Enthalpy is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure. The unit of measurement for enthalpy in the International System of Units (SI) is the joule, but other historical, conventional units are still in use, such as the BTU and the calorie.

The enthalpy is the preferred expression of system energy changes in many chemical, biological, and physical measurements, because it simplifies certain descriptions of energy transfer. This is because a change in enthalpy takes account of energy transferred to the environment through the expansion of the system under study.

The total enthalpy, H, of a system cannot be measured directly. Thus, change in enthalpy, ΔH, is a more useful quantity than its absolute value. The change ΔH is positive in endothermic reactions, and negative in heat-releasing exothermic processes. ΔH of a system is equal to the sum of non-mechanical work done on it and the heat supplied to it.

For quasistatic processes under constant pressure, ΔH is equal to the change in the internal energy of the system, plus the work that the system has done on its surroundings. This means that the change in enthalpy under such conditions is the heat absorbed (or released) by a chemical reaction. Enthalpies for chemical substances at constant pressure assume standard state: most commonly 1 bar (0.1 MPa) pressure. Standard state does not, strictly speaking, specify a temperature, but expressions for enthalpy generally reference the standard enthalpy of formation at 25^oC (298.15 K).

While H denotes the total enthalpy of a system, h is specific enthalpy, i.e. the enthalpy per mole or unit mass of a substance. We will simply refer to both as enthalpy, since the context will make it clear which is applicable.

The enthalpy of formation, or heat or formation, h^o_f , is the enthalpy of a substance at the base conditions of 25^oC and 0.1 MPa pressure. The enthalpy in any other state, relative to this base, would be found by adding the change of enthalpy between 25^oC and 0.1 MPa and the given state to the enthalpy of formation. That is, the enthalpy at any temperature and pressure, h_T,P, is

where the term (Δh)_{298,0.1 MPa --> T,P} represents the difference in enthalpy between any given state and the enthalpy at 298.15 K, 0.1 MPa. For convenience we usually drop the subscripts in the examples that follow.

In most cases the substances that comprise the reactants and products in a chemical reaction are not at a temperature of 25^oC and a pressure of 0.1 MPa. Therefore the change of enthalpy between 25^oC and 0.1 MPa and the given state must be known. In the case of a solid or liquid, this change of enthalpy can usually be found from a table of thermodynamic properties or from specific heat data. In the case of gases, this change of enthalpy can be found as follows:

Assume ideal gas behavior between 25^oC, 0.1 MPa, and the given state. In this case, the enthalpy is a function of the temperature only, and can be found by use of an equation, from tabulated values of enthalpy as a function of temperature (which assumes ideal gas behavior), or calculated from specific heat capacity, C_p^o, where

Consider a given combustion process that takes place adiabatically and with no work or changes in kinetic or potential energy involved. For such a process the temperature of the products is referred to as the adiabatic flame temperature. With the assumptions of no work and no changes in kinetic or potential energy, this is the maximum temperature that can be achieved for the given reactants because any heat transfer from the reacting substances and any incomplete combustion would tend to lower the temperature of the products. For a given fuel and given pressure and temperature of the reactants, the maximum adiabatic temperature that can be achieved is with a stoichiometric mixture.

The following example shows how the adiabatic flame temperature may be found. The dissociation that takes place in the combustion products, which has a significant effect on the adiabatic flame temperature, will be considered in the next section.

Suppose we react a stoichiometric mixture of gaseous methane and gaseous oxygen at an initial temperature of 25^oC and 4 MPa pressure. Our reaction equation is

CH₄ + 2 O₂ CO₂ + 2 H₂O

We assume ideal gas behavior for all the constituents. In this case the first law of thermodynamics reduces to

_R n_i[h^o_f + Δh]_i = _P n_e[h^o_f + Δh]_e

where H_R is the enthalpy of the reactants, H_P the enthalpy of the products, h^o_f the enthalpy of formation of each constituent, Δh_i the change in enthalpy of each constituent in the reactants at the initial temperature, and Δh_e the change in enthalpy of each constituent in the products at the adiabatic flame temperature.

Below is a table giving enthalpy of formation and ideal gas enthalpy for several of the products that will result from a methane/oxygen reaction. Note that in the simplified reaction considered above, we have only CO₂ and H₂O on the right side of the equation. The additional substances will appear when we consider dissociation in the next section.

Knowing the temperature of the reactants, we can easily calculate H_R. Since the reactant temperature in our example is equal to the base temperature of 25^oC, Δh_i = 0 and we need only to sum the enthalpies of formation of the reactants. From the table above we see that for diatomic oxygen (h_f^o)₂₉₈ = 0, which is typical for diatomic gases, and for methane we find (h_f^o)₂₉₈ = -74.873 kJ/mol. Therefore, we have

H_R = _i  n_i [h^o_f + Δh]_i = 1 × [-74.873 + 0] + 2 × [0 + 0] = -74.873 kJ

For the products, we have

H_P = _e  n_e [h^o_f + Δh]_e = 1 × [-393.522 + Δh_CO2] + 2 × [-241.827 + Δh_H2O] = -74.873 kJ

By trial-and-error solution, a temperature of the products is found that satisfies this equation. Suppose we've already performed several trials and have begun to hone in on a solution. For the next trial, assume T_P = 5,400 K.

Since H_P > H_R, our assumed temperature is too high. Let's now try T_P = 5,200 K.

We can now interpolate to find the adiabatic flame temperature.

Cryogenic Reactants

From the preceding example it can be seen that the initial temperature of the reactants has an effect on the final temperature of the products. For storable propellants, temperature variations within the normal liquid range have only a small effect on engine performance. For hypergols, a 10^oC change in reactant temperature results in about 0.1% change in specific impulse. We can, therefore, safely assume that the initial temperature of storable propellants is the standard enthalpy of formation temperature of 25^oC. This assumption has the added benefit of simplifying the calculations as we need only look up the enthalpies of formations for the reactants without making any further adjustments for temperature.

Cryogenic propellants are another story, as their extremely low temperatures have a significant effect on performance. We must further take into consideration that the readily available thermochemical data for these substances is gaseous state rather than liquid. An adjustment must be made to account for the change of state.

Let's consider the example of cryogenic oxygen, which has a boiling point of -183^oC (90 K). We start with gaseous oxygen at 298 K, which has an enthalpy of formation of 0 kJ/mol. We then add the change in enthalpy required to lower the temperature from 298 K to 90 K. From the enthalpy tables we estimate that this adjustment is -6.06 kJ/mol. Next we deduct the enthalpy of vaporization to account for the change of state from gas to liquid. The enthalpy of vaporization of oxygen is 6.82 kJ/mol. Therefore, the enthalpy of liquid oxygen reactant is

We now turn our attention to chemical equilibrium. Here we will consider a chemical reaction involving only one phase, which is referred to as a homogeneous chemical reaction. In our case we will be considering a gaseous phase, but the basic considerations apply to any phase.

Consider a vessel that contains four compounds, A, B, C, and D, which are in chemical equilibrium at a given pressure and temperature. Let the number of moles of each component be designated n_A, n_B, n_C, and n_D. Further, let the chemical reaction that takes place between these four constituents be

v_AA + v_BB v_CC + v_DD

where the v's are the stoichiometric coefficients. It should be emphasized that there is a very definite relation between the v's (the stoichiometric coefficients) whereas the n's (the number of moles present) for any constituent can be varied simply by varying the amount of that component in the reaction vessel.

To determine the equilibrium composition for a chemical reaction, we must be able to determine the activity of the various constituents in the mixture. We do so by applying the concept of the equilibrium constant, K. Assuming the ideal gas model, which will be appropriate for our examples and applications, the chemical equilibrium equation can be written as

where y is the equilibrium mole fraction of the A, B, C and D constituents, v is the stoichiometric coefficient for each constituent, and the term P/P^o represents the ratio of pressure at which the reaction occurs to the standard state pressure.

The equation is written in this form because it demonstrates quite clearly the influence of various factors on the equilibrium composition (the y's). That is, we know that temperature and pressure both influence the equilibrium composition. It can be seen that the influence of temperature enters through the value of K (which is a function of temperature only) and the influence of pressure though the term (P/P^o)^{V_C+V_D-V_A-V_B}.

It should be noted that the equilibrium of combustion gases is very sensitive to temperature. Products existing at a high combustion temperature are very different from those existing at a lower combustion temperature. At high temperatures, dissociation of the products occurs, as the thermal energy causes the break up of molecules into simpler and monatomic constituents. Dissociation reactions are reversible (as indicated by the double arrow in the equations), so as the gases cool, recombination can occur.

Let's now revisit our previous example and employ a procedure for determing the equilibrium composition of the combustion products.

It is convenient to view the over-all process as though it occurred in two separate steps, a combustion process followed by a heating and dissociation of the combustion products. The combustion reaction is

CH₄ + 2 O₂ CO₂ + 2 H₂O

There are several dissociation reactions that will come into play, but we'll consider the two most dominant in this particular example.

CO₂ CO + 1/2 O₂

H₂O 1/2 H₂ + OH

That is, the energy released by the combustion of CH₄ and O₂ heats the CO₂ and H₂O to high temperature, resulting in the dissociation of part of the CO₂ to CO and O₂, and part of the H₂O to H₂ and OH. Thus, the overall reaction can be written

CH₄ + 2 O₂ a CO₂ + b CO + c O₂ + d H₂O + e H₂ + f OH

where the unknown coefficients a, b, c, d, e, and f must be found by solution of the equilibrium equations associated with the dissociation reactions.

From the combustion reaction we find that the initial compositions for the dissociation reactions are 1 mole CO₂ and 2 moles H₂O. Therefore, letting z' be the number of moles of CO₂ dissociated, and z" be the number of moles of H₂O dissociated, we find

                          CO2    CO   +   1/2 O2

     Initial:              1        0        0
     Change:               -z'      +z'      +z'/2
     _____________________________________________

     At equilibrium:     (1-z')     z'       z'/2



                          H2O    1/2 H2  +  OH

     Initial:              2        0         0
     Change:               -z"      +z"/2     +z"
     _____________________________________________

     At equilibrium:     (2-z")     z"/2      z"

Therefore the overall reaction is

CH₄ + 2 O₂ (1-z') CO₂ + z' CO + z'/2 O₂ + (2-z") H₂O + z"/2 H₂ + z" OH

And the total number of moles at equilibrium for each reaction is

The equilibrium mole fractions are

Substituting these quantities along with P = 4 MPa, we have the equilibrium equations.

Values of the equilibrium constants are found in the following table:

By trial-and-error solution, a temperature of the products must be found that satisfies the equations. The steps are

• Select a trial value for T_P.
• Obtain ln(K) values from the table and calculate values of K.
• By trial-and-error, calculate the values of z.
• Calculate the number of moles of each product.
• Calculate the total enthalpy of the products.
• Compare the product enthalpy to the reactant enthalpy.
• If H_P ≠ H_R, select a new trial value of T_P and repeat.
• When close, interpolate to find final T_P.

Let's assume T_P = 3,600 K, we have

CH₄ + 2 O₂ 0.5691 CO₂ + 0.4309 CO + 0.2154 O₂ + 1.5413 H₂O + 0.2294 H₂ + 0.4587 OH

Assume T_P = 3,800 K and repeat.

CH₄ + 2 O₂ 0.4715 CO₂ + 0.5285 CO + 0.2643 O₂ + 1.4706 H₂O + 0.2647 H₂ + 0.5294 OH

Calculate adiabatic flame temperature by interpolation.

Knowing the temperature, we can now interpolate to find the corresponding values of ln(K) and calculate the equilibrium mixture.

CH₄ + 2 O₂ 0.5186 CO₂ + 0.4814 CO + 0.2407 O₂ + 1.5051 H₂O + 0.2474 H₂ + 0.4949 OH

For a more accurate solution, there are additional dissociation reactions that have to be considered, such as

H₂O H₂ + 1/2 O₂

H₂ 2 H

O₂ 2 O

The average molecular weight, M, of a mixture is simply

where n is the total number of moles in the mixture, and m is the total mass of the mixture. The total moles and total mass are found by

n = _i n_i

m = _i (n_i × M_i)

where n_i is the number of moles of each constituent, M_i is the molecular weight of each constituent.

Referring back to the last iteration of our combustion equation, we have

CH₄ + 2 O₂ 0.5186 CO₂ + 0.4814 CO + 0.2407 O₂ + 1.5051 H₂O + 0.2474 H₂ + 0.4949 OH

Calculating the molecular weight of the mixture, we have

Alternatively, if the mole fraction, y_i, of each constituent is given, where y_i = n_i / n, the average molecular weight is found by

M = _i (y_i × M_i)

Referring to the STANJAN run above, we can see that both the mole fraction and the gas molecular weight is given, where M = 22.519 kg/kmol. This is lower than our derived unit because our example failed to take into consideration all the applicable dissociations.

The constant-pressure specific heat, or heat capacity, C_p, and the constant-volume specific heat, C_v, are useful functions for thermodynamic calculations, particularly for gases, and are defined by the following relationships

where h is specific enthalpy, u is specific internal energy, and T is temperature.

A very important relationship between the constant-pressure and constant-volume specific heats of an ideal gas is found in the equation

where C_p and C_v are on a mole basis, and R' is the universal gas constant (8.31446 J/mol-K). This tells us that the difference between the constant-pressure and constant-volume specific heats of an ideal gas is always constant, though both are a function of temperature.

Many thermodynamic solutions use the ratio of the specific heats, k or , defined as

Since it is common for thermodynamic references to provide the value of C_p only, from which C_v is calculated, we can combine equations and write

Values of constant-pressure specific heat are typically found by use of an equation, or from tabulated values of C_p as a function of temperature. The Shomate equation is commonly used, which is shown below along with the Shomate coefficients of the products found in our combustion example.

Calculating the specific heat ratio of a single substance is straightforward. Say we have carbon monoxide at 3,000 K, the specific heat ratio is

When we have a mixture of gases, we must determine the specific heat of the mixture, where

C_pgas = 1 / n × _i (n_i × Cp_i)   , or

C_pgas = _i (y_i × Cp_i)

Let's calculate the specific heat ratio for the mixture of gases found in the STANJAN solution to our methane/oxygen problem, noting that the temperature is 3,553.87 K.

C_pgas = _i (y_i × Cp_i) = 49.0786 J/mol-K

Although some computer programs calculate the value of specific heat ratio, STANJAN does not. If STANJAN is used, the user will have to calculate k separately, though this can easily be done through the use of spreadsheet software.

Everything so far has assumed the combustion products consist of a homogenous mixture of gases. When dealing with liquid rocket propellants, this is a fair assumption, as any condensed species (either liquids or solids) will appear in trace quantities only. However, when dealing with solid propellants, a significant portion of the exhaust products will be in a condensed phase, which is evident as visible smoke in the exhaust plume. When the products exist in more than one phase, the mixture is said to be heterogeneous.

Here we present two important equations that relate the pressure, temperature, and volume that a gas occupies during reversible compression or expansion. Such a process occurs during the compression and power strokes for an internal combustion engine. The same equations describe the conditions across the compressor and turbine of a gas turbine, or across the nozzle of a rocket engine. The resulting compression and expansion are reversible processes in which the entropy of the system remains constant. Such a process is called an isentropic process.

For isentropic compression and expansion, the effect of pressure on temperature is described by

and the effect of volume on pressure is

where T₁, p₁ and v₁ are the initial temperature, pressure and volume; T₂, p₂ and v₂ are the final temperature, pressure and volume; and k is the specific heat ratio.

We can also combine equations to find the effect of volume on temperature

Since the expansion of exhaust gases through a rocket nozzle is an isentropic process, the above equations are applicable. Say, for instance, we want to know the temperature at the nozzle exit for a rocket engine having a chamber temperature of 3,250 K, a chamber pressure of 6 MPa, a nozzle exit pressure of 0.07 MPa, and k is known to be 1.23. We have