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Chapter 8 and 9 – Energy Balances
Reference States. Recall that enthalpy and internal energy are always defined
relative to a reference state (Chapter 7). When solving energy balance problems, it
is therefore necessary to define a reference state for each chemical species in the
energy balance (the reference state may be predefined if a tabulated set of data is
used such as the steam tables).
Example. Suppose water vapor at 300 oC and 5 bar is chosen as a reference state at
which Ĥ is defined to be zero. Relative to this state, what is the specific enthalpy of
liquid water at 75 oC and 1 bar? What is the specific internal energy of liquid water
at 75 oC and 1 bar? (Use Table B. 7).
Calculating changes in enthalpy and internal energy. Ĥ and Û are state
functions, meaning that their values only depend on the state of the system, and
not on the path taken to arrive at that state.
IMPORTANT: Given a state A (as characterized by a set of variables such as
pressure, temperature, composition) and a state B, the change in enthalpy of the
system as it passes from A to B can be calculated along any path that leads from A
to B, whether or not the path is the one actually followed.
Example. 18 g of liquid water freezes to 18 g of ice while the temperature is held
constant at 0 oC and the pressure is held constant at 1 atm. The enthalpy change for
the process is measured to be ∆ Ĥ = - 6.01 kJ.
What would the ∆ Ĥ for the process be if, instead, the 18 g of water is first heated
from 0 oC to 100 oC and entirely vaporized to steam at 100 oC and 1 atm, then
liquified by compression from 1 atm to 10 atm at 100 oC, than cooled to - 200 oC
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(during which step it freezes to ice) while being decompressed to a pressure of 1
atm, and finally thus formed 18 g of ice is heated from -200 oC and 1 atm to ice at
0 oC and 1 atm?
Types of Paths. There are five types of paths for which we will learn to calculate
enthalpy changes ∆ Ĥ :
1). Changes in pressure (p) at constant temperature (T) and state of aggregation
(i.e. no phase changes).
2). Changes in T at constant p and state of aggregation.
3). Phase changes (i.e. melting, condensation, evaporation, solidification,
sublimation) at constant T and p.
4). Mixing steps (two liquids, gas in a liquid, solid in a liquid) at constant T and p.
5). Chemical reactions taking place at constant T and p.
The overall path from a state A to a state B will be able to be expressed as a
combination of the above five types of steps. Because enthalpy is a state function,
the total change ∆ Ĥ for passing from state A to state B can be calculated as the sum
of the enthalpy changes ∆ Ĥ j for the individual steps,
∆ Ĥ = ∆ Ĥ 1 + ∆ Ĥ 2 + ∆ Ĥ 3 + ∆ Ĥ 4…
j = 1, 2, …k
(1)
where k is the total number of steps used, for purposes of the calculation, to take
the system from the initial state A to the final state B. Note that the steps used for
the calculation do not need to correspond to the actual path taken by the system
from A to B.
Example. Construct a process path, consisting of the types of steps listed above,
that could be used to calculate the enthalpy change for the following processes.
(i) Cyclohexane vapor at 180 oC and 5 atm is cooled and condensed to liquid
cyclohexane at 25 oC and 5 atm. The enthalpy change for the condensation of
cyclohexane at 80.7 oC and 1 atm is known.
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(ii) O2 at 170 oC and 1 atm and CH4 at 25 oC and 1 atm are mixed and react
completely to form CO2 and H2O at 300 oC and 1 atm. The enthalpy change for the
reaction occurring at 25 oC and 1 atm is known.
Paths of type 1
changes in pressure (p) at constant temperature (T) and state of aggregation
Liquids and solids: Û and Vˆ are only weak functions of p for liquids and solids.
Thus, for a path of type 1 performed on a system that is a liquid or a solid, ∆ Û ≈ 0
and ∆ Vˆ ≈ 0. This in turn implies that
∆ Ĥ = ∆ Û + ∆(p Vˆ ) ≈ Vˆ ∆p
(solids or liquids, change in p only)
(2)
Gases: for gases that can be modeled as ideal, internal energy and enthalpy only
depend on T. Thus, for a type 1 path, for an ideal gas ∆ Û and ∆ Ĥ are both zero
since T is not changing. Note that Ĥ = Û + p Vˆ = Û + RT/M for an ideal gas, where
M is the mass of 1 mole of the gas. This expression assumed that specific
quantities are expressed per unit mass of substance. If specific quantities are
expressed per mole of substance, then, for an ideal gas, Ĥ = Û + p Vˆ = Û + RT. For
gases behaving non-ideally, either tables of Û and Ĥ (e.g. like the steam tables)
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must be used or more complex calculations (not covered in this course) are
required.
Example. Given the process
C2H6 (g, 25 oC, 1 atm) C2H6 (g, 25 oC, 30 atm)
How would you use the compressibility charts to determine whether it is
reasonable to assume ∆ Ĥ = 0?
Paths of type 2
changes in T at constant p and state of aggregation
Sensible heat: Sensible heat is heat added to a system that results in a change in
temperature (as opposed to latent heat, see below).
Internal energy changes. The rate of change of specific internal energy of a
substance with T, while its volume V is kept fixed (i.e. no PV work), is called the
heat capacity at constant volume and is given the symbol CV,
 ∂Uˆ
CV (T ) = 
 ∂T



V
(3)
Note that CV is in general a function of T. Given the function CV(T), the change in
specific internal energy as the system temperature is raised from T1 to T2 at
constant volume is
T2
∆Uˆ = ∫ CV (T ) dT
(constant V)
(4)
T1
However, a type 2 path is in general not a constant volume process. Rather, as T
changes, in order to maintain V strictly constant one would have to adjust p as well,
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violating the requirement p = constant for a type 2 path. How can we resolve this
difficulty?
Fortunately, the volume of liquids and solids in general does not change very much
with T. Therefore, it is usually a very good approximation to assume that, for a
type 2 path, one can still use
T2
∆Uˆ ≈ ∫ CV (T )dT
(solids and liquids, type 2 path)
(5)
T1
Moreover, for ideal gases, since Û is a function of temperature only, equation (4) is
exact even though the volume will change for a type 2 path,
T2
∆Uˆ = ∫ CV (T ) dT
(ideal gases, type 2 path)
(4)
T1
For gases behaving nonideally, equation (4) is exact only if V is maintained
constant during the temperature change, what in general will not be true for a type
2 path. Since V will in general change during the performance of a type 2 path with
a nonideal gas, one has to resort to more complex calculations (not covered in this
course) or use tabulated information (e.g. steam tables).
Enthalpy changes. The rate of change of specific enthalpy Ĥ of a substance with
T, while the pressure p on the substance is kept fixed (a type 2 process), is called
the heat capacity at constant pressure and is given the symbol Cp,
 ∂Hˆ 

C p (T ) = 

∂
T

p
(6)
Cp is in general a function of T. Given the function Cp(T), the change in enthalpy of
a substance as its temperature is raised from T1 to T2 at constant pressure is
T2
∆Hˆ = ∫ C p (T )dT
T1
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(type 2 process)
(7)
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Evaluation of heat capacities (also known as "specific heats"). Cp and CV have
units of energy per amount per temperature interval, where the amount of material
may be measured in molar or mass units (e.g. units of heat capacity could be J/(kg
o
C), J/(mol oC), etc). Table B.2 provides polynomial expressions for heat capacities
Cp
Cp = a + bT + cT2 + dT3
(8)
where a, b, c and d are constants. Most often we will use these expressions.
NOTE: in some cases Cp may be available at equally spaced T intervals (e.g. from
a table), rather than in the form of equation 8. In those instances, equation 7 can be
numerically integrated with respect to T to obtain ∆ Ĥ , for example using
Simpson's Rule (see Appendix A.3).
Kopp's Rule. Kopp's rule is used to roughly estimate Cp values for materials based
on their atomic composition. See text for discussion.
Cp values for mixtures. When dealing with heat capacities of mixtures, the most
accurate method is to look up Cp values from tables or other experimental
information, if available. Lacking such data, if one is willing to neglect corrections
to Cp that derive from mixing of the different components of the mixture, the
overall heat capacity of the mixture Cp,mix can be approximated as the sum of heat
capacity contributions from the separate components of the mixture,
Cp,mix(T) =
∑x C
i
all
mixture
components
p ,i
(T )
(9)
In equation 9, the index i ranges over all of the components of the mixture. xi is the
mass fraction (if using heat capacities expressed per mass of material) or mole
fraction (if using heat capacities expressed per mole of material) of component i in
the mixture, and Cp,i is the heat capacity of species i in its pure form. The enthalpy
change ∆ Ĥ for a type 2 path performed on a mixture is then calculated by inserting
the expression for Cp,mix from equation 9 into equation 7.
Example. For ideal gases, Cp = CV + R. How can one prove this expression?
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Example. For liquids and solids, Cp ≈ CV. How can we justify this approximation?
Example 8.3-2. Assuming ideal gas behavior, calculate the heat transferred in the
following situations:
1). A stream of nitrogen flowing at 100 mol/min is heated from 20 oC to 100 oC.
2). Nitrogen in a 5 L flask, at an initial pressure of 3 bar, is cooled from 90 oC to 30
o
C.
For nitrogen at a constant pressure of 1 atm, the heat capacity Cp is
Cp(kJ/(mol oC)) = 0.02900 + 0.2199×10-5T + 0.5723×10-8T2 - 2.871×10-12T3
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Example 8.3-5. A stream containing 10 % CH4 and 90 % air by volume is to be
heated from 20 oC to 300 oC. Calculate the required rate of heat input in kilowatts
if the flow rate of the gas is 2.00 × 103 liters (STP)/min. Assume ideal gas
behavior. Table B.8 provides specific enthalpy of air, yielding Ĥ air,in = -0.15
kJ/mol and Ĥ air,out = 8.17 kJ/mol. The heat capacity of methane is given by
Cp (kJ/(mol oC)) = 0.03431 + 5.469 × 10-5T + 0.3661 × 10-8T2 - 11.0 × 10-12T3
Also, in solving this problem, do you think we can neglect the enthalpy of mixing
of methane and air? Why yes or why not?
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Example 8.3-6. A gas stream containing 8.0 mole % CO and 92.0 mole% CO2 at
500 oC is fed to a waste heat boiler. In the boiler, heat is transferred from this gas
stream to a water stream, which is fed to the boiler at 25 oC in a ratio of 0.200 mole
feed water per mole of hot gas. The water is heated by the gas so that it forms
saturated steam at 5.0 bar. The boiler operates adiabatically, with all heat lost by
the gas stream going to heat the water stream.
What is the temperature of the exiting gas?
The heat capacities for CO and CO2 are:
Cp,CO = 0.02895 + 0.4110×10-5T + 0.3548×10-8T2 - 2.220×10-12T3
Cp,CO2 = 0.03611 + 4.223×10-5T - 2.887×10-8T2 + 7.464×10-12T3
Outline the solution to this problem by setting up all the necessary equations,
including equations from which values of needed enthalpies are evaluated.
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Paths of type 3
phase changes at constant T and p
Latent heat: Latent heat is the enthalpy change that accompanies a change in
phase of a substance at a constant T and p. Latent heat of vaporization ∆ Ĥ v (which
is the negative of the latent heat of condensation) refers to the enthalpy change per
unit amount of liquid when that liquid is vaporized (units: energy/amount, where
amount could be given in a mass or a molar basis). Similarly, latent heat of melting
∆ Ĥ m (equivalently, heat of fusion, which is also the negative of the heat of
solidification) refers to the amount of heat (change in enthalpy) that must be added
to a unit amount of material to cause it to melt. Contributions to latent heats
include changes in molecular level interactions as well as changes in specific
volume that accompany the phase change.
Latent heats are, in general, functions of p and T. However, they depend much
more strongly on T than on p. Therefore, when calculating heat associated with a
change of phase, it is important to ensure that the latent heat value used for the
calculation is that for the actual T at which the phase transformation occurs (e.g.
don't assume that ∆ Ĥ v at 30 oC is the same as at 100 oC).
Example. Imagine that a liquid is to be vaporized at 130 oC, but that ∆ Ĥ v is only
known at 90 oC. How could you calculate ∆ Ĥ v (130 oC) from ∆ Ĥ v (90 oC)? What
additional information would you need to perform this calculation?
Obtaining values of latent heats. Latent heats are often available from a table
(e.g. Table B.1 and the steam tables). However, they can also be estimated or
calculated in various ways (one was already outlined above).
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Estimation of ∆ Ĥ v . For a very rough approximation of ∆ Ĥ v at the boiling point of a
liquid at 1 atm (i.e. at the normal boiling point), one can use Trouton's Rule,
∆ Ĥ v (kJ/mol) ≈ 0.088 Tb(K)
∆ Ĥ v (kJ/mol) ≈ 0.109 Tb(K)
(nonpolar liquids)
(water, low molar weight alcohols)
(10)
(11)
where Tb is the boiling point of the liquid in degrees Kelvin at 1 atm. A more
accurate estimate of ∆ Ĥ v at the normal boiling point is provided by Chen's
equation,
Tb [0.0331(Tb / Tc ) − 0.0327 + 0.0297 log10 Pc ]
∆ Ĥ v (kJ/mol) ≈
1.07 − (Tb / Tc )
(12)
where Tc and Pc are the critical temperature (in oK) and pressure (in atm) of the
liquid.
Also, as discussed in Chapter 6, ∆ Ĥ v values can be obtained from vapor
pressure-temperature data of the liquid, i.e. from p*(T). If ∆ Ĥ v does not vary
appreciably with T and the gas behaves ideally, then one can use the ClausiusClapeyron equation for this purpose. When ∆ Ĥ v depends on T (the usual case), the
Clapeyron equation needs to be used instead. If ideal gas behavior applies, the
Clapeyron equation can be expressed as (see Chapter 6 notes for a full derivation)
∆Hˆ v
d ln p *
=−
d (1 / T )
R
(13)
Finally, Watson's correlation can be used to estimate ∆ Ĥ v at one temperature
from its known value at another temperature,
 Tc − T2
∆ Ĥ v (T2) = ∆ Ĥ v (T1) 
 Tc − T1



0.38
(14)
Estimation of ∆ Ĥ m . ∆ Ĥ m can be roughly estimated from
∆ Ĥ m (kJ/mol) ≈ 0.0092Tm(K)
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(metallic elements)
(15)
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∆ Ĥ m (kJ/mol) ≈ 0.0025Tm(K)
∆ Ĥ m (kJ/mol) ≈ 0.050Tm(K)
(inorganic compounds)
(organic compounds)
(16)
(17)
Psychrometric charts. These charts provide values of many properties of humid
air (i.e. air containing water vapor), a system that is very common in
humidification and other processes. The properties include specific enthalpy,
specific volume, absolute humidity, dry-bulb and wet-bulb temperatures, and dew
point data. For the present purposes, we are most interested in values of specific
enthalpy. For a temperature of interest, the psychrometric chart allows one to look
up the enthalpy Ĥ sat of air saturated with water vapor (in units of energy per
amount of dry air). Moreover, lines of "enthalpy deviation" Ĥ dev, plotted on the
chart, can be used to calculate enthalpy of air that is not fully saturated with water
vapor. The enthalpy Ĥ of air that is not saturated with water vapor is obtained by
summing the saturation enthalpy and the enthalpy deviation, Ĥ = Ĥ sat + Ĥ dev.
Enthalpies obtained from psychrometric charts can be used in energy balance
calculations that involve condensation or evaporation of water from air; for
example, in humidification processes.
Example. Given the accompanying psychrometric chart, what is the enthalpy of air
at 25 oC and 10 % relative humidity (kJ per kg dry air)?
Example 8.4-4. An equimolar mixture of benzene (B) and toluene (T) is fed to a
heater in which the temperature is raised from 10 oC to 50 oC. The liquid product is
40 mole % B, and the vapor product is 68.4 mole % B. How much heat is needed
per g-mole of feed?
In choosing reference states for this problem, do we need to specify states of
aggregation (phase)?
How do we handle the fact that no pressure information is given?
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Outline the solution to this problem by setting up all the necessary equations,
including equations from which values of the enthalpies needed in the energy
balance can be found.
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Paths of type 4
mixing operations at constant T and p
Ideal mixtures. For an ideal mixture, the enthalpy of mixing is zero. In this case, if
the mixing itself does not contribute to a change in enthalpy, the specific enthalpy
of the mixture Ĥ mix is simply equal to the sum of the enthalpies of the mixture
components,
Ĥ mix =
∑ x Hˆ
i
all
mixture
components
(ideal mixtures)
i
(18)
where xi is the fraction of component i in the mixture, and Ĥ i is the specific
enthalpy of pure component i. If the specific enthalpy is expressed per unit mass,
then xi is the mass fraction; if instead Ĥ i is expressed per mole, then xi is the mole
fraction. The assumption that a mixture behaves ideally nearly always works well
for mixtures of gases, and it also works well for liquid mixtures when the species
being mixed are chemically similar (such as two aromatic species, or two linear
hydrocarbons).
Nonideal mixtures. Generally, mixtures are not ideal. This is because the mixing
of two, or more, species alters the molecular interactions experienced by the
molecules being mixed; thus, the internal energy and hence enthalpy will change.
There may also be a change in volume when two species are mixed. In this case,
the final volume of the mixture V does not equal the sum of the volumes of the
components that were mixed together. Therefore, in general, the enthalpy of
mixing for a type 4 process (mixing at constant T and p) for a nonideal system can
be expressed as
∆ Ĥ mix = ∆ Û mix+ ∆(p Vˆ )mix = ∆ Û mix+ p∆ Vˆ mix
(19)
where ∆ Û mix and ∆ Vˆ mix are the changes in specific internal energy and specific
volume associated with the mixing process. ∆ Û mix is the internal energy (per
amount of the mixture) minus the internal energy of the species that were mixed,
∆ Û mix = Û mixture -
19
∑ x Uˆ
i
all
mixture
components
i
(20)
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where Û i is the specific internal energy for pure species i, whose fraction (mass or
mole fraction, depending on units being used) in the mixture is xi. Similarly,
∆ Vˆ mix = Vˆ mixture -
∑ x Vˆ
i i
all
mixture
components
(21)
where Vˆ i is the specific volume of pure species i.
Enthalpies of mixing are often expressed in terms of heat of solution ∆ Hˆ s (T , r ) .
∆ Hˆ s (T , r ) is the change in enthalpy that results from dissolving one mole of solute
in r moles of liquid solvent at constant T. In the limit when 1 mole of solute is
dissolved in an infinitely large amount of solvent, ∆ Hˆ s (T , r ) approaches a limiting
value known as the heat of solution at infinite dilution. Values of ∆ Hˆ s (T , r ) are
usually found from tables, such as Table B.11 in our text. Note that the values of
∆ Hˆ s (T , r ) are expressed per mole of solute, not per mole of solution.
Enthalpy-Concentration ( Ĥ -x) Charts. These charts plot the specific enthalpy of
a binary solution (i.e. a single solute in a solvent) as a function of the solute
concentration. Several curves, corresponding to enthalpy of mixing for different
temperatures, can be drawn on the same chart. The enthalpy values are measured
relative to reference states that typically are the pure solute and the pure solvent;
i.e. the plotted enthalpies of mixing are for a process that starts with the pure
solvent and the pure solute species at the mixing temperature T, and makes them
into a solution of the given solute concentration.
Example. What is ∆ Hˆ s (T , r ) for a solution of HCl in water at 25 oC and for r = 10,
if the reference state for HCl is pure HCl gas at 25 oC and if the reference state for
water is pure water liquid at 25 oC? Use Table B.11 to come up with the answer.
What is ∆ Hˆ s (T , r ) for a solution of HCl in water at 25 oC and for r = 10, if the
reference state for HCl is highly dilute HCl in water at 25 oC and if the reference
state for water is pure water liquid at 25 oC? Use Table B.11 to come up with the
answer.
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Example 8.5-1. Hydrochloric acid is produced by absorbing gaseous HCl in water.
Calculate the heat that must be transferred to or from an absorption process unit if
HCl gas at 100 oC and water liquid at 25 oC are mixed to produce 1000 kg/h of
20.0 wt % HCl solution in water at 40 oC. The heat capacity Cp of the aqueous HCl
solution product can be taken to be constant at 0.557 kJ/(mol HCl oC).
Outline the solution to this problem by setting up all the necessary equations,
including equations from which values of the enthalpies needed in the energy
balance can be found.
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Paths of type 5
chemical reactions at constant T and p
Heat of reaction (Enthalpy of reaction). The heat of reaction, ∆ Hˆ r (T , p) , is the
enthalpy change for a process in which stoichiometric ratios of reactants at a given
T and p are completely consumed and converted to products at the same T and p.
Note that this process starts with reactants and no products, and ends with all
products (as all reactants are consumed). An exothermic reaction is one for which
∆ Hˆ r (T , p) < 0; that is, exothermic reactions carried out at constant T and p release
heat. An endothermic reaction is one for which ∆ Hˆ r (T , p) > 0; thus, heat must be
input into the system in order to sustain an endothermic reaction at constant T and
p. The heat of reaction is always reported per stoichiometric quantity of a reactant
consumed or product formed. For example, for the reaction in which species A and
B react to form C according to
2A + 3B C
∆ Hˆ r (T , p) is the heat released per 2 moles of A consumed, which is the same as the
heat released per 3 moles of B consumed, which is the same as the heat released
per one mole of C produced. If we write the reaction instead as
4A + 6B 2C
then ∆ Hˆ r (T , p) is the heat released per 4 moles of A consumed, which is the same
as the heat released per 6 moles of B consumed, which is the same as the heat
released per 2 moles of C produced. In this case, ∆ Hˆ r (T , p) would be twice as large
in magnitude than when we wrote the reaction as 2A + 3B C.
If the reaction has proceeded to a point when the extent of reaction is ξ, the total
amount of heat released or consumed so far, i.e. the total enthalpy change ∆H
realized due to the reaction, is calculated from
∆H = ξ∆ Hˆ r (T , p)
(22)
Note that, as always, units must be consistent. Thus if units of ∆ Hˆ r (T , p) are
energy/mole, ξ must be in units of moles; if instead ∆ Hˆ r (T , p) is given in units of
energy/mass, then ξ must be in units of mass.
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Comments:
1). The heat of reaction ∆ Hˆ r (T , p) often does not depend strongly on pressure, if
pressures are not too high. Why is this so? Recall the definition
∆ Ĥ r = ∆ Û r+ ∆(p Vˆ )r = ∆ Û r+ p∆ Vˆ r
(23)
where the last equality on the right follows because p is kept constant for a type 5
path. Here, ∆ Û r and ∆ Vˆ r are the changes in specific internal energy and specific
volume of the system when stoichiometric amounts of reactants are fully converted
to the products, at same T and p. How do these quantities approximately depend on
the system pressure p?
2). The value of ∆ Hˆ r (T , p) depends on the states of aggregation of the reactants
and products. For example, for the two reactions
A(g) + B(g) C(g);
∆ Ĥ r,1
A(l) + B(g) C(g)
∆ Ĥ r,2
and
the value of ∆ Ĥ r for the second expression would be increased by ∆ Ĥ v,A, the
latent heat of vaporization of A; that is, ∆ Ĥ r,2 = ∆ Ĥ r,1 + ∆ Ĥ v,A.
3). The standard heat of reaction, written ∆ Hˆ r0 , is the value of the heat of
reaction when both the reactants and the products are in their reference states. The
textbook uses reference temperature and pressure of 25 oC and 1 atm.
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CBE2124, Levicky
Example 9.1-1. The standard heat of the combustion reaction of n-butane vapor is
C4H10(g) + 13/2 O2(g) 4 CO2(g) + 5 H2O(l);
∆ Hˆ r0 = -2878 kJ/mol
a). Calculate the rate of enthalpy change, ∆ H& (kJ/s), if 2400 mol/s of CO2 is
produced, and the reaction is carried out at 25 oC.
b). Calculate the standard heat of the reaction
2C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l)
Calculate the rate of enthalpy change, ∆ H& (kJ/s), if 2400 mol/s of CO2 is produced
in this reaction, and the reaction is carried out at 25 oC.
c). The heats of vaporization of n-butane and water at 25 oC are 19.2 kJ/mol and
44.0 kJ/mol, respectively. What is the standard heat of the reaction
C4H10(l) + 13/2 O2(g) 4 CO2(g) + 5 H2O(v)
Calculate the rate of enthalpy change, ∆ H& (kJ/s), if 2400 mol/s of CO2 is produced
in this reaction, and the reaction is carried out at 25 oC.
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Calculation of heats of reaction from Hess's Law. Hess's Law states: If the
stoichiometric equation of reaction #1 can be expressed through a set of
summation (or difference) operations on the stoichiometric equations of
independent reactions #2..k, then the enthalpy change for reaction #1 can be
calculated by applying the same set of operations to the enthalpy changes of the
#2..k reactions. Note that all reactions are understood to be carried out at the same
T and p. Hess's Law is just another manifestation of the state function nature of
enthalpy.
Example. Given
C + O2 CO2;
CO + 1/2O2 CO2;
∆ Hˆ r0 = -393.51 kJ/mol
∆ Hˆ r0 = -282.99 kJ/mol
what is the heat of reaction for
C + 1/2O2 CO ?
Calculation of heats of reaction from standard heats of formation. The
standard heat of formation ∆ Hˆ 0f of a chemical species is the enthalpy change
associated with the reaction in which 1 mole of the species is formed from its
elemental constituents as they are normally found in nature (C(s), O2(g), N2(g),
H2(g) are the most important elemental constituents). The standard conditions for
the formation reaction are most often taken to be 25 oC and 1 atm. For example,
the formation reaction for ammonium nitrate would be written
N2(g) + 2H2(g) + 3/2 O2(g) NH4NO3(s)
∆ Hˆ 0f = -365.14 kJ/mol
The standard heat of formation of elemental species, in their naturally occurring
state, is zero (can you see why?). Important: If ∆ Hˆ 0f is known for all reactants and
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products that participate in a reaction, then the standard heat for that reaction can
be calculated from the known standard heats of formation using Hess's Law.
Example 9.3-1. Determine the standard heat of reaction for
C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(l)
given the standard heats of formation (Table B.1)
∆ Hˆ 0f (C5H12(l)) = -173.0 kJ/mol
∆ Hˆ 0f (CO2(g)) = -393.5 kJ/mol
∆ Hˆ 0f (H2O(l)) = -285.84 kJ/mol
Heats of combustion. The standard heat of combustion ∆ Hˆ c0 of a substance is
the heat of reaction when 1 mole of the substance is fully reacted with O2(g) to
yield the products CO2(g) and H2O(l), with the combustion reaction carried out at
25 oC and 1 atm. Standard heats of combustion can be used to calculate ∆ Hˆ r0 for
reactions involving only combustible substances and combustion products. This is
another application of Hess's Law.
The above concludes discussion of how to calculate enthalpy changes associated
with the five types of subprocesses considered in this course. The remaining pages
consider energy balances on reactive processes.
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Energy Balances on Reactive Processes
Our text covers two methods that may be used to account for the occurrence
of a chemical reaction in a process: (i) heat of reaction method and (ii) heat of
formation method. The heat of reaction approach is more straightforward if there is
only a single reaction whose ∆ Hˆ r0 is known. Problems that have multiple reactions
occurring simultaneously, and for which the ∆ Hˆ r0 may not be known, are better
treated with the heat of formation method. Below we outline the main differences
between the two methods.
Selection of Reference Conditions.
Heat of reaction method. For all reactive species, reactants and products, the most
convenient choice of reference conditions is that at which the reaction is imagined
to be carried out; i.e. usually that T and p at which the heat of reaction is known.
Typically, this would be standard conditions (25 oC, 1 atm) and a state of
aggregation for each reactive species as specified by the reaction equation.
Heat of formation method. The reference conditions refer to the elemental species
found in the reactants and products, at standard conditions (25 oC, 1 atm) and as
naturally occurring at these conditions.
Reference conditions for nonreactive species. Nonreactive species are those that do
not participate in chemical reactions. The best reference conditions for nonreactive
species, regardless of how the reactive species are treated, are usually the
conditions for one of the process streams (enthalpy of the species in that stream
then becomes zero) or, if a table is used to look up enthalpy, the reference state
conditions used by the table.
Example (text section 9.5). The below "propane" problem will be used to illustrate
the heat of reaction and heat of formation methods. Propane is combusted to
CO2(g) and H2O(v) as shown in the diagram.
Also given is the standard enthalpy change for the reaction
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C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
∆ Hˆ r0 = -2220 kJ/mol
Question: What reference conditions would one choose for
(i) the heat of reaction method
and (ii) the heat of formation method?
Enthalpy change associated with a reaction. For the usual case of a continuous
process, the quantity of interest is ∆ H& reactor, the rate of enthalpy change due to the
occurrence of the reaction. This rate is the difference between the enthalpy outflow
in the product stream at the outlet conditions from the reactor and the enthalpy
inflow with the reactants at the inlet conditions. The approach to calculation of
∆ H& reactor uses different paths for the heat of reaction and the heat of formation
methods.
Heat of reaction method. The path used to calculate the total enthalpy change
associated with the reaction is to (1) bring all reactants from inlet stream conditions
to the given (standard) conditions for the heat of reaction ∆ Hˆ r0 , (2) carry out the
reaction under the conditions given for ∆ Hˆ r0 , and (3) bring the products from the
standard conditions to the actual conditions in the outlet stream; i.e. the conditions
at which they leave the reactor. Therefore, the equation for ∆ H& reactor is
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∆ H& reactor =
∑ n& ∆Hˆ
j
reactants
j
+
∑ ξ& ∆Hˆ
j
reactions
0
r, j
+
∑ n& ∆Hˆ
j
j
(24)
products
What do the terms in equation 24 represent? ∆ H& reactor is the rate of enthalpy change
associated with the chemical reaction, in units of energy/time. Equation 24 states
that this rate can be thought to consist of three contributions, one for each of the
three subpaths mentioned above. The first term on the right is the part contributed
by the subpath in which the reactants are taken from their inlet conditions to the
conditions for which ∆ Hˆ r0 is given. In this term, ∆ Ĥ j is the enthalpy change
associated with bringing one mole of reactant j from the inlet conditions to the
conditions for which the heat of reaction is given. n& j is the molar flowrate of
reactant j, so the product n& j ∆ Ĥ j is the rate of enthalpy change that results from
taking species j from the inlet conditions to the reaction conditions at a rate of
n& j moles/time. How would we calculate ∆ Ĥ j for each of the reactants in the
propane example?
The second term on the right of equation 24 is the rate of enthalpy change
due to carrying out the chemical reactions at the standard conditions for which the
∆ Hˆ r0, j are given. It corresponds to the second subpath mentioned above. The
summation allows for the possibility of multiple reactions taking place in the
reactor simultaneously. Here, ∆ Hˆ r0, j is the standard enthalpy of reaction j in units of
energy/mole. ξ& j is the rate of progress being experienced by reaction j, in units of
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moles/time (refer to Chapter 4 for an introduction to ξ& j ). How would we calculate
the
∑ ξ& ∆Hˆ
j
0
r, j
term for the example problem?
reactions
The third term on the right of equation 24 is analogous to the first term, except now
the summation extends over the product species and the enthalpy changes
correspond to taking product species from the standard conditions of the reaction
to the actual conditions of the outlet stream. This term corresponds to the 3rd
subpath above. How would we calculate ∆ Ĥ j for each of the products in the
example problem?
Heat of formation method. This method is usually followed when ∆ Hˆ r0 is not given
or when multiple reactions occur simultaneously. The main goal, to calculate ∆ H& ,
does not change; that is, we again need to calculate the difference between the
enthalpy outflow in the product stream at the reactor outlet conditions and the
enthalpy inflow with the reactants at the inlet conditions. In the heat of formation
method, this difference is found by using a path that consists of the following
steps: (1) bringing of the reactants from the inlet stream conditions to 25 oC and 1
atm, (2) decomposition of the reactants into elemental species at 25 oC and 1 atm
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(note that these conditions of T and p are those used to define standard heats of
formation, ∆ Hˆ 0f ), (3) formation of the products from the elemental species at 25 oC
and 1 atm, and (4) bringing of the products from 25 oC and 1 atm to the outlet
stream conditions. The sum of these enthalpy changes gives the rate of enthalpy
change ∆ H& reactor,
∆ H& reactor =
∑ n&
reactants
j
(∆Hˆ j − ∆Hˆ 0f , j ) +
∑ n&
j
(∆Hˆ 0f , j + ∆Hˆ j )
(25)
products
In equation 25, the ∆ Ĥ j term in the first summation on the right corresponds to the
change in specific enthalpy of reactant j when it is brought from the inlet (process)
conditions to 25 oC and 1 atm; in other words, this term corresponds to step 1 of
the path specified above. The -∆ Hˆ 0f , j in the first summation corresponds to
decomposition of reactant j into elemental species at 25 oC and 1 atm; this is step 2
of the path. The negative sign is present because we are decomposing the reactant,
not forming it. The ∆ Hˆ 0f , j term in the second summation on the right corresponds
to formation of product j from the elemental species at 25 oC and 1 atm, which is
the third step of the path. Finally, the ∆ Ĥ j term in the second summation
represents bringing of product j from 25 oC and 1 atm to the outlet (process) stream
conditions; this is the 4th step of the overall path. Each specific enthalpy term is
multiplied by the molar flowrate of that species j in the inlet (reactants) or outlet
(products) streams to yield the corresponding rate of enthalpy change.
The ∆ Ĥ j terms for the products and reactants are calculated in the same
manner as in the heat of reaction method. For each reactant and product, they may
include changes in T, changes in p, and changes in the state of aggregation. The
heat of formation terms are obtained from Table B.1. In the propane example, how
0
would we calculate ∆Hˆ j − ∆Hˆ f , j for the propane reactant?
0
How would we calculate ∆Hˆ j − ∆Hˆ f , j for the O2 reactant?
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0
How would we calculate ∆Hˆ j + ∆Hˆ f , j for the H2O product?
Example 9.5-1. 100 mol/s of ammonia, NH3, and 200 mol/s of O2 at 25 oC are fed
into a reactor. In the reactor, the ammonia is completely consumed according to
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(v); ∆ Hˆ r0 = -904.7 kJ/mol
The product gas leaves at 300 oC. What is the rate at which heat must be
transferred to/from the reactor, assuming the operation is at approximately 1 atm?
Use the heat of reaction method to outline the solution to the problem by setting up
all the necessary equations.
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Example 9.5-4. Ethanol can be dehydrogenated to acetaldehyde according to the
reaction
C2H5OH(v) CH3CHO(v) + H2(g)
The above reaction is carried out with a feed that consists of 90.0 mole % ethanol
and 10.0 mole % acetaldehyde. The feed enters the reactor at a flowrate of 150
mol/s and at a temperature of 300 oC. To maintain the reaction at the desired rate of
progress heat is added to the reactor at a rate of 2440 kW. Under these conditions,
the temperature of the outlet product stream is determined to be 253 oC.
What is the fractional conversion of ethanol achieved?
Outline the solution to this problem by setting up all the necessary equations. Use
the heat of formation method to calculate the standard enthalpy change of the
dehydrogenation reaction.
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