Concentration or Partial Pressure and Temperature Differences

When integrated over the film thickness L for steady-state diffusion, with Na = constant, and with an average constant value for DAm, (3.2.3.A-e) gives 

TEMPERATURE DIFFERENCES BETWEEN BULK FLUID AND SURFACE OF A CATALYST PARTICLE 

One of the most important uses of the above mass and heat transfer relationships is in determining external mass and heat transfer resistances for catalyst particles. Here, the rate is usually expressed in terms of catalyst mass (kmol / kg cat. s), and using = external surface per weight of catalyst (m p / kg cat.) gives 

In experimental kinetic studies in particular, the question often arises if the partial pressure drop. pA over the so-called external film may be neglected. One has to check whether or not it is allowed to substitute Pa, the partial pressure of A in the bulk fluid stream, into the rate equation for the reaction. The value of Ag is determined from a correlation, such as (3.2.1-5) with (3.2.1-2) and (3.2.1-3). 

The calculation of is not straightforward, since the calculation of the film pressure factor pfA requires the knowledge of. iteration is required. 

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Concentration or Partial Pressure and Temperature Differences Between Bulk Fluid and Surface of a Catalyst Particle ... 

CONCENTRATION OR PARTIAL PRESSURE AND TEMPERATURE DIFFERENCES 163 BETWEEN BULK FLUID AND SURFACE OF A CATALYST PARTICLE... 

Thus the activation energy, in principle, depends on whether the rate equation is expressed in terms of concentrations or partial pressures. Also, the difference between and Ep depends on the temperature. In practice this difference is not significant. In this example, at a temperature of ITC,... 

It is evident that CL is a function of Henry s constant and the concentration of the gas (or its partial pressure). It is clear from Figure 1.1 that Henry s constant increases till the temperature of 100 °C and then decreases. As a result, under constant gas-phase concentration (or partial pressure), the solubility of oxygen decreases up to 100 °C and then increases. Consequently, the common belief of continuous decrease in the solubility of gases in water by increasing temperature is true up to a point (Debellefontaine et al., 1996, 2000). There is a certain temperature, different for each gas species, above which this picture is reversed. Moreover, above a certain temperature, the volatilization of water decreases the partial pressure of the gas, and thus the solubility could be further decreased even if Henry s constant is decreased at the same time. 

Two important ways in which heterogeneously catalyzed reactions differ from homogeneous counterparts are the definition of the rate constant k and the form of its dependence on temperature T. The heterogeneous rate equation relates the rate of decline of the concentration (or partial pressure) c of a reactant to the fraction / of the catalytic surface area that it covers when adsorbed. Thus, for a first-order reaction,... 

The driving forces, or driving potentials, for transport phenomena are (i) the temperature difference for heat transfer (ii) the concentration or partial pressure difference for mass transfer and (iii) the difference in momentum for momentum transfer. When the driving force becomes negligible, then the transport phenomenon will cease to occur, and the system will reach equilibrium. 

The amount of adsorption is limited by the available surface and pore volume, and depends also on the chemical natures of the fluid and solid. The rate of adsorption also depends on the amount of exposed surface but, in addition, on the rate of diffusion to the external surface and through the pores of the solid for accessing the internal surface which comprises the bulk of the surface. Diffusion rates depend on temperature and differences in concentration or partial pressures. The smaller the particle size, the greater is the utilization of the internal surface, but also the greater the pressure drop for flow of bulk fluid through a mass of the particles. 

The factor introduces into equation (9) an explicit dependence of m on the concentration of species 1 in the gas adjacent to the interface [see equation (B-78)]. Except for this difference, equation (9) contains the same kinds of parameters as does equation (6), since the coefficient a can be analyzed from the viewpoint of transition-state theory. Although a may depend in general on and the pressure and composition of the gas at the interface, a reasonable hypothesis, which enables us to express a in terms of kinetic parameters already introduced and thermodynamic properties of species 1, is that a is independent of the pressure and composition of the gas [a = a(7])]. Under this condition, at constant 7] the last term in equation (9) is proportional to the concentration j and the first term on the right-hand side of equation (9) is independent of. Therefore, by increasing the concentration (or partial pressure) of species 1 in the gas, the surface equilibrium condition for species 1—m = 0—can be reached. If Pi e(T denotes the equilibrium partial pressure of species 1 at temperature 7], then when m = 0, equation (9) reduces to... 

Hence, if no experimental value for AuHi is available (i.e., from measurements of Ka( ) at different temperatures), it can be obtained from experimental (or estimated) AvapH, and Hfe values. Finally, we should note that Eq. 6-8 applies in a strict sense only if we express the amount of the compound in the gas and liquid phase as partial pressure and mole fraction, respectively. However, if we assume that the molar volume of the liquid, Vi, is not significantly affected by temperature changes, we may also apply Eq. 6-8 to describe the temperature dependence of K,n ( ) (Eq. 6-5) with a constant term that is given by constant + In Ve Furthermore, if we express the amount of the compound in the gas phase in molar concentrations (Eq. 6-6), then we have to add the term RTm to Aa< // where rav (in K) is the average temperature of the temperature range considered (see Section 3.4) ... 

For these reactions, the equilibrium mixture will not have a lot of products present mostly reactants are present at equilibrium. If we define tbe change that must occur in terms of x as the amount (molarity or partial pressure) of a reactant that must react to reach equilibrium, then x must be a small number because fC is a very small number. We want to know the value of x in order to solve the problem, so we don t assume = 0. Instead, we concentrate on the equilibrium row in the ICE table. Those reactants (or products) that have equilibrium concentrations in the form of 0.10 — x or 0.25 + or 3.5 — 3x, etc., is where an important assumption can be made. The assumption is that because K 1, x will be small (x 1) and when we add x or subtract x from some initial concentration, it will make little or no difference. That is, we assume that 0.10 — X 0.10 or 0.25 + x 0.25 or 3.5 — 3x 3.5 we assume that the initial concentration of a substance is equal to the equilibrium concentration. This assumption makes the math much easier and usually gives a value of x that is well within 5% of the true value of x (we get about the same answer with a lot less work). When the 5% rule fails, the equation must be solved exactly or by using the method of successive approximations (see Appendix A1.4). 39. [CO2] = 0.39 M [CO] = 8.6 X 10 M [O2] = 4.3 x 10 M 41. 66.0% 43. a. 1.5 X 10 atmb. Pco = Pci = 1-8 X 10 atm Pcoci2 = 5.0 atm 45. Only statement d is correct. Addition of a catalyst has no effect on the equilibrium position the reaction just reaches equilibrium more quickly. Statement a is false for reactants that are either solids or liquids (adding more of these has no effect on the equilibrium). Statement b is false always. If temperature remains constant, then the value of K is constant. Statement c is false for exothermic reactions where an increase in temperature decreases the value of K. 47. a. no effect b. shifts left c. shifts right 49. H " + OH — H2O sodium hydroxide (NaOH) will react with the H " on the product side of the reaction. This effectively removes H " from the equilibrium, which will shift the reaction... 

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