Chapter 8 and 9 Energy Balances

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 o C 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 o C and 1 bar? What is the specific internal energy of liquid water at 75 o C and 1 bar? (Use Table B. 7). Calculating changes in enthalpy and internal energy. Ĥ and Uˆ 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 o C 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 o C to 100 o C and entirely vaporized to steam at 100 o C and 1 atm, then liquified by compression from 1 atm to 10 atm at 100 o C, than cooled to - 200 o C 1

(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 o C and 1 atm to ice at 0 o C 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 o C and 5 atm is cooled and condensed to liquid cyclohexane at 25 o C and 5 atm. The enthalpy change for the condensation of cyclohexane at 80.7 o C and 1 atm is known. 2

(ii) O 2 at 170 o C and 1 atm and CH 4 at 25 o C and 1 atm are mixed and react completely to form CO 2 and H 2 O at 300 o C and 1 atm. The enthalpy change for the reaction occurring at 25 o C and 1 atm is known. Paths of type 1 changes in pressure (p) at constant temperature (T) and state of aggregation Liquids and solids: Uˆ 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, Uˆ 0 and Vˆ 0. This in turn implies that Ĥ = Uˆ + (pvˆ ) 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 Uˆ and Ĥ are both zero since T is not changing. Note that Ĥ = Uˆ + pvˆ = Uˆ + 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 = Uˆ + pvˆ = Uˆ + RT. For gases behaving non-ideally, either tables of Uˆ and Ĥ (e.g. like the steam tables) 3

must be used or more complex calculations (not covered in this course) are required. Example. Given the process C 2 H 6 (g, 25 o C, 1 atm) C 2 H 6 (g, 25 o C, 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 C V, C V Uˆ ( T ) = T (3) V Note that C V is in general a function of T. Given the function C V (T), the change in specific internal energy as the system temperature is raised from T 1 to T 2 at constant volume is T 2 U ˆ = CV ( T ) dt (constant V) (4) T 1 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, 4

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 T 2 U ˆ CV ( T ) dt (solids and liquids, type 2 path) (5) T 1 Moreover, for ideal gases, since Uˆ is a function of temperature only, equation (4) is exact even though the volume will change for a type 2 path, T 2 U ˆ = CV ( T ) dt (ideal gases, type 2 path) (4) T 1 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 C p, C p Hˆ ( T ) = T (6) p C p is in general a function of T. Given the function C p (T), the change in enthalpy of a substance as its temperature is raised from T 1 to T 2 at constant pressure is T 2 H ˆ = C p ( T ) dt (type 2 process) (7) T 1 5

Evaluation of heat capacities (also known as "specific heats"). C p and C V 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 o C), etc). Table B.2 provides polynomial expressions for heat capacities C p C p = a + bt + ct 2 + dt 3 (8) where a, b, c and d are constants. Most often we will use these expressions. NOTE: in some cases C p 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 C p values for materials based on their atomic composition. See text for discussion. C p values for mixtures. When dealing with heat capacities of mixtures, the most accurate method is to look up C p values from tables or other experimental information, if available. Lacking such data, if one is willing to neglect corrections to C p that derive from mixing of the different components of the mixture, the overall heat capacity of the mixture C p,mix can be approximated as the sum of heat capacity contributions from the separate components of the mixture, C p,mix (T) = i all mixture components x C ( T ) p i, (9) In equation 9, the index i ranges over all of the components of the mixture. x i 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 C p,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 C p,mix from equation 9 into equation 7. Example. For ideal gases, C p = C V + R. How can one prove this expression? 6

Example. For liquids and solids, C p C V. 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 o C to 100 o C. 2). Nitrogen in a 5 L flask, at an initial pressure of 3 bar, is cooled from 90 o C to 30 o C. For nitrogen at a constant pressure of 1 atm, the heat capacity C p is C p (kj/(mol o C)) = 0.02900 + 0.2199 10-5 T + 0.5723 10-8 T 2-2.871 10-12 T 3 7

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Example 8.3-5. A stream containing 10 % CH 4 and 90 % air by volume is to be heated from 20 o C to 300 o C. Calculate the required rate of heat input in kilowatts if the flow rate of the gas is 2.00 10 3 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 C p (kj/(mol o C)) = 0.03431 + 5.469 10-5 T + 0.3661 10-8 T 2-11.0 10-12 T 3 Also, in solving this problem, do you think we can neglect the enthalpy of mixing of methane and air? Why yes or why not? 10

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Example 8.3-6. A gas stream containing 8.0 mole % CO and 92.0 mole% CO 2 at 500 o C 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 o C 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 CO 2 are: C p,co = 0.02895 + 0.4110 10-5 T + 0.3548 10-8 T 2-2.220 10-12 T 3 C p,co2 = 0.03611 + 4.223 10-5 T - 2.887 10-8 T 2 + 7.464 10-12 T 3 Outline the solution to this problem by setting up all the necessary equations, including equations from which values of needed enthalpies are evaluated. 12

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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 at 30 o C is the same as at 100 o C). v Example. Imagine that a liquid is to be vaporized at 130 o C, but that Ĥ v is only known at 90 o C. How could you calculate Ĥ v (130 o C) from Ĥ v (90 o C)? 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). 14

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 T b (K) (nonpolar liquids) (10) Ĥ (kj/mol) 0.109 T b (K) (water, low molar weight alcohols) (11) v where T b 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, Ĥ (kj/mol) v Tb [ 0.0331( Tb / Tc ) 0.0327 + 0.0297log10 1.07 ( T / T ) b c P ] c (12) where T c and P c are the critical temperature (in o K) 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 Clausius- Clapeyron 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) d ln p * d(1/ T ) Hˆ R v = (13) Finally, Watson's correlation can be used to estimate Ĥ v at one temperature from its known value at another temperature, Ĥ (T 2 ) = Ĥ (T 1 ) v v T T c c T2 T1 0.38 (14) Estimation of Ĥ m. Ĥ m can be roughly estimated from Ĥ (kj/mol) 0.0092T m (K) (metallic elements) (15) m 15

Ĥ m (kj/mol) 0.0025T m (K) (inorganic compounds) (16) Ĥ (kj/mol) 0.050T m (K) (organic compounds) (17) m 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 o C 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 o C to 50 o C. 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? 16

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. 17

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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 = i H i all mixture components x ˆ (ideal mixtures) (18) where x i 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 x i is the mass fraction; if instead Ĥ i is expressed per mole, then x i 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 = Uˆ mix+ (pvˆ ) mix = Uˆ mix+ p Vˆ mix (19) where Uˆ mix and Vˆ mix are the changes in specific internal energy and specific volume associated with the mixing process. Uˆ mix is the internal energy (per amount of the mixture) minus the internal energy of the species that were mixed, Uˆ mix = Uˆ mixture - all mixture components x ˆ (20) iu i 19

where Uˆ 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 x i. Similarly, Vˆ mix = Vˆ mixture - all mixture components where Vˆ i is the specific volume of pure species i. x ˆ (21) iv i Hˆ s. Enthalpies of mixing are often expressed in terms of heat of solution ( T, r) ˆ ( T, r) H s 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 H 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. ˆ Example. What is H s ( T, r) for a solution of HCl in water at 25 o C and for r = 10, if the reference state for HCl is pure HCl gas at 25 o C and if the reference state for water is pure water liquid at 25 o C? Use Table B.11 to come up with the answer. ˆ ˆ What is H s ( T, r) for a solution of HCl in water at 25 o C and for r = 10, if the reference state for HCl is highly dilute HCl in water at 25 o C and if the reference state for water is pure water liquid at 25 o C? Use Table B.11 to come up with the answer. 20

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 o C and water liquid at 25 o C are mixed to produce 1000 kg/h of 20.0 wt % HCl solution in water at 40 o C. The heat capacity C p of the aqueous HCl solution product can be taken to be constant at 0.557 kj/(mol HCl o C). 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. 21

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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, H 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 ˆ ( T, p) 2A + 3B C H r 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 ˆ then H 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 ˆ H = ξ H r ( T, p) (22) Note that, as always, units must be consistent. Thus if units of H 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. ˆ ˆ 23

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 = Uˆ r+ (pvˆ ) r = Uˆ r+ p Vˆ r (23) where the last equality on the right follows because p is kept constant for a type 5 path. Here, Uˆ 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? ˆ 2). The value of H r ( T, p) depends on the states of aggregation of the reactants and products. For example, for the two reactions and A(g) + B(g) C(g); Ĥ r,1 A(l) + B(g) C(g) Ĥ r,2 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 0 ˆ r H, 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 o C and 1 atm. 24

Example 9.1-1. The standard heat of the combustion reaction of n-butane vapor is C 4 H 10 (g) + 13/2 O 2 (g) 4 CO 2 (g) + 5 H 2 O(l); H ˆ 0 r = -2878 kj/mol a). Calculate the rate of enthalpy change, H & (kj/s), if 2400 mol/s of CO 2 is produced, and the reaction is carried out at 25 o C. b). Calculate the standard heat of the reaction 2C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) Calculate the rate of enthalpy change, H & (kj/s), if 2400 mol/s of CO 2 is produced in this reaction, and the reaction is carried out at 25 o C. c). The heats of vaporization of n-butane and water at 25 o C are 19.2 kj/mol and 44.0 kj/mol, respectively. What is the standard heat of the reaction C 4 H 10 (l) + 13/2 O 2 (g) 4 CO 2 (g) + 5 H 2 O(v) Calculate the rate of enthalpy change, H & (kj/s), if 2400 mol/s of CO 2 is produced in this reaction, and the reaction is carried out at 25 o C. 25

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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 + O 2 CO 2 ; H ˆ 0 r = -393.51 kj/mol CO + 1/2O 2 CO 2 ; H 0 = -282.99 kj/mol what is the heat of reaction for C + 1/2O 2 CO? ˆ r Calculation of heats of reaction from standard heats of formation. The standard heat of formation H ˆ 0 f 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), O 2 (g), N 2 (g), H 2 (g) are the most important elemental constituents). The standard conditions for the formation reaction are most often taken to be 25 o C and 1 atm. For example, the formation reaction for ammonium nitrate would be written N 2 (g) + 2H 2 (g) + 3/2 O 2 (g) NH 4 NO 3 (s) H ˆ 0 f = -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 H 0 is known for all reactants and ˆ f 27

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 C 5 H 12 (l) + 8O 2 (g) 5CO 2 (g) + 6H 2 O(l) given the standard heats of formation (Table B.1) H ˆ 0 f (C 5 H 12 (l)) = -173.0 kj/mol H ˆ 0 f (CO 2 (g)) = -393.5 kj/mol H ˆ 0 f (H 2 O(l)) = -285.84 kj/mol Heats of combustion. The standard heat of combustion H ˆ 0 c of a substance is the heat of reaction when 1 mole of the substance is fully reacted with O 2 (g) to yield the products CO 2 (g) and H 2 O(l), with the combustion reaction carried out at 25 o C and 1 atm. Standard heats of combustion can be used to calculate H ˆ 0 r 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. 28

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 H ˆ 0 r is known. Problems that have multiple reactions occurring simultaneously, and for which the H ˆ 0 r 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 o C, 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 o C, 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 CO 2 (g) and H 2 O(v) as shown in the diagram. Also given is the standard enthalpy change for the reaction 29

C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) H ˆ 0 r = -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 H ˆ 0 r, (2) carry out the reaction under the conditions given for H ˆ 0 r, 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 30

H & reactor = reactants n j H ˆ j + j H ˆ 0 & & ξ r j + n& j H ˆ, j (24) reactions 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 0 r 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& 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? j 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 0 r, 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 0 r, 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 31

moles/time (refer to Chapter 4 for an introduction to & ξ j ). How would we calculate the & ξ j H ˆ 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 0 ˆ r H 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 o C and 1 atm, (2) decomposition of the reactants into elemental species at 25 o C and 1 atm 32

(note that these conditions of T and p are those used to define standard heats of formation, H ˆ 0 f ), (3) formation of the products from the elemental species at 25 o C and 1 atm, and (4) bringing of the products from 25 o C and 1 atm to the outlet stream conditions. The sum of these enthalpy changes gives the rate of enthalpy change H & reactor, H & reactor = reactants n j Hˆ j Hˆ 0 f j + n j Hˆ 0 & ( ) & ( f 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 o C and 1 atm; in other words, this term corresponds to step 1 of the path specified above. The - H ˆ 0 f, j in the first summation corresponds to decomposition of reactant j into elemental species at 25 o C 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 H ˆ 0 f, j term in the second summation on the right corresponds to formation of product j from the elemental species at 25 o C 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 o C 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 ˆ for the propane reactant? j H f, j 0 How would we calculate Hˆ ˆ for the O 2 reactant? j H f, j 33

How would we calculate Hˆ + ˆ 0 j H f, j for the H 2 O product? Example 9.5-1. 100 mol/s of ammonia, NH 3, and 200 mol/s of O 2 at 25 o C are fed into a reactor. In the reactor, the ammonia is completely consumed according to 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(v); H ˆ 0 r = -904.7 kj/mol The product gas leaves at 300 o C. 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. 34

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Example 9.5-4. Ethanol can be dehydrogenated to acetaldehyde according to the reaction C 2 H 5 OH(v) CH 3 CHO(v) + H 2 (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 o C. 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 o C. 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. 36

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