Volumetric Properties of Pure Fluids
Chapter 4 Heat Effects
4.7 HEAT EFFECTS OF INDUSTRIAL REACTIONS
Example 4.6
Calculate the standard heat of the following methanol-synthesis reaction at 800°C:
CO(g) + 2 H 2 (g) → CH 3 OH(g)
Solution 4.6
Apply Eq. (4.16) to this reaction for reference temperature T0 = 298.15 K and with heat-of-formation data from Table C.4:
Δ H 0 ° = Δ H 298° = − 200,660 − (−110,525) = − 90,135 J
Evaluation of the parameters in Eq. (4.20) is based on data taken from Table C.1:
i ν i A 10 3 B 10 6 C 10 −5 D _______
CH 3 OH
1
2.211
12.216
− 3.450
0.000 CO − 1 3.376 0.557 0.000 − 0.031 H 2
− 2
3.249
0.422
0.000
0.083 ____________
From its definition,
ΔA = (1)(2.211) + (−1)(3.376) + (− 2)(3.249) = − 7.663 Similarly,
ΔB = 10.815 × 10 −3 ΔC = − 3.450 × 10 −6 ΔD = − 0.135 × 10 5 The value of the integral of Eq. (4.20) for T = 1073.15 K is represented by:
IDCPH(298.15, 1073.15; − 7.663, 10.815 ×10 −3, − 3.450 ×10 −6, − 0.135 ×10 5 ) The value of this integral is −1615.5 K, and by Eq. (4.19),
ΔH° = − 90,135 + 8.314 ( −1615.5 ) = −103,566 J
4.7. Heat Effects of Industrial Reactions 157
Solution 4.7
The reaction is CH4 + 2O2 → CO2 + 2H2O(g) for which,
Δ H 298° = −393,509 + (2)(−241,818) − (−74,520) = −802,625 J Because the maximum attainable temperature (called the theoretical flame temperature) is sought, assume that the combustion reaction goes to completion adiabatically (Q = 0). If the kinetic- and potential-energy changes are negligible and if Ws = 0, the overall energy balance for the process reduces to ΔH = 0.
For purposes of calculating the final temperature, any convenient path between the initial and final states may be used. The path chosen is indicated in the diagram.
HP H
0
Products at 1 bar andTK
Reactants at 1 bar
and 25 C H298
1 mol CH4
2.4 mol O2
9.03 mol N2
0.4 mol O2
1 mol CO2
2 mol H2O 9.03 mol N2
When one mole of methane burned is the basis for all calculations, the follow- ing quantities of oxygen and nitrogen are supplied by the entering air:
Moles O 2 required = 2.0
Moles excess O 2 = (0.2)(2.0) = 0.4 Moles N 2 entering = (2.4)(79/21) = 9.03
The mole numbers ni of the gases in the product stream leaving the burner are 1 mol CO2, 2 mol H2O(g), 0.4 mol O2, and 9.03 mol N2. Because the enthalpy change must be independent of path,
Δ H 298° + Δ H P ° = ΔH = 0 (A) where all enthalpies are on the basis of 1 mol CH4 burned. The enthalpy change of the products as they are heated from 298.15 K to T is:
Δ H P ° = ⟨ C P ° ⟩ H (T − 298.15) (B) where we define ⟨ C P ° ⟩ H as the mean heat capacity for the total product stream:
⟨ C P ° ⟩ H ≡ ∑
i n i ⟨ C Pi ° ⟩ H
The simplest procedure here is to sum the mean-heat-capacity equations for the products, each multiplied by its appropriate mole number. Because C = 0 for each product gas (Table C.1), Eq. (4.9) yields:
⟨ C P ° ⟩ H = ∑
i n i ⟨ C Pi ° ⟩ H = R [ ∑
i n i A i + ∑ _i n i B i
2 (T − T 0 ) + _∑ i n i D i T T 0 ] Data from Table C.1 are combined as follows:
A = ∑
i n i A i = (1)(5.457) + (2)(3.470) + (0.4)(3.639) + (9.03)(3.280) = 43.471 Similarly, B = ∑
i n i B i = 9.502 × 10 −3 and D = ∑
i n i D i = − 0.645 × 10 5 . For the product stream ⟨ C P ° ⟩ H / R is therefore represented by:
MCPH(298.15, T; 43.471, 9.502 ×10 −3, 0.0, − 0.645 ×10 5) Equations (A) and (B) are combined and solved for T:
T = 298.15 − ___________Δ H 298° ⟨ C P ° ⟩ H
Because the mean heat capacities depend on T, one first evaluates ⟨ C P ° ⟩ H for an assumed value of T > 298.15, then substitutes the result in the preceding equation.
This yields a new value of T for which ⟨ C P ° ⟩ H is reevaluated. The procedure con- tinues to convergence on the final value,
T= 2066 K or 1793° C
Again, solution can be easily automated with the Goal Seek or Solver function in a spreadsheet or similar solve routines in other software packages.
Example 4.8
One method for the manufacture of “synthesis gas” (a mixture of CO and H2) is the catalytic reforming of CH4 with steam at high temperature and atmospheric pressure:
CH 4 (g) + H 2 O(g) → CO(g) + 3H 2 (g)
The only other reaction considered here is the water-gas-shift reaction:
CO(g) + H 2 O(g) → CO 2 (g) + H 2 (g)
Reactants are supplied in the ratio 2 mol steam to 1 mol CH4, and heat is added to the reactor to bring the products to a temperature of 1300 K. The CH4 is completely con- verted, and the product stream contains 17.4 mol-% CO. Assuming the reactants to be preheated to 600 K, calculate the heat requirement for the reactor.
4.7. Heat Effects of Industrial Reactions 159
Solution 4.8
The standard heats of reaction at 25°C for the two reactions are calculated from the data of Table C.4:
CH 4 (g) + H 2 O(g) → CO(g) + 3H 2 (g) Δ H 298° = 205,813 J CO(g) + H 2 O(g) → CO 2 (g) + H 2 (g) Δ H 298° = −41,166 J These two reactions may be added to give a third reaction:
CH 4 (g) + 2H 2 O(g) → CO 2 (g) + 4H 2 (g) Δ H 298° = 164,647 J Any pair of the three reactions constitutes an independent set. The third reaction is not independent; it is obtained by combination of the other two. The reactions most convenient to work with here are the first and third:
CH 4 (g) + H 2 O(g) → CO(g) + 3H 2 (g) Δ H 298° = 205,813 J (A) CH 4 (g) + 2H 2 O(g) → CO 2 (g) + 4H 2 (g) H 298° = 164,647 J (B) First one must determine the fraction of CH4 converted by each of these reactions.
As a basis for calculations, let 1 mol CH4 and 2 mol steam be fed to the reactor. If x mol CH4 reacts by Eq. (A), then 1 − x mol reacts by Eq. (B). On this basis the products of the reaction are:
CO: x
H2: 3x + 4(1 − x) = 4 − x CO2: 1 − x
H2O: 2 − x − 2(1 − x) = x Total: 5 mol products
The mole fraction of CO in the product stream is x/5 = 0.174; whence x = 0.870.
Thus, on the basis chosen, 0.870 mol CH4 reacts by Eq. (A) and 0.130 mol reacts by Eq. (B). Furthermore, the amounts of the species in the product stream are:
Moles CO = x = 0.87 Moles H 2 = 4 − x = 3.13 Moles CO 2 = 1 − x = 0.13 Moles H 2 O = x = 0.87
We now devise a path, for purposes of calculation, to proceed from reactants at 600 K to products at 1300 K. Because data are available for the standard heats of reaction at 25°C, the most convenient path is the one that includes the reactions at 25°C (298.15 K). This is shown schematically in the accompanying diagram. The dashed line represents the actual path for which the enthalpy change is ΔH. Because this enthalpy change is independent of path,
ΔH = Δ H R ° + Δ H 298° + Δ H P °
Po
∆H
Ro
∆H
298o
∆H
∆H
Products at 1 bar and 1300 K 0.87 mol CO 3.13 mol H2 0.13 mol CO2
0.87 mol H2O
Reactants at 1 bar and 600 K 1 mol CH4
2 mol H2O
For the calculation of Δ H 298° , reactions ( A) and (B) must both be taken into account. Because 0.87 mol CH4 reacts by (A) and 0.13 mol reacts by (B),
Δ H 298° =
(
0.87)
(
205,813)
+(
0.13)
(
164,647)
= 200,460 JThe enthalpy change of the reactants when they are cooled from 600 K to 298.15 K is:
Δ H R ° = ( ∑ i n i ⟨ C Pi ° ⟩ H ) (298.15 − 600) where subscript i denotes reactants. The values of ⟨ C Pi ° ⟩ H / R are:
CH 4 : MCPH(298.15, 600; 1.702, 9.081 ×10 −3, − 2.164 ×10 −6, 0.0) = 5.3272 H 2 O : MCPH(298.15, 600; 3.470, 1.450 ×10 −3, 0.0, 0.121 ×10 5) = 4.1888 and Δ H R ° = (8.314) [ (1)(5.3272)+(2)(4.1888) ] (298.15 − 600) = − 34,390 J The enthalpy change of the products as they are heated from 298.15 to 1300 K is calculated similarly:
Δ H P ° = ( ∑ i n i ⟨ C Pi ° ⟩ H ) (1300 − 298.15) where subscript i here denotes products. The ⟨ C Pi ° ⟩ H / R values are:
CO:
MCPH(298.15, 1300; 3.376, 0.557 ×10 −3, 0.0, − 0.031 ×10 5 )=3.8131 H 2 :
MCPH(298.15, 1300; 3.249, 0.422 ×10 −3, 0.0, − 0.083 ×10 5 )=3.6076 CO 2 :
MCPH(298.15, 1300; 5.457, 1.045 ×10 −3, 0.0, − 1.157 ×10 5 )=5.9935 H 2 O:
MCPH(298.15, 1300; 3.470, 1.450 ×10 −3, 0.0, 0.121 ×10 5 )=4.6599
Whence,
Δ H P ° =
(
8.314)
[(
0.87)
(
3.8131)
+(
3.13)
(
3.6076)
+
(
0.13)
(
5.9935)
+(
0.87)
(
4.6599)
] ×(
1300 − 298.15)
= 161, 940 J
4.7. Heat Effects of Industrial Reactions 161 Therefore,
ΔH = −34,390 + 200,460 + 161,940 = 328,010 J
The process is one of steady flow for which Ws, Δz, and Δu2/2 are presumed negligible. Thus,
Q = ΔH = 328, 010 J This result is on the basis of 1 mol CH4 fed to the reactor.
Example 4.9
Solar-grade silicon can be manufactured by thermal decomposition of silane at moder- ate pressure in a fluidized-bed reactor, in which the overall reaction is:
SiH 4 (g) → Si(s) + 2 H 2 (g)
When pure silane is preheated to 300°C, and heat is added to the reactor to promote a reasonable reaction rate, 80% of the silane is converted to silicon and the products leave the reactor at 750°C. How much heat must be added to the reactor for each kilogram of silicon produced?
Solution 4.9
For a continuous-flow process with no shaft work and negligible changes in kinetic and potential energy, the energy balance is simply Q = ΔH, and the heat added is the enthalpy change from reactant at 300°C to products at 750°C. A convenient path for calculation of the enthalpy change is to (1) cool the reactant to 298.15 K, (2) carry out the reaction at 298.15 K, and (3) heat the products to 750°C.
On the basis of 1 mol SiH4, the products consist of 0.2 mol SiH4, 0.8 mol Si, and 1.6 mol H2. Thus, for the three steps we have:
Δ H 1 = ∫573.15K298.15K C P ° ( SiH 4 ) dT Δ H 2 = 0.8 × Δ H 298°
Δ H 3 = ∫298.15K1023.15K [ 0.2 × C P ° ( SiH 4 ) + 0.8 × C P ° (Si) + 1.6 × C P ° ( H 2 ) ] dT
Data needed for this example are not included in App. C, but are readily obtained from the NIST Chemistry WebBook (http://webbook.nist.gov). The reaction here is the reverse of the formation reaction for silane, and its standard heat of reaction at 298.15 K is Δ H 298° = − 34,310 J . Thus, the reaction is mildly exothermic.
Heat capacity in the NIST Chemistry WebBook is expressed by the Shomate equation, a polynomial of different form from that used in this text. It includes a T3 term, and is written in terms of T/1000, with T in K:
C P ° = A + B ( _T
1000 ) + C ( _____T
1000 ) 2 + D ( _____T
1000 ) 3 + E ( _____T 1000 ) −2 Formal integration of this equation gives the enthalpy change:
ΔH = ∫T 0T C P ° dT
ΔH = 1000 [ A ( _____T
1000 ) + __B 2 ( _____T
1000 ) 2 + C__
3 ( _____T 1000 ) 3 + __D
4 ( _____T
1000 ) 4 − E ( _____T 1000 ) −1 ]
T 0 T
The first three rows in the accompanying table give parameters, on a molar basis, for SiH4, crystalline silicon, and hydrogen. The final entry is for the collec- tive products, represented for example by:
A(products) = (0.2) (6.060) + (0.8)(22.817) + (1.6)(33.066) = 72.3712 with corresponding equations for B, C, D, and E.
Species A B C D E
SiH4(g) 6.060 139.96 −77.88 16.241 0.1355
Si(s) 22.817 3.8995 −0.08289 0.04211 −0.3541
H2(g) 33.066 −11.363 11.433 −2.773 −0.1586
Products 72.3712 12.9308 2.6505 −1.1549 −0.5099
For these parameters, and with T in kelvins, the equation for ΔH yields values in joules. For the three steps making up the solution to this problem, the following results are obtained:
1. Substitution of the parameters for 1 mol of SiH4 into the equation for ΔH leads upon evaluation to: ΔH1 = −14,860 J.
2. Here, Δ H 2 = (0.8)(− 34,310) = − 27,450 J .
3. Substitution of the parameters for the total product stream into the equation for ΔH leads upon evaluation to: ΔH3 = 58,060 J.
For the three steps the sum is:
ΔH = − 14,860 − 27,450 + 58,060 = 15,750 J
This enthalpy change equals the heat input per mole of SiH4 fed to the reactor. A kilogram of silicon, with a molar mass of 28.09, is 35.60 mol. Producing a kilogram of silicon therefore requires a feed of 35.60/0.8 or 44.50 mol of SiH4. The heat require- ment per kilogram of silicon produced is therefore (15,750)(44.5) = 700,900 J.
4.7. Heat Effects of Industrial Reactions 163
Example 4.10
A boiler is fired with a high-grade fuel oil (consisting only of hydrocarbons) having a standard heat of combustion of −43,515 J·g−1 at 25°C with CO2(g) and H2O(l) as prod- ucts. The temperature of the fuel and air entering the combustion chamber is 25°C.
The air is assumed dry. The flue gases leave at 300°C, and their average analysis (on a dry basis) is 11.2% CO2, 0.4% CO, 6.2% O2, and 82.2% N2. Calculate the fraction of the heat of combustion of the oil that is transferred as heat to the boiler.
Solution 4.10
Take as a basis 100 mol dry flue gases, consisting of:
CO2 11.2 mol
CO 0.4 mol
O2 6.2 mol
N2 82.2 mol
Total 100.0 mol
This analysis, on a dry basis, does not take into account the H2O vapor present in the flue gases. The amount of H2O formed by the combustion reaction is found from an oxygen balance. The O2 supplied in the air represents 21 mol-% of the air stream. The remaining 79% is N2, which goes through the combustion process unchanged. Thus the 82.2 mol N2 appearing in 100 mol dry flue gases is supplied with the air, and the O2 accompanying this N2 is:
Moles O2 entering in air = (82.2)(21 / 79) = 21.85 and
Total moles O 2 in the dry flue gases = 11.2 + 0.4 / 2 + 6.2 = 17.60 The difference between these figures is the moles of O2 that react to form H2O.
Therefore on the basis of 100 mol dry flue gases,
Moles H 2 O formed = (21.85 − 17.60) (2) = 8.50 Moles H 2 in the fuel = moles of water formed = 8.50 The amount of C in the fuel is given by a carbon balance:
Moles C in flue gases = moles C in fuel = 11.2 + 0.4 = 11.60 These amounts of C and H2 together give:
Mass of fuel burned = ( 8.50 ) ( 2 ) + ( 11.6 ) ( 12 ) = 156.2 g
If this amount of fuel is burned completely to CO2(g) and H2O(l) at 25°C, the heat of combustion is:
Δ H 298° = (− 43,515)(156.2) = − 6,797,040 J
However, the reaction actually occurring does not represent complete combustion, and the H2O is formed as vapor rather than as liquid. The 156.2 g of fuel, consist- ing of 11.6 mol of C and 8.5 mol of H2, is represented by the empirical formula C11.6H17. Omit the 6.2 mol O2 and 82.2 mol N2 which enter and leave the reactor unchanged, and write the reaction:
C 11.6 H 17 (l) + 15.65 O 2 (g) → 11.2 CO 2 (g) + 0.4 CO(g) + 8.5 H 2 O(g) This result is obtained by addition of the following reactions, for each of which the standard heat of reaction at 25°C is known:
C 11.6 H 17 (l) + 15.85 O 2 (g)
→ 11.6 CO 2 (g) + 8.5 H 2 O(l) 8.5 H 2 O(l) → 8.5 H 2 O(g)
0.4 CO 2 (g)
→ 0.4CO(g) + 0.2 O 2 (g)
The sum of these reactions yields the actual reaction, and the sum of the Δ H 298° values gives the standard heat of the reaction occurring at 25°C:
Δ H 298° = −6,797,040 + (44,012)(8.5) + (282,984)(0.4) = − 6,309,740 J The actual process leading from reactants at 25°C to products at 300°C is rep- resented by the dashed line in the accompanying diagram. For purposes of calcu- lating ΔH for this process, we may use any convenient path. The one drawn with solid lines is a logical one: Δ H 298° has already been calculated and Δ H P ° is easily evaluated.
HP H
Products at 1 bar and 300 C
Reactants at 1 bar
and 25 C H298
11.2 mol CO2 0.4 mol CO 8.5 mol H2O 6.2 mol O2 82.2 mol N2
156.2 g fuel 21.85 mol O2
82.2 mol N2
The enthalpy change associated with heating the products of reaction from 25 to 300°C is:
Δ H P ° = ( ∑ i n i ⟨ C Pi ° ⟩ H ) (573.15 − 298.15) where subscript i denotes products. The ⟨ C Pi ° ⟩ H / R values are:
4.7. Heat Effects of Industrial Reactions 165
CO 2 :
MCPH(298.15, 573.15; 5.457, 1.045 ×10 −3, 0.0, −1.157 ×10 5) =5.2352 CO:
MCPH(298.15, 573.15; 3.376, 0.557 ×10 −3, 0.0, −0.031 ×10 5) =3.6005 H 2 O: MCPH(298.15, 573.15; 3.470, 1.450 ×10 −3, 0.0, 0.121 ×10 5) =4.1725 O 2 :
MCPH(298.15, 573.15; 3.639, 0.506 ×10 −3, 0.0, − 0.227 ×10 5) =3.7267 N 2 :
MCPH(298.15, 573.15; 3.280, 0.593 ×10 −3, 0.0, 0.040 ×10 5) =3.5618
Therefore,
Δ H P ° = (8.314)[(11.2)(5.2352) + (0.4)(3.6005) + (8.5)(4.1725) + (6.2)(3.7267) + (82.2)(3.5618)](573.15 − 298.15) = 940,660 J
and
ΔH = Δ H 298° + Δ H P ° = − 6, 309,740 + 940,660 = − 5,369,080 J Because the process is one of steady flow for which the shaft work and kinetic- and potential-energy terms in the energy balance [Eq. (2.32)] are zero or negligi- ble, ΔH = Q. Thus, Q = −5369 kJ, and this amount of heat is transferred to the boiler for every 100 mol dry flue gases formed. This represents
5,369,080
_________
6,797,040 (100) = 79.0%
of the higher heat of combustion of the fuel.
In the foregoing examples of reactions that occur at approximately 1 bar, we have tacitly assumed that the heat effects of reaction are the same whether gases are mixed or pure, an acceptable procedure for low pressures. For reactions at elevated pressures, this may not be the case, and it may be necessary to account for the effects of pressure and of mixing on the heat of reaction. However, these effects are usually small. For reactions occurring in the liquid phase, the effects of mixing are generally more important. They are treated in detail in Chapter 11.
For biological reactions, occurring in aqueous solution, the effects of mixing are partic- ularly important. The enthalpies and other properties of biomolecules in solution usually depend not only on temperature and pressure, but also on the pH, ionic strength, and concen- trations of specific ions in solution. Table C.5 in App. C provides enthalpies of formation of a variety of molecules in dilute aqueous solution at zero ionic strength. These can be used to estimate heat effects of enzymatic or biological reactions involving such species. However, corrections for effects of pH, ionic strength, and finite concentration may be significant.16 Heat capacities are often unknown for such species, but in dilute aqueous solution the overall specific heat is usually well approximated by the specific heat of water. Moreover, the tem- perature range of interest for biological reactions is quite narrow. The following example illus- trates estimation of heat effects for a biological reaction.
16For analysis of these effects, see Robert A. Alberty, Thermodynamics of Biochemical Reactions, John Wiley &
Sons, Hoboken, NJ, 2003.
Example 4.11
A dilute solution of glucose enters a continuous fermentation process, where yeast cells convert it to ethanol and carbon dioxide. The aqueous stream entering the reac- tor is at 25°C and contains 5 wt-% glucose. Assuming this glucose is fully converted to ethanol and carbon dioxide, and that the product stream leaves the reactor at 35°C, estimate the amount of heat added or removed per kg of ethanol produced. Assume that the carbon dioxide remains dissolved in the product stream.
Solution 4.11
For this constant pressure process with no shaft work, the heat effect is simply equal to the enthalpy change from the feed stream to the product stream. The fer- mentation reaction is:
C 6 H 12 O 6 (aq) → 2 C 2 H 5 OH(aq) + 2 CO 2 (aq)
The standard enthalpy of reaction at 298 K obtained using the heats of forma- tion in dilute aqueous solution from Table C.5 is:
Δ H 298° = (2)(−288.3) + (2)(−413.8) − (−1262.2) = −142.0 kJ·mol −1 One kg of ethanol is 1/(0.046069 kg·mol−1) = 21.71 mol ethanol. Each mole of glucose produces two moles of ethanol, so 10.85 mol of reaction must occur to produce 1 kg of ethanol. The standard enthalpy of reaction per kg ethanol is then (10.85)(−142.0) = −1541 kJ·kg−1. The mass of glucose required to produce 1 kg ethanol is 10.85 mol × 0.18016 kg·mol−1 = 1.955 kg glucose. If the feed stream is 5 wt-% glucose, then the total mass of solution fed to the reactor per kg ethanol produced is 1.955/0.05 = 39.11 kg. Assuming that the product stream has the spe- cific heat of water, about 4.184 kJ·kg−1·K−1, then the enthalpy change per kg eth- anol for heating the product stream from 25°C to 35°C is:
4.184 kJ· kg −1 ·K −1 × 10 K × 39.11 kg = 1636 kJ
Adding this to the heat of reaction per kg ethanol gives the total enthalpy change from feed to product, which is also the total heat effect:
Q = ΔH = − 1541 + 1636 = 95 kJ ·(kg ethanol) −1
This estimate leads to the conclusion that a small amount of heat must be added to the reactor because the reaction exothermicity is not quite sufficient to heat the feed stream to the product temperature. In an actual process, the glucose would not be fully converted to ethanol. Some fraction of the glucose must be directed to other products of cellular metabolism. This means that somewhat more than 1.955 kg glucose will be needed per kg of ethanol produced. The heat release from other reactions may be some- what higher or lower than that for the production of ethanol, which would change the estimate. If some of the CO2 leaves the reactor as a gas, then the heat requirement will be slightly higher because the enthalpy of CO2(g) is higher than that of aqueous CO2.
4.9. Problems 167
4.9 167