1、化 学 反 应 工 程,Chapter 9,Temperature and Pressure Effects,In our search for favorable conditions for reaction we have considered how reactor type and size influence the extent of conversion and distribution of products. The reaction temperature and pressure also influence the progress of reactions, and
2、 it is the role of these variable that we now consider.,化 学 反 应 工 程,We follow a three-step procedure: First, we must find how equilibrium composition, rate of reaction, and product distribution are affected by change in operating temperatures and pressures. This will allow us to determine the optimu
3、m temperature progression, and it is this that we strive(努力,奋斗) to approximate with a real design.,化 学 反 应 工 程,Second, chemical reactions are usually accompanied by heat effects, and we must know how these will change the temperature of the reacting mixture. With this information we are able to prop
4、ose a number of favorable reactor and heat exchange systems those which closely approach the optimum conditions and then seeing how economic considerations will select one of these favorable systems as the best.,化 学 反 应 工 程,So, with the emphasis on finding the optimum conditions and then seeing how
5、best to approach them in actual design rather than determining what specific reactors will do, let us start with discussion of single reactions and follow this with the special considerations of multiple reactions.,化 学 反 应 工 程,9.1 SINGLE REACTIONS,With single reactions we are concerned with conversi
6、on level and reactor stability. Questions of product distribution do not occur.Thermodynamics gives two important pieces of information, the first being the heat liberated(释放) or absorbed for a given extent of reaction, the second being the maximum possible conversion. Let us briefly summarize these
7、 findings.,化 学 反 应 工 程,Heats of Reaction from Thermodynamics,The heat liberated or absorbed reaction at temperature T depends on the nature of the reacting system, the amount of material reacting, and the temperature and pressure of the reacting system, and is calculated from the heat of reaction Hr
8、 (反应热), for the reaction in question. When this is not known, it can in most cases be calculated from known and tabulated thermochemical data on heats of formation Hf (生成热) or heats of combustion Hc (燃烧热) of the reacting materials.,化 学 反 应 工 程,aA rR + sS,These are tabulated at some standard temperat
9、ure, T1, usually 25. As a brief reminder, consider the reaction By convention(习俗,惯例) we define the heat of reaction at temperature T as the heat transferred to the reacting system from the surroundings when a moles of A disappear to produce r moles of R and s moles of S with the system measured at t
10、he same temperature and pressure before and after the change. Thus,aA rR + sS HrT,(1),化 学 反 应 工 程,Heat of Reaction and Temperature. The first problem is to evaluate the heat of reaction at temperature T2 knowing the heat of reaction at temperature T1. This is found by the law of conservation(守恒,保存)
11、of energy as follows:,化 学 反 应 工 程,=,+,+,(2),化 学 反 应 工 程,In term of enthalpies(焓) of reactants and products this becomes,Hr2 = -(H2 H1)reactants +Hr1 + (H2 H1)products,Where subscripts 1 and 2 refer to quantities measured at temperature T1 and T2 , respectively. In term of specific heats (比热，CP),Wher
12、e,(4),(3),(5),化 学 反 应 工 程,CpA = A + AT + A T 2 CpR = R + R T + R T 2 CpS = S + S T + S T 2,When the molar specific heats are functions of temperatures as follows,(6),化 学 反 应 工 程,We obtain,Hr2 = Hr1 + + + T 2)dT= Hr1 + (T2 - T1) + +,Where, = rR + sS - aA = rR + sS - aA = rR + sS - aA,(7),(8),化 学 反 应
13、工 程,EXAMPLE 9.1 Hr AT VARIOUS TEMPERATURES,From the Hc and Hf tables, Ive calculated that the standard heat of my gas-phase reaction at 25 is as follows:,A + B 2R Hr, 298K = -50 000 J,At 25 the reaction is strongly exothermic. But this doesnt interest me because I plan to run the reaction at 1025. W
14、hat is the Hr at that temperature, and is the reaction still exothermic at that temperature?,Data. Between 25 and 1025 the average Cp values for the various reaction components are,= 35J/molK = 45 J/molK = 70 J/molK,化 学 反 应 工 程,SOLUTION,First, prepare a reaction map as shown in Fig.E9.1. Then an ent
15、halpy balance for 1 mol A, 1 mol B, 2 mol R gives,H1 = H2 + H3 +H4= (n T)reactants + Hr, 25 + (n T)products,= 1(35)(25-1025) + 1(45)(25-1025) + (-50 000) +2(70)(1025 25) = 10 000 J,Hr. 1025 = 10 000 J,or,化 学 反 应 工 程,Figure E9.1,The reaction is,化 学 反 应 工 程,Equilibrium Constants from Thermodynamics,Fr
16、om the second law of thermodynamics equilibrium constants, hence equilibrium compositions of reacting systems, may be calculated. We must remember, however, that real systems do not necessarily achieve this conversion; therefore, the conversions calculated from thermodynamics are only suggested atta
17、inable values.,As a brief reminder, the standard free energy G0 for the reaction of Eq.1 at temperature T is defined as,化 学 反 应 工 程,(9),where f is the fugacity(逸度) of the component at equilibrium conditions; f 0 is the fugacity of the component at the arbitrarily selected standard state at temperatu
18、re T, the same one used in calculating G 0 ;,化 学 反 应 工 程,G0 is the standard free energy of a reacting component, tabulated for many compounds; and K is the thermodynamic equilibrium constant for the reaction. Standard states at given temperature are commonly chosen as follow:,Gas pure component at o
19、ne atmosphere, at which pressure ideal gas behavior is closely approximated； Solid pure solid component at unit pressure Liquid pure liquid at its vapor pressure Solute in liquid 1 molar solution; or at such dilute concentrations that the activity is unity.,化 学 反 应 工 程,For convenience define,Kf = ，K
20、p = ，Ky = ，Kc =,and,n = r + s a,Simplified forms of Eq.9 can be obtained for various systems. For gas reactions standard states are usually chosen at a pressure of 1 atm. At this low pressure the deviation from ideality invariably is small: hence fugacity and pressure are identical and f 0 = p0 = 1
21、atm.,(10),化 学 反 应 工 程,K = = Kp,Thus,The term in braces in this equation and in Eq.13 is always unity but is retained to keep the equations dimensionally correct.For any component i of an ideal gas,fi = pi = yi= CiRT,Hence,Kf = Kp,(11),(12),化 学 反 应 工 程,For a solid component taking part in a reaction,
22、 fugacity variation with pressure are small and can usually be ignored. Hence,(14),and,(13),化 学 反 应 工 程,Equilibrium Conversion. The equilibrium composition, as governed by the equilibrium constant, changes with temperature, and from thermodynamics the rate of change is given by,(15),On integrating E
23、q.15, we see how the equilibrium constant changes with temperature. When the heat of reaction can be considered to be constant in the temperature interval, integration yields,(16),化 学 反 应 工 程,When the variation of Hr must be accounted for in the integration we have,(17),where is given by a special f
24、orm of Eq.4 in which subscript 0 refers to the base temperature,(18),化 学 反 应 工 程,Replacing Eq.18 in Eq.17 and integrating, while using the temperature dependency for given by Eq.18, gives,These expressions allow us to find the variation of the equilibrium constant, hence, equilibrium conversion, wit
25、h temperature.,(19),化 学 反 应 工 程,The following conclusions may be drawn from thermodynamics. These are illustrated in part by Fig.9.1.,1. The thermodynamics equilibrium constant is unaffected by the pressure of the system, by the pressure or absence of inerts, or by the kinetics of the reaction, but
26、is affected by the temperature of the system.2. Though the thermodynamic equilibrium constant is unaffected by pressure or inerts, the equilibrium concentration of materials and equilibrium conversion of reactants can be influenced by these variables.,化 学 反 应 工 程,3. K 1 indicates that practically co
27、mplete conversion may be possible and that the reaction can be considered to be irreversible. K1 indicates that reaction will not proceed to any appreciable extent. 4. For an increase in temperature, equilibrium conversion rises for endothermic reactions and drops for exothermic reactions. 5. For an
28、 increase in pressure in gas reactions, conversion rises when the number of moles decreases with reaction; conversion drops when the number of moles increase with reaction. 6. A decrease in inerts for all reactions acts in the way that an increase in pressure acts for gas reactions.,化 学 反 应 工 程,Figu
29、re 9.1 Effect of temperature on equilibrium conversion as predicted by thermodynamics (pressure fixed.),化 学 反 应 工 程,EXAMPLE 9.2 EQUILIBRIUM CONVERSION AT DIFFERENT TEMPERATURES,(a) Between 0 and 100 determine the equilibrium conversion for the elementary aqueous reaction,Present the results in the f
30、orm of a plot of temperature versus conversion.(b) What restrictions should be placed on the reactor operating isothermally if we are to obtain a conversion of 75% or higher?,化 学 反 应 工 程,SOLUTION,(a) With all specific heats alike, . Then from Eq.4 the heat of reaction is independent of temperature a
31、nd is given by,(),From Eq.9 the equilibrium constant at 25 is given by,(),Since the heat of reaction does not change with temperature, the equilibrium constant K at any temperature T is now found from Eq.16.,化 学 反 应 工 程,Rearranging gives,Replacing K298 and Hr from Eqs. () and () gives on rearranging
32、,(),Thus,化 学 反 应 工 程,But at equilibrium,or,(),Putting T values into Eq. (), then K into Eq.() , as shown in Table E9.2 gives the changing equilibrium conversion as a function of temperature in the range of 0 to 100. This results is displayed in Fig.E9.2.,化 学 反 应 工 程,Table E9.2 Calculating from Eqs.(
33、) and (),5 278 2700 0.999+ 15 288 860 0.999 25 298 300 0.993 35 308 110 0.991 45 318 44.2 0.978 55 328 18.4 0.949 65 338 8.17 0.892 75 348 3.79 0.791 85 358 1.84 0.648 95 368 0.923 0.480,化 学 反 应 工 程,Figure E9.2,(b) From the graph we see that the temperature must stay below 78 if conversion of 75% or
34、 higher may be expected.,化 学 反 应 工 程,General Graphical Design Procedure,Temperature, composition, and reaction rate are uniquely related for any single homogeneous reaction, and this may be represented graphically in one of three ways, as shown in Fig.9.2. The first of these, the composition-tempera
35、ture plot, is the most convenient so we will use it throughout to represent data, to calculate reactor sizes, and to compare design alternatives.,For a given feed (fixed ) and using conversion of key component as a measure of the composition and extent of reaction, the XA versus T plot has the gener
36、al shape shown in Fig.9.3.,化 学 反 应 工 程,Reaction rate surface for a reversible exothermic reaction,化 学 反 应 工 程,Figure 9.2 Different ways of representing the relationship of temperature, composition, and rate for a single homogeneous reaction.,化 学 反 应 工 程,This plot can be prepared either from a thermo
37、dynamically consistent rate expression for the reaction (the rate must be zero at equilibrium) or by interpolating(插入) from a given set of kinetic data in conjunction with thermodynamic information on the equilibrium. Naturally, the reliability(可靠性) of all the calculations and predictions that follo
38、w are directly dependent on the accuracy of this chart. Hence, it is imperative(必要的) to obtain good kinetic data to construct this chart.,化 学 反 应 工 程,Figure 9.3 General shape of the temperature-conversion plot for different reaction types.,化 学 反 应 工 程,The size of reactor required for a given duty an
39、d for a given temperature progression is found as follows:Draw the reaction path on the XA versus T plot. This is the operating line for the operation.2. Find the rates at various XA along this path.3. Plot the 1/(- rA) versus for this path.4. Find the area under this curve. This gives V/FA0.,化 学 反
40、应 工 程,For exothermic reactions we illustrate this procedure in Fig.9.4 for three paths: path AB for plug flow with an arbitrary temperature profile. path CD for nonisothermal plug flow with 50% recycle, and point E for mixed flow. Note that for mixed flow the operating line reduces to a single point
41、.,This procedure is quite general, applicable for any kinetics, any temperature progression, and any reactor type or any series of reactors. So, once the operating line is known, the reactor size can be found.,化 学 反 应 工 程,Figure 9.4 Finding the reactor size for different types of flow and for a feed
42、 temperature .,化 学 反 应 工 程,EXAMPLE 9.3 CONSTRUCTION OF THE RATE- CONVERSION-TEMPERATURE CHART FROM KINETIC DATA,With the system of Example 9.2 and starting with an R-free solution, kinetic experiments in a batch reactor give 58.1% conversion in 1 min at 65, 60% conversion in 10 min at 25. Assuming r
43、eversible first-order kinetics, find the rate expression for this reaction and prepare the conversion-temperature chart with reaction rate as parameter.,化 学 反 应 工 程,SOLUTION,Integrate the Performance Equation. For a reversible first-order reaction, the performance equation for a batch reactor is,Acc
44、ording to Eq.3.54 this integrates to give,(),化 学 反 应 工 程,Calculate the Forward Rate Constant. From the batch run at 65, noting from Example 9.2 that XAe = 0.89, we find with Eq. (),or,(),Similarly, for the batch run at 25 we find,(),化 学 反 应 工 程,Assuming an Arrhenius temperature dependency, the ratio
45、 of forward rate constants at these two temperature gives,(iv),from which activation energy of the forward reaction can be evaluated, giving,化 学 反 应 工 程,Note that there are two activation energies for this reaction, one of the forward reaction, another for the reverse. Now for the complete rate cons
46、tant for the forward reaction. From either the numerator(分子) or the denominator(分母) of Eq.(iv) first evaluate k10, then use it as shown below:,Noting that , thus , where K is given by Eq.(iii) of Example 9.2, we can find the value of k2 .,化 学 反 应 工 程,Summary. For the reversible first-order reaction
47、of Example 9.2 we have,化 学 反 应 工 程,From these values the XA versus T chart for any specific CA0 can be prepared and for this purpose the electronic computer is a great timesaver(省时的东西). Figure E9.3 is such a plot prepared for CA0 = 1 mol/ liter and CR0= 0.,Since we are dealing with first-order reactions this plot can be used for any CA0 value by properly relabeling(贴标签, 标明) the rate curves. Thus, for CA0 = 10 mol/ liter simply multiply(乘) all the rate values on this graph by a factor of 10.,化 学 反 应 工 程,Figure E9.3,化 学 反 应 工 程,Optimum Temperature Progression,
