttling. It cannot be said that the enthalpy of the fluid volumes is constant along the entire throttle path, since non-equilibrium states are generally present at the throttle point for which no enthalpy is defined. If the kinetic energies on both sides of the throttle point are not negligibly small, it is also sufficient if their difference is negligibly small, because they are then reduced on both sides of the energy balance.

The working fluid in particular an ideal gas , are , , and thus only dependent on the temperature. In this case, equality of enthalpies also means equality of temperatures before and after the throttling: An ideal gas does not experience any temperature change due to the throttling. In the case of non-ideal fluids, depending on the type of fluid and the process conditions, there may be an increase or decrease in temperature (see → Joule-Thomson effect ). ${\ displaystyle U}$${\ displaystyle p}$${\ displaystyle V}$${\ displaystyle H}$

## Enthalpy in isobaric physical and chemical processes

Numerous processes from physics (e.g. phase transitions ) or from chemistry (e.g. chemical reactions ) take place under constant pressure. In these cases, the enthalpy allows a simple description and calculation of the heat conversion.

### Standard enthalpy of formation

As already mentioned, any process that connects the two states can be used to calculate the enthalpy difference between two states. For example, in the case of a chemical reaction, the starting materials can be broken down into their elements and then reassembled into the product materials. The enthalpy to be expended or removed is the so-called enthalpy of formation of the substance in question. The enthalpies of formation are temperature and pressure dependent. The enthalpies of formation that are implemented under standard conditions are the standard enthalpies of formation.

The molar standard enthalpy of formation (usually for short standard enthalpy of formation) is the enthalpy that is released (negative sign) or is required for formation during the formation of one mole of a substance from the most allotropically stable form of the pure elements under standard conditions (100 kPa and 25 ° C) (positive sign). It is usually given in kilojoules per mole. The size is formulated ( by Engl. Formation is therein for formation; the superscript zero for standard conditions, the delta for the difference). ${\ displaystyle \ Delta H_ {f} ^ {0}}$ ${\ displaystyle f}$

If it is negative, is an exothermic reaction and is energy in the formation of the substance from the elements released ( heat of formation ). If, on the other hand, it is positive, it is an endothermic reaction and energy has to be expended to form the substance from its starting elements. Strongly negative values ​​of the standard enthalpy of formation are a characteristic of chemically particularly stable compounds (i.e. a lot of energy is released during their formation and a lot of energy has to be expended to destroy the bonds). The standard enthalpy of formation of the chemical elements in their most stable state ( H 2 , He , Li , ...) is set to 0 kJ / mol by definition.

If the standard enthalpy of formation of the substances involved in a chemical reaction is known, the reaction enthalpy of this reaction can easily be calculated under standard conditions. It is the difference between the standard enthalpies of formation of the reaction products ("products") on the one hand and the starting materials (reactants; "starting materials") on the other hand ( Hess' s proposition ):

${\ displaystyle \ Delta H _ {\ mathrm {reaction}} ^ {0} = \ sum \ Delta H_ {f, \ mathrm {Products}} ^ {0} - \ sum \ Delta H_ {f, \ mathrm {Educts} } ^ {0}}$

All values ​​refer to the thermodynamic equilibrium , otherwise the temperature would not be defined.

Conversely, the standard enthalpy of formation can be determined with the aid of Hess's theorem from the enthalpies of reactions in which the respective substance participates as a starting material or product. If no experimental data are available, an estimate of the standard enthalpies of formation can also be estimated using group contribution methods. The Benson increment method is particularly suitable for this .

#### Inorganic substances

chemical formula material ${\ displaystyle \ Delta H_ {f} ^ {0}}$ (kJ / mol) source
H 2 O (g) Water (gaseous) −241.83
H 2 O (l) Water (liquid) −285.83
CO 2 (g) Carbon dioxide −393.50
NH 3 (g) ammonia −45.94

#### Organic substance

Molecular formula material ${\ displaystyle \ Delta H_ {f} ^ {0}}$ (kJ / mol) source
CH 4 (g) methane −74.87
C 2 H 4 (g) Ethylene +52.47
C 2 H 6 (g) Ethane −83.8

### Enthalpy in physics (thermodynamics)

In a narrower sense, thermodynamics only describes the intermolecular forces, i.e. the energetic relationships (phase states or their changes) between the individual molecules of a substance.

#### Enthalpy of evaporation / enthalpy of condensation

Temperature dependence of the enthalpy of vaporization of water, methanol , benzene and acetone .

The molar enthalpy of vaporization is the enthalpy that is required to bring one mole of a substance isothermally and isobarically from the liquid to the gaseous state. The enthalpy of vaporization depends on the substance as well as on temperature and pressure, whereby it is always positive and its sign is therefore usually not given. ${\ displaystyle \ Delta _ {V} H}$

The molar enthalpy of condensation is the enthalpy that is released when a mole of a substance condenses , whereby this again changes isothermally and isobarically from the gaseous to the liquid state of aggregation. It always has a negative sign. ${\ displaystyle \ Delta _ {K} H}$

The enthalpy of evaporation decreases as the temperature rises and becomes zero when the critical point is reached, since there it is no longer possible to differentiate between liquid and gas. As a rule, in tables, enthalpy of vaporization data is either related to 25 ° C or tabulated for the various temperature-pressure combinations along the boiling point curve, this always applies . ${\ displaystyle \ Delta _ {V} H = - \ Delta _ {K} H}$

In the case of mixtures or solutions of substances, the enthalpies add up in the ratio of their mixture components .

If no enthalpy of vaporization values are available for a substance, these can be calculated for any temperature using the Clausius-Clapeyron equation if the temperature dependence of the vapor pressure curve at the temperature under consideration is known.

In rare cases, values ​​for enthalpies of vaporization have been tabulated. The enthalpy of evaporation can always be derived from the thermodynamic data by forming the difference if standard enthalpy of formation values ​​for the liquid and gaseous state are known, e.g. B. for water, carbon disulfide , methanol, ethanol, formic acid, acetic acid, bromine in the table above.

Example table salt:

• Enthalpy of evaporation ∆ V H = +170 kJ / mol (at 1465 ° C, table value)
• The following table values ​​for the enthalpy of formation are used for the practical calculation of the heat of vaporization:
 NaCl (melt) ${\ displaystyle \ longrightarrow}$ NaCl (g) −386 kJ / mol −182 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$Enthalpy of evaporation ∆ V H = +204 kJ / mol

The difference between the two values ​​170 and 204 is within the usual range.

#### Enthalpy of sublimation

The sublimation describes the transition of a solid to the gas phase by-passing the liquid melt phase (technical application in the freeze-drying ). The enthalpy of sublimation is partly listed in tables. In principle, the enthalpy of melting and evaporation may also be combined with the same reference temperature:

Enthalpy of sublimation = enthalpy of fusion + enthalpy of evaporation.

The enthalpy of sublimation is not to be confused with the heat of sublimation . This corresponds to the sum of the heat of fusion and the heat of evaporation .

The enthalpy of sublimation can always be derived from the thermodynamic data if standard enthalpy of formation values ​​are known for the solid and gaseous state of aggregation.

• Enthalpy of sublimation table salt: 211 kJ / mol (25 ° C, table value)
• Calculation:
 NaCl (s) ${\ displaystyle \ longrightarrow}$ NaCl (g) −411 kJ / mol −182 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$ Enthalpy of sublimation +229 kJ / mol

Note: The example shows that, in principle, it is also possible to calculate processes that can hardly be carried out in practice. The “enthalpy of sublimation of elemental carbon” was “determined” in this way.

Sublimation enthalpies of some substances
Fabric signs Fabrics Enthalpy of sublimation (kJ / mol)
N / A sodium 108
K potassium 89
Rb Rubidium 82
Cs Cesium 78
Mg magnesium 150
Approx Calcium 192
Sr strontium 164
Ba barium 176
I 2 Iodine 62.4
C 10 H 8 naphthalene 72.6
CO 2 carbon dioxide 26.1

#### Enthalpy of fusion / enthalpy of crystallization

After a solid substance has been heated to its melting point temperature, heat of fusion is absorbed at this temperature without the temperature rising any further. This form of heat is called latent heat because it does not change the temperature. In the case of ionic solids, the solid / liquid phase transition results in molten salts with easily mobile ions (technical application in melt flow electrolysis ) . Table salt melts at 800 ° C.

Enthalpies of fusion are rarely recorded in tables.

The enthalpy of fusion can always be derived from the thermodynamic data if standard enthalpy of formation values ​​for the solid and liquid state are known.

• Enthalpy of fusion for table salt: 28; 30.2 kJ / mol (800 ° C, table values)
• The following table values ​​are used for the practical calculation of the heat of fusion:
 NaCl (s) ${\ displaystyle \ longrightarrow}$ NaCl (melt) −411 kJ / mol −386 kJ / mol Enthalpy of formation (25 ° C) NaCl (s) ${\ displaystyle \ longrightarrow}$ Na + (melt) + Cl - (melt) −411 kJ / mol approx. −215 kJ / mol approx. −170 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$ Melting enthalpy +25 kJ / mol (NaCl, 25 ° C)

When this process is reversed, crystallization from the melt, the ions of a salt can combine directly to form their solid crystal lattice. During the precipitation of common salt crystals from the melt, −25.2 kJ / mol crystallization enthalpy (or 29 ± 1 kJ / mol at 800 ° C) is released.

Experience has shown that “supercooled molten salts” can release considerable amounts of heat through spontaneous crystallization. (Application: heating pad ).

#### Lattice enthalpy

According to a common definition, the lattice energy is the energy that has to be expended in a vacuum (i.e. at external pressure ) in order to convert a crystalline ionic solid into the gas phase (i.e., sublimation energy), i.e. to separate it into the gaseous ions. The lattice energy and the lattice enthalpy differ qualitatively. The lattice energy is an internal energy, while the lattice enthalpy is an enthalpy. The lattice enthalpy also takes into account the volume work to be performed against a constant external pressure. If a molar lattice enthalpy has been determined for the separation of the components of the solid , then it is the molar lattice energy . Here is the change in volume related to the amount of substance. ${\ displaystyle p = 0}$${\ displaystyle p \ Delta V}$${\ displaystyle \ Delta _ {G} H}$${\ displaystyle \ Delta _ {G} U = \ Delta _ {G} Hp \ Delta V _ {\ mathrm {m}}}$${\ displaystyle \ Delta V _ {\ mathrm {m}}}$

The lattice enthalpy of NaCl is composed as follows:

 NaCl (solid) ${\ displaystyle \ longrightarrow}$ Na + (g) + Cl - (g) −411 kJ / mol 611 kJ / mol −244 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$ Lattice enthalpy +778 kJ / mol

Comparison: This is roughly twice the energy required that would be released in the strongly exothermic reaction of sodium metal and chlorine gas. The formation of gaseous ions is therefore extremely endothermic.

The lattice enthalpy ΔH 0 L depends on the size and charge of the ions involved and is always positive with this type of definition, since the lattice would otherwise not be stable. Aluminum oxide Al 2 O 3 (Al 3+ and O 2− ) has a very high lattice enthalpy with 15157 kJ / mol. The high lattice enthalpy is used in aluminothermic processes; These include, for example, aluminothermic welding and the representation of elements from their oxides and aluminum by means of aluminothermic . In the latter case, the high lattice enthalpy of the aluminum oxide is a main driving force for the reaction, as it is reflected directly in the Gibbs energy .

The lattice energy is often defined as the enthalpy of reaction in the formation of the solid salt lattice based on ions in the gas phase. If the lattice energy is defined in this way, the process is exothermic and the associated change in enthalpy must be given as negative. The lattice enthalpy of aluminum oxide would then be −15157 kJ / mol, for example.

The lattice enthalpy depends on the one hand on the size of the ions involved: the larger the ions, the smaller the lattice energy released, since the forces of attraction decrease as the distance between the positive nuclei and the negative electron shell of the binding partner increases.

Examples: Molar lattice enthalpy of alkali fluorides at 25 ° C in kJ / mol :

Surname formula Ionic radius of the monovalent
alkali metal cations X + in pm
Lattice enthalpy in kJ per mol
Lithium fluoride LiF 74 1039
Sodium fluoride NaF 102 920
Potassium fluoride Theatrical Version 138 816
Rubidium fluoride RbF 149 780
Cesium fluoride CsF 170 749

On the other hand, the lattice energy depends on the electrical charge of the ions involved: the larger the charges, the greater the forces of attraction and the greater the lattice energy.

Examples: Molar lattice enthalpy at 25 ° C in kJ per mol (in the examples the ion radius changes only slightly):

Surname formula Cations Anions Lattice enthalpy in kJ per mol
Sodium chloride NaCl Na + Cl - 780
Sodium sulfide Na 2 S Na + S 2− 2207
Magnesium chloride MgCl 2 Mg 2+ Cl - 2502
Magnesium sulfide MgS Mg 2+ S 2− 3360

For a similar effect on graphite under neutron radiation see Wigner-Energie .

##### Enthalpy of solvation, enthalpy of hydration

It indicates what energy is released when gaseous ions attach to solvents, i.e. when solvated ions form. The most common case of solvent = water is called hydration enthalpy .

 Na + (g) ${\ displaystyle \ longrightarrow}$ Na + (hydrated) 611 kJ / mol −240 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$Enthalpy of hydration Na + −851 kJ / mol (i.e. extremely exothermic) Cl - (g) ${\ displaystyle \ longrightarrow}$ Cl - (hydrated) −244 kJ / mol −167 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$Enthalpy of hydration Cl - +77 kJ / mol (i.e. weakly endothermic)

The enthalpy of hydration of the gaseous ions of common salt is strongly exothermic at −774 kJ / mol.

##### Enthalpy of solution / enthalpy of crystallization

The enthalpy of solution of salts includes 1) the separation of the ion lattice into single ions and 2) the solvation of the single ions. Sub-step 1) is very endothermic, sub-step 2) is very exothermic.

Enthalpy of solution = enthalpy of lattice + enthalpy of solvation.

Enthalpy of solution NaCl in water = (+778 kJ / mol) + (−851 + 77 kJ / mol) = +4 kJ / mol (25 ° C).

This value is in good agreement with tables of +3.89 kJ / mol for the heat of sodium chloride solution. When dissolving, there is therefore a very slight cooling of the solution.

Of course, for the practical calculation of the heat of solution, one does not go the detour via the lattice energy, but one calculates directly and with only a few table values (occasionally one finds the value (NaCl) hydrate = −407 kJ / mol instead of the individual ions) :

 NaCl (s) ${\ displaystyle \ longrightarrow}$ NaCl (hydrated) −411 kJ / mol −407 kJ / mol Enthalpy of formation (25 ° C) NaCl (s) ${\ displaystyle \ longrightarrow}$ Na + (hydrated) + Cl - (hydrated) −411 kJ / mol −240 kJ / mol −167 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$ Enthalpy of solution +4 kJ / mol (water, 25 ° C)

The enthalpy of solvation can always be derived from the thermodynamic data if standard enthalpy of formation values ​​are found for the solid and dissolved state, e.g. B. formic acid and carbon dioxide in the table above. It applies to “infinite dilution”.

When this process is reversed, crystallization from solution, the dissolved ions of a salt 1) release their solvation shell and 2) combine in a solid crystal lattice. During the precipitation of common salt crystals from water, −3.89 kJ / mol crystallization enthalpy is released.

#### Intermolecular enthalpy contributions

Differently strong interactions between the molecules are the reason why substance groups have similar or very different enthalpies of sublimation.

• London forces, dipole-dipole interactions and ion-dipole interactions with 1–15 kJ / mol bond make weak contributions. They are summarized as the van der Waals interaction . See as examples the enthalpies of melting and evaporation of hydrogen, carbon monoxide and methane.
• Hydrogen bonds with 20–40 kJ / mol bond (depending on polarization) make greater contributions . See as examples the enthalpies of melting and evaporation of water, methanol and formic acid and also enthalpies of hydration. Hydrogen bonds are also responsible for the fact that the boiling point of water is 100 ° C, while that of hydrogen sulfide is only −83 ° C (see boiling point anomaly ).
• Ion-ion interactions in crystals make very strong contributions. In the crystal, sodium chloride does not consist of discrete NaCl molecules, but of an equal number of sodium cations and chloride anions, which are exactly arranged in the crystal lattice according to the Coulomb forces .

### Enthalpy in chemistry (thermochemistry)

In a narrower sense, thermochemistry only describes the intramolecular forces, i.e. the energetic relationships between the individual atoms of a molecule. Covalent bonds contain approx. 150–1000 kJ / mol binding energy, ionic bonds approx. Five times as large amounts.

With knowledge of the standard enthalpies of formation of educts and products, a possible chemical reaction can be roughly balanced in terms of energy. The most important question is often whether a process is endothermic or exothermic and how strong it is.

${\ displaystyle \ Delta H _ {\ mathrm {reaction}} ^ {0} = \ sum \ Delta H_ {f, \ mathrm {Products}} ^ {0} - \ sum \ Delta H_ {f, \ mathrm {Educts} } ^ {0}}$

Through details such as evaporation, melting, solvation or crystallization enthalpies, partial steps within the chem. Reaction are energetically specified.

#### Enthalpy of reaction

The enthalpy of reaction is the energy that is released or required when new chemical bonds are formed between the molecules of two substances . It depends on the reactants (reactants) and the type of chemical bond in the product. For the calculation, the sum of the enthalpies of formation of the products is compared with that of the educts. The difference is the enthalpy of reaction, which can then be standardized by referring to the amount of substance of the product of interest:

 2 Na (s) + Cl 2 (g) ${\ displaystyle \ longrightarrow}$ 2 NaCl (s) 2 x 0 kJ / mol 0 kJ / mol 2 × −411 kJ / mol Enthalpy of formation (25 ° C) ${\ displaystyle \ Rightarrow}$Enthalpy of reaction = 2 mol × (−411) kJ / mol - 2 mol × 0 kJ / mol - 1 mol × 0 kJ / mol = −822 kJ .
So the reaction is exothermic. Division by the amount of substance obtained, in this case 2 mol of sodium chloride, yields its molar enthalpy of formation of −411 kJ / mol NaCl (25 ° C) (which in this example is already assumed at the beginning ).

#### Standard enthalpy of combustion

Combustion is also a chemical reaction. The enthalpy of reaction of the combustion reaction or the standard enthalpy of combustion of a substance is the change in enthalpy that occurs when a substance burns completely under O 2 excess (O 2 excess pressure) and standard conditions (101.325 kPa and 25 ° C ). By definition, this heat of combustion relates to the formation of gaseous carbon dioxide and liquid water (or N 2 ) as end products; No gaseous water can form under excess oxygen pressure. It is denoted by Δ V H 0 and its absolute value is identical to the calorific value H s .

The following reaction is carried out in an autoclave tube with excess oxygen pressure:

 C 3 H 8 (g) + 5 O 2 (g) ${\ displaystyle \ longrightarrow}$ 3 CO 2 (g) + 4 H 2 O ( l ) −103.2 kJ / mol 5 x 0 kJ / mol 3 × −393.5 kJ / mol 4 × −285.8 kJ / mol
${\ displaystyle \ Rightarrow}$Enthalpy of reaction = 3 mol × (−393.5) kJ / mol + 4 mol × (−285.8) kJ / mol - 1 mol × (−103.2) kJ / mol - 5 mol × (0) kJ / mol = -2.22 MJ .
Division by the amount of substance used, in this case 1 mol propane, yields its molar enthalpy of combustion of −2.22 MJ / mol propane (25 ° C)

The same reaction in an open burner flame; only gaseous combustion products are created:

 C 3 H 8 (g) + 5 O 2 (g) ${\ displaystyle \ longrightarrow}$ 3 CO 2 (g) + 4 H 2 O ( g ) −103.2 kJ / mol 5 x 0 kJ / mol 3 × −393.5 kJ / mol 4 × −241.8 kJ / mol
${\ displaystyle \ Rightarrow}$Enthalpy of reaction = 3 mol × (−393.5) kJ / mol + 4 mol × (−241.8) kJ / mol - 1 mol × (−103.2) kJ / mol - 5 mol × (0) kJ / mol = -2.04 MJ .
Division by the amount of substance used, in this case 1 mol of propane, yields its molar enthalpy of combustion of −2.04 MJ / mol propane (25 ° C)

It is pointless to look for the standard enthalpies of formation of the educts and products for every reaction, in addition to that in the correct aggregate state. In addition, with larger molecules you quickly run into a “data vacuum”. The following simplified considerations have proven themselves in practice:

1. It is irrelevant whether a long-chain alkyl-substituted ethylene is brominated or ethylene itself, the heat of reaction per (C = C) double bond is largely the same.
2. It is irrelevant whether you calculate a reaction completely in the liquid phase or completely in the gas phase, the heat of reaction hardly influences this.
3. It is insignificant (some discrepancies) when calculating reactions performed at 150 ° C for 25 ° C standard conditions. (The enthalpy of reaction can be calculated for any temperature if the temperature dependence of the molar heats of all reactants is known.)

Therefore, normal organic-chemical reactions such as halogen additions, cycloadditions, esterifications with acids or anhydrides, hydrolyses etc. can be calculated with the help of numerous tabulated increments for gaseous molecules according to Benson .

In the following example, the reaction energy of the addition of bromine to ethylene is calculated with Benson increments and, for comparison, estimated from the bond energies of the bonds involved. (Note: binding energies are averaged dissociation energies, not standard enthalpies of formation!)

Bromine addition to an alkene , enthalpy of reaction calculated with standard enthalpies of formation:

 H 2 C = CH 2 + Br 2 ${\ displaystyle \ longrightarrow}$ H 2 BrC-CBrH 2 52 kJ / mol 0 kJ / mol −39 kJ / mol
${\ displaystyle \ Rightarrow}$Enthalpy of reaction (25 ° C, all values ​​for the gaseous state) = (−39 - 52) kJ / mol = −91 kJ / mol double bond

Bromine addition to an alkene , enthalpy of reaction calculated with Benson increments :

 H 2 C = CH 2 + Br 2 ${\ displaystyle \ longrightarrow}$ H 2 BrC-CBrH 2 2 C d - (H) 2 : 2 x +28.1 kJ / mol 2 C - (H) 2 (Br) (C): 2 × −22.6 kJ / mol
${\ displaystyle \ Rightarrow}$Enthalpy of reaction (calculated according to Benson ) = (−45.2 - (+56.2)) kJ / mol = −101 kJ / mol double bond

Bromine addition to an alkene , enthalpy of reaction estimated with binding energies:

 H 2 C = CH 2 + Br 2 ${\ displaystyle \ longrightarrow}$ H 2 BrC-CBrH 2 4 CH: 4 x> 455 kJ / mol 4 CH: 4 x 380 ± 50 kJ / mol 1 C = C: 1 x 614 kJ / mol 1 CC: 1 x 348 kJ / mol 1 Br-Br: 193 kJ / mol 2 C-Br: 2 x 260 ± 30 kJ / mol
${\ displaystyle \ Rightarrow}$Enthalpy of reaction (estimated from bond energies) = (2388 ± 200 - 2627) kJ / mol = -239 ± 200 kJ / mol double bond

The best estimate for reaction enthalpies is achieved with standard enthalpies of formation or increments according to Benson, when using "binding energies" the uncertainty of ± 200 kJ / mol is much too high.

#### Technical applicability

The enthalpies of reaction of organic reactions are in the range −160 to +100 kJ per mole of “reactive centers”. All addition reactions with epoxides, anhydrides and halogens prove to be very exothermic. These thermochemical considerations do not provide any information about how quickly these heats of reaction are released. They only make the statement that this heat is released until the end of the reaction. Each reaction increases its speed by two to three times with a temperature increase of 10 K. Conversely, a two-fold dilution of the reactants often means halving the reaction speed or heat output of the reaction. Calculated reaction enthalpies are used to calculate the temperature profile in a system of reactants and solvents via their heat capacities . Large-scale technical systems only have limited cooling capacities (heat / time); this is often neglected in laboratory tests.

### Binding energy / dissociation energy

The binding energy or binding strength indicates the “stability” of the covalent bond . The determination is only possible in the case of diatomic symmetrical molecules such as B. hydrogen or halogens directly possible. In these cases the dissociation energy for the formation of two identical radicals can easily be measured / calculated. In the case of “element radicals”, the standard enthalpy of formation of radicals is also referred to as the enthalpy of atomization .

In all other cases, values ​​for the “binding energy” are only possible indirectly by comparing several dissociation energy measurements on homologous molecules. The values ​​fluctuate depending on the substitution pattern at the radical centers.

The standard enthalpy of formation of gaseous

• Bromine radicals contain the enthalpy of vaporization (31 kJ / mol) which is necessary to convert liquid bromine into gaseous bromine.
• Iodine radicals contain the enthalpy of sublimation (62 kJ / mol), which is necessary to convert crystalline iodine into gaseous iodine.
• Carbon radicals are identical to the standard enthalpy of formation of gaseous carbon vapor.

## Thermodynamic properties of enthalpy

### Enthalpy 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} ... N_ {r}}$${\ displaystyle r}$${\ displaystyle U}$

${\ displaystyle U = U (S, V, N_ {1}, ..., N_ {r})}$

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. For example, if the volume by the pressure to be replaced could be from the functions and be eliminated to a function of the shape to be obtained. However, since the pressure is thermodynamically defined as the partial derivative of the internal energy according to the volume ${\ displaystyle p}$${\ displaystyle V}$${\ displaystyle U (S, V, N_ {i})}$${\ displaystyle p (S, V, N_ {i})}$${\ displaystyle U (S, p, N_ {i})}$

${\ displaystyle p = - \ left ({\ frac {\ partial U} {\ partial V}} \ right) _ {S, N_ {1}, ..., N_ {r}}}$

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: ${\ displaystyle (S, p, N_ {1}, ..., N_ {r})}$

${\ displaystyle H (S, p, N_ {1}, ..., N_ {r}): = U (S, V, N_ {1}, ..., N_ {r}) - \ left ({ \ frac {\ partial U} {\ partial V}} \ right) _ {S, N_ {1}, ..., N_ {r}} \ cdot V = U + p \ V}$.

The Legendre transform is called enthalpy. It is in turn a fundamental function if it is given as a function of the variables - these are the natural variables of enthalpy. It can also be expressed as a function of other variables, but is then no longer a fundamental function. ${\ displaystyle H = U + p \ V}$${\ displaystyle (S, p, N_ {1}, ..., N_ {r})}$

The origin of the enthalpy from a Legendre transformation explains the additive term : It compensates for the loss of information that would otherwise be associated with the change in variables. ${\ displaystyle p \, V}$

Fundamental functions, which have the dimension energy, are also called thermodynamic potentials . The enthalpy is therefore a thermodynamic potential.

### Derivatives of enthalpy

If you start from the internal energy as a function of its natural variables and form its total differential, you get:

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

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 {\ begin {aligned} T &: = \ left ({\ frac {\ partial U (S, V, N_ {1}, ..., N_ {r})} {\ partial S}} \ right ) _ {V, N_ {1}, ..., N_ {r}}, \\ p &: = - \ left ({\ frac {\ partial U (S, V, N_ {1}, ..., N_ {r})} {\ partial V}} \ right) _ {S, N_ {1}, ..., N_ {r}}, \\\ mu _ {i} &: = \ left ({\ frac {\ partial U (S, V, N_ {1}, ..., N_ {r})} {\ partial N_ {i}}} \ right) _ {S, V, N_ {j \ neq i} }, \ end {aligned}}}

so that the differential can also be written as

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

On the one hand, the total differential of enthalpy as a function of its natural variables is formal

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

and on the other hand, using their definition

{\ displaystyle {\ begin {aligned} \ mathrm {d} H & = \ mathrm {d} (U + pV) \\ & = \ mathrm {d} U + \ mathrm {d} (pV) \\ & = T \ , \ mathrm {d} Sp \, \ mathrm {d} V + \, \ sum \ mu _ {i} \, \ mathrm {d} N_ {i} \, + \ mathrm {d} (pV) \\ & = T \, \ mathrm {d} Sp \, \ mathrm {d} V + \, \ sum \ mu _ {i} \, \ mathrm {d} N_ {i} \, + p \, \ mathrm {d} V + V \ mathrm {d} p \\ & = T \, \ mathrm {d} S + V \ mathrm {d} p \, + \, \ sum \ mu _ {i} \, \ mathrm {d} N_ {i} \ quad (*) \ end {aligned}}}

so that it follows from the comparison of the coefficients in the marked equations

${\ displaystyle \ left ({\ frac {\ partial H} {\ partial S}} \ right) _ {p, N_ {1}, ..., N_ {r}} = T}$,

such as

${\ displaystyle \ left ({\ frac {\ partial H} {\ partial p}} \ right) _ {S, N_ {1}, ..., N_ {r} ..} = V}$

and

${\ displaystyle \ left ({\ frac {\ partial H} {\ partial N_ {i}}} \ right) _ {S, p, N_ {j \ neq i}} = \ mu _ {i}}$.

At the same time, the derivation shows how the addition of the term changes the list of independent variables from in by removing the term dependent on and adding a dependent term in the total differential . ${\ displaystyle p \, V}$${\ displaystyle (S, V, ...)}$${\ displaystyle (S, p, ...)}$${\ displaystyle \ mathrm {d} V}$${\ displaystyle \ mathrm {d} p}$

The second of the marked equations is a "differential fundamental function", namely the differential enthalpy as a function of its natural variables:

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

### Minimum principle of enthalpy

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, a minimum principle of internal energy can be derived: 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 enthalpy: A system that is kept at constant pressure, of all attainable states with this pressure, assumes the state of equilibrium in which the enthalpy has the lowest possible value.

As a proof, consider a system whose pressure is kept at a constant value. The pressure regulation can be done, for example, in that the system under consideration is in contact via a movable adiabatic wall with a second system which invariably has the desired pressure (in thermodynamic terminology: a pressure reservoir). By moving the contact wall, the system under consideration can, if necessary, “exchange volume” with the pressure reservoir until it has adjusted its pressure to that of the reservoir again. The overall system formed from the system under consideration and the pressure reservoir assumes the lowest possible internal energy with the entropy kept constant according to the energy minimum principle , and the following applies in the energy minimum: ${\ displaystyle U _ {\ mathrm {Ges}}}$

${\ displaystyle \ mathrm {d} U _ {\ mathrm {Ges}} = \ mathrm {d} U + \ mathrm {d} U _ {\ mathrm {Res}} = 0}$.

Since only volume is exchanged with the reservoir according to its definition, the internal energy of the reservoir can only be changed by doing volume change work on it:, and thus is ${\ displaystyle \ mathrm {d} U _ {\ mathrm {Res}} = - p _ {\ mathrm {Res}} \, \ mathrm {d} V _ {\ mathrm {Res}}}$

${\ displaystyle \ mathrm {d} U-p _ {\ mathrm {Res}} \, \ mathrm {d} V _ {\ mathrm {Res}} = 0}$.

During the displacement of the partition is considered so that ${\ displaystyle \ mathrm {d} V _ {\ mathrm {Res}} = - \ mathrm {d} V}$

${\ displaystyle \ mathrm {d} U + p _ {\ mathrm {Res}} \, \ mathrm {d} V = 0}$.

Since the pressure of the reservoir is constant according to the assumption, it is possible to write ${\ displaystyle p _ {\ mathrm {Res}}}$

${\ displaystyle \ mathrm {d} (U + p _ {\ mathrm {Res}} \, V) = 0}$.

The extremal properties of the function should now be examined more closely. However, since pressure equalization is expected between the system under consideration (pressure ) and the pressure reservoir (pressure ), it is obvious to limit the further investigation to the states with the property . However, on the subset of these states is identical to . So it applies ${\ displaystyle U + p _ {\ mathrm {Res}} \, V}$${\ displaystyle p}$${\ displaystyle p _ {\ mathrm {Res}}}$${\ displaystyle p = p _ {\ mathrm {Res}}}$${\ displaystyle U + p _ {\ mathrm {Res}} \, V}$${\ displaystyle U + p \, V}$

${\ displaystyle \ mathrm {d} H = \ mathrm {d} (U + p \, V) = 0}$, provided that .${\ displaystyle p = p _ {\ mathrm {Res}}}$

As can be shown, the second derivative of is positive in this state, so that the extremum found for the enthalpy is actually a minimum. ${\ displaystyle H}$

The minimum principle for the internal energy of the overall system with constant entropy thus leads to the enthalpy of the system under consideration assuming a minimum on the subset of states with constant pressure . If the system is not yet in equilibrium, it will move (under isobaric conditions) voluntarily into states of lower enthalpy. Equilibrium is reached when the enthalpy has the lowest possible value under the given conditions. ${\ displaystyle p = p _ {\ mathrm {Res}}}$

If one wants to determine the state of equilibrium with the help of the (general and always valid) entropy criterion, the maximum of the total entropy must be determined, i.e. the sum of the entropies of the system under investigation and its environment. Therefore, not only the change in the system entropy in the event of a change of state has to be considered, but also the entropy change that the system generates by reacting to the environment there. The enthalpy 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 isobaric conditions). When using the enthalpy criterion, the determination of the (isobaric) equilibrium state can be limited to the consideration of the system, which considerably facilitates the investigations.

If the enthalpy is expressed as a function of its natural variables , the constant pressure condition reduces the list of variables by and the enthalpy only depends on the variables . This simplification demonstrates one of the possible advantages of using enthalpy in appropriate cases. The state of equilibrium could also be determined by looking for the minimum of the internal energy of the system under consideration,, but the condition of constant pressure would in this case not lead to a simplification of the function, but to complicated dependencies between the variables . Additional problems would arise if the equation of state is not explicitly known. ${\ displaystyle (S, p, N_ {i})}$${\ displaystyle p}$${\ displaystyle (S, N_ {i})}$${\ displaystyle U (S, V, N_ {i})}$${\ displaystyle (S, V, N_ {i})}$${\ displaystyle p = p (S, V, N_ {i})}$

The atmosphere can often serve as a pressure reservoir for a real physical or chemical process. Because of its large volume, its pressure does not change significantly when a system does volume change work on it. The prerequisites for the applicability of the minimum principle of enthalpy are therefore fulfilled when a system is adiabatically isolated from the environment (to prevent heat exchange with the environment and thus to keep the entropy constant) and to the atmospheric via a movable piston or a similar device Ambient pressure is coupled (to keep the system pressure constant).

However, such adiabatic systems rarely occur in laboratory practice. Chemical reactions usually do not take place in adiabatically insulated vessels, so that heat can be exchanged with the environment. The atmosphere then serves not only as a pressure, but also as a heat reservoir : It keeps pressure and temperature constant. The thermodynamic potential that takes on a minimum under these conditions is the Gibbs energy . ${\ displaystyle G = U + p \, VT \, S}$

### Enthalpy and heat capacity

If the specific (i.e. mass-related) amount of heat is supplied to a system and thereby causes a temperature change , then the specific heat capacity of the system is defined by the equation ${\ displaystyle \ mathrm {d} q}$${\ displaystyle \ mathrm {d} T}$ ${\ displaystyle c}$

${\ displaystyle \ mathrm {d} q = c \ cdot \ mathrm {d} T}$

or

${\ displaystyle c = {\ frac {\ mathrm {d} q} {\ mathrm {d} T}}}$.

The specific heat capacity is not only dependent on the material, but also on the process management. If the heat supply is isochoric (i.e. at constant volume), then the total amount of heat supplied contributes to the increase in the specific internal energy : ${\ displaystyle \ mathrm {d} u}$

${\ displaystyle \ mathrm {d} q = \ mathrm {d} u}$

and the process is described by the specific heat capacity at constant volume, ${\ displaystyle c_ {v}}$

${\ displaystyle c_ {v} = \ left ({\ frac {\ partial u} {\ partial T}} \ right) _ {v}}$,

which is the derivative of the specific internal energy according to the temperature at constant volume.

If the heat supply is isobaric, the specific amount of heat supplied is equal to the increase in the specific enthalpy : ${\ displaystyle \ mathrm {d} h}$

${\ displaystyle \ mathrm {d} q = \ mathrm {d} h}$

and the process is described by the specific heat capacity at constant pressure, ${\ displaystyle c_ {p}}$

${\ displaystyle c_ {p} = \ left ({\ frac {\ partial h} {\ partial T}} \ right) _ {p}}$,

which is the derivative of the specific enthalpy according to the temperature at constant pressure.

Since the volume change work to be performed by the system must also be added to achieve a desired temperature increase in the isobaric case, it is to be expected that the isobaric specific heat capacity of a substance will be greater than the isochoric. A closer look must take into account that the expansion of the system usually also changes its internal energy. However, it can be shown that in general:

${\ displaystyle c_ {p} -c_ {v} \, = \, \ left ({\ frac {\ alpha ^ {2}} {\ kappa}} \ right) Tv}$.

On the right-hand side, the temperature , the specific volume , the square of the isobaric thermal expansion coefficient and (for reasons of thermodynamic stability) the isothermal compressibility cannot become negative, so that always ${\ displaystyle T}$${\ displaystyle v}$ ${\ displaystyle \ alpha}$ ${\ displaystyle \ kappa}$

${\ displaystyle c_ {p} \ geq c_ {v}}$

is. In some cases it can be 30% larger than . ${\ displaystyle c_ {p}}$${\ displaystyle c_ {v}}$

### Temperature and pressure dependence of the enthalpy

For reasons of space, tabulations of the enthalpy usually relate to a specific temperature and a specific pressure. If the enthalpy is to be determined for other conditions, formulas are desirable that allow the transition from the reference state to other temperatures and pressures. It is advantageous if only the knowledge of directly measurable quantities is required for the conversion.

For a closed system that is in equilibrium and in which no chemical reactions take place ( ), as stated above, an infinitesimal change in enthalpy is given by ${\ displaystyle \ mathrm {d} N_ {i} = 0}$

${\ displaystyle \ mathrm {d} H = T \ \ mathrm {d} S + V \ \ mathrm {d} P}$.

The differential can be - understood as a function of the variables of interest here and - developed as follows: ${\ displaystyle \ mathrm {d} S}$${\ displaystyle T}$${\ displaystyle p}$

{\ displaystyle {\ begin {aligned} \ mathrm {d} H & = T \ left \ {\ left ({\ frac {\ partial S} {\ partial T}} \ right) _ {p} \ mathrm {d} T + \ left ({\ frac {\ partial S} {\ partial p}} \ right) _ {T} \ mathrm {d} p \ right \} + V \ mathrm {d} p \\ & = T \ left ({\ frac {\ partial S} {\ partial T}} \ right) _ {p} \ mathrm {d} T + \ left \ {V + T \ left ({\ frac {\ partial S} {\ partial p }} \ right) _ {T} \ right \} \ mathrm {d} p \ end {aligned}}}

With the help of identities

${\ displaystyle T \ left ({\ frac {\ partial S} {\ partial T}} \ right) _ {p} = \ left ({\ frac {\ partial Q} {\ partial T}} \ right) _ {p} = \ left ({\ frac {\ partial H} {\ partial T}} \ right) _ {p}}$(because of the Second Law )${\ displaystyle \ mathrm {d} Q = T \ \ mathrm {d} S}$

and

${\ displaystyle \ left ({\ frac {\ partial S} {\ partial p}} \ right) _ {T} = - \ left ({\ frac {\ partial V} {\ partial T}} \ right) _ {p}}$(one of the Maxwell relationships )

follows

${\ displaystyle \ mathrm {d} H = \ left ({\ frac {\ partial H} {\ partial T}} \ right) _ {p} \ mathrm {d} T + \ left \ {VT \ left ({\ frac {\ partial V} {\ partial T}} \ right) _ {p} \ right \} \ mathrm {d} p}$,

What can be expressed using the isobaric heat capacity and the isobaric thermal expansion coefficient exclusively through the measurable quantities , and : ${\ displaystyle C_ {p} = \ left ({\ frac {\ partial H} {\ partial T}} \ right) _ {p}}$${\ displaystyle \ alpha = {\ frac {1} {V}} \ left ({\ frac {\ partial V} {\ partial T}} \ right) _ {p}}$${\ displaystyle V}$${\ displaystyle C_ {p}}$${\ displaystyle \ alpha}$

${\ displaystyle \ mathrm {d} H = C_ {p} \ \ mathrm {d} T + V (1- \ alpha \ T) \ \ mathrm {d} p}$.

For a finitely large change of state from state to state , this infinitesimal state change has to be integrated: ${\ displaystyle (T_ {1}, p_ {1})}$${\ displaystyle (T_ {2}, p_ {2})}$

${\ displaystyle H_ {2} -H_ {1} = \ int _ {T_ {1}} ^ {T_ {2}} C_ {p} \ \ mathrm {d} T + \ int _ {p_ {1}} ^ {p_ {2}} V (1- \ alpha \ T) \ \ mathrm {d} p}$.

If the change of state is associated with a phase transition, the relevant latent heat must also be taken into account.

The sizes , and are usually themselves dependent on temperature and pressure. These dependencies must be known so that the integrals can be carried out. ${\ displaystyle V}$${\ displaystyle C_ {p}}$${\ displaystyle \ alpha}$

Since the enthalpy is a state variable, the integrals do not depend on the choice of the path along which integration takes place. A convenient choice of path is to first run the temperature integral on a constant pressure path that connects the initial state with an intermediate state . To do this, the temperature range from to must be known at fixed pressure . Then the pressure integral is executed on the path of constant temperature that connects with . To do this, in the pressure range from to must be known at the fixed temperature . ${\ displaystyle (T_ {1}, p_ {1})}$${\ displaystyle (T_ {2}, p_ {1})}$${\ displaystyle C_ {p}}$${\ displaystyle T_ {1}}$${\ displaystyle T_ {2}}$${\ displaystyle p_ {1}}$${\ displaystyle (T_ {2}, p_ {1})}$${\ displaystyle (T_ {2}, p_ {2})}$${\ displaystyle V \ alpha}$${\ displaystyle p_ {1}}$${\ displaystyle p_ {2}}$${\ displaystyle T_ {2}}$

Alternatively, the integration can only be carried out on a path of constant temperature from to , which must be known in the pressure range from to at the fixed temperature . Then the path leads from to at constant pressure , which must be known in the temperature range from to at fixed pressure . ${\ displaystyle (T_ {1}, p_ {1})}$${\ displaystyle (T_ {1}, p_ {2})}$${\ displaystyle V \ alpha}$${\ displaystyle p_ {1}}$${\ displaystyle p_ {2}}$${\ displaystyle T_ {1}}$${\ displaystyle (T_ {1}, p_ {2})}$${\ displaystyle (T_ {2}, p_ {2})}$${\ displaystyle C_ {p}}$${\ displaystyle T_ {1}}$${\ displaystyle T_ {2}}$${\ displaystyle p_ {2}}$

If the process is carried out isobarically , the pressure integral is omitted and only knowledge of the temperature-dependent heat capacity at the relevant fixed pressure is necessary: ${\ displaystyle p_ {1} = p_ {2}}$${\ displaystyle C_ {p}}$

${\ displaystyle H_ {2} -H_ {1} = \ int _ {T_ {1}} ^ {T_ {2}} C_ {p} \ \ mathrm {d} T \ quad \ mathrm {(isobaric)}}$.

Wiktionary: enthalpy  - explanations of meanings, word origins, synonyms, translations

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