Perry´s chemical engeneers handbook
(Parte 6 de 15)
Gaseous equilibria are expressed in terms of fugacitiesor fugacity coefficients. In terms of partial pressures, p=yp,
Pressure affects the composition of an equilibrium mixture, but not the equilibrium constant itself.
Although the equilibrium constant can be evaluated in terms of kinetic data, it is usually found independently so as to simplify finding the other constants of the rate equation. With Kknown, the correct exponents of Eq. (7-64) can be found by choosing trial sets until k comes out approximately constant. When the exponents are small integers or simple fractions, this process is not overly laborious.
Example 4: Reaction between Methane and SteamAt 600°C the principal reactions between methane and steam are p p
D H + E D Cp dT dln K }
TABLE 7-3Adsorption-rate Controlling (Rapid Surface Reaction) ReactionSpecial conditionBasic rate equationDriving forceAdsorption term
1.A fiM +Nr =kpqp1 ++Kp+Kp r =
Adsorption rate of substance A is controlling in each case. When an inert substance I is adsorbed, the term Kpis to be added to the adsorption term. SOURCE:From Walas, Reaction Kinetics for Chemical Engineers,McGraw Hill, 1959; Butterworths, 1989.
K p p }
K p p }
K p p }
K p p }
K p p }
Simultaneous solution by the Newton-Raphson method yields x=0.9121, y= 0.6328. Accordingly, the fractional compositions are:
Approach to EquilibriumAs equilibrium is approached the rate of reaction falls off, and the reactor size required to achieve a specified conversion goes up. At some point, the cost of increased reactor size will outweigh the cost of discarded or recycled unconverted material. No simple rule for an economic appraisal is really possible, but sometimes a basis of 95 percent of equilibrium conver- sion is taken. For adiabatic operation, a certain approach to equilibrium temperature is common practice, say within 10 to 20°C (18 to 36°F), a number possibly based on experience with a particular process.
The volume escalates rapidly at high percent approaches.
In either batch or flow systems, many single-rate equations lead to integrands that are ratios of low-degree polynomials that can be integrated by inspection or with the briefest of integral tables. Some of the cases of frequent occurrence are summarized in Table 7-4. When the problem is to relate Cand t,the constants are known, and the polynomials are of second degree or higher, numerical integration may save dA }} kV }
7-14 REACTION KINETICS
1. A fiProducts: - kA
3. Reversible reaction A A B:
4. Reversible reaction, second order, A+B A R + S:
5. The reaction nAfinR+nSbetween ideal gases at constant Tand P:
-= kn dn }
2a A +b- q
2a A +b+ q
2a A +b dA } dA } dA } dA }
E dn,in general
6. Equations readily solvable by Laplace transforms. For example:
B fi C
Rate equations are -=kA -kB
Laplace transformations are made and rearranged to
Inversion of the transforms can be made to find the concentrations A, B,and C as functions of the time t.
dC } dB } dA }
TABLE 7-4Some Isothermal Rate Equations and Their Integrals
SOURCE:Adapted from Walas, Chemical Process Equipment Selection and Design,Butterworth-Heinemann, 1990.
time and preserve reliability. Some 40 cases of integrations at constant volume are developed by Capellos and Bielski (Kinetic Systems Mathematical Descriptions of Chemical Kinetics,Wiley, 1972).
Sets of first-order rate equations are solvable by Laplace transform
(Rodiguin and Rodiguina, Consecutive Chemical Reactions,Van Nostrand, 1964). The methods of linear algebra are applied to large sets of coupled first-order reactions by Wei and Prater (Adv. Catal.,13,203 ). Reactions of petroleum fractions are examples of this type.
Example 6: Laplace Transform ApplicationFor the reaction A ÞB ÛC
When even second-order reactions are included in a group to be analyzed, individual integration methods may be needed. Three cases of coupled first- and second-order reactions will be touched on. All of them are amenable only with difficulty to the evaluation of specific rates from kinetic data. Numerical integrations are often necessary.
dB } dA }
1.The reactions are 2AÞBÞC. The partial solutions are A =
Although the differential equation is first-order linear, its integration requires evaluation of an infinite series of integrals of increasing difficulty. 2.The reactions are AÞB and A +BÞC. After A is expressed in terms of B by elimination of t, but this cannot be integrated analytically.
3.For the reactions AÞB, 2BÞC; 2AÞBÞC; 2AÞB, 2BÞC; the rate equations are solved in terms of higher transcendental functions by Chien (J. Am. Chem. Soc.,76,2256 ). For the first case, with B=0:
wheret=exp (-kt) K =kkA
(Parte 6 de 15)