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Equilibrium values, thermodynamic

Next we consider how to evaluate the factor 6p. We recognize that there is a local variation in the Gibbs free energy associated with a fluctuation in density, and examine how this value of G can be related to the value at equilibrium, Gq. We shall use the subscript 0 to indicate the equilibrium value of free energy and other thermodynamic quantities. For small deviations from the equilibrium value, G can be expanded about Gq in terms of a Taylor series ... [Pg.681]

The equilibrium ratios of hydrogen-to-hydrogen sulfide for the reaction, derived (34) from available thermodynamic data (35), are plotted in Figure 10 as a function of temperature. When Ph2/Ph2s over the catalyst is less than the equilibrium value, the nickel can be sulfided and hence poisoned. Conversely, when this ratio is greater than the equi-... [Pg.25]

Many authors have observed that the cis-trans ratio of the products of the metathesis reaction is equal to the thermodynamic equilibrium value. This suggests that the reaction is not highly stereoselective. However, under certain conditions the product distribution is influenced by kinetic factors. For instance, it proves to be possible to prepare from cyclopentene... [Pg.157]

Figure 2.19 provides the thermodynamic equilibrium data for methane decomposition reaction. At temperatures above 800°C, molar fractions of hydrogen and carbon products approach their maximum equilibrium value. The effect of pressure on the molar fraction of H2 at different temperatures is shown in Figure 2.20. It is evident that the H2 production yield is favored by low pressure. The energy requirement per mole of hydrogen produced (37.8 kj/mol H2) is significantly less than that for the SMR reaction (68.7 kj/mol H2). Owing to a relatively low endothermicity of the process, <10% of the heat of methane combustion is needed to drive the process. In addition to hydrogen as a major product, the process produces a very important by-product clean carbon. Because no CO is formed in the reaction, there is no need for the WGS reaction and energy-intensive gas separation stages. Figure 2.19 provides the thermodynamic equilibrium data for methane decomposition reaction. At temperatures above 800°C, molar fractions of hydrogen and carbon products approach their maximum equilibrium value. The effect of pressure on the molar fraction of H2 at different temperatures is shown in Figure 2.20. It is evident that the H2 production yield is favored by low pressure. The energy requirement per mole of hydrogen produced (37.8 kj/mol H2) is significantly less than that for the SMR reaction (68.7 kj/mol H2). Owing to a relatively low endothermicity of the process, <10% of the heat of methane combustion is needed to drive the process. In addition to hydrogen as a major product, the process produces a very important by-product clean carbon. Because no CO is formed in the reaction, there is no need for the WGS reaction and energy-intensive gas separation stages.
Thermodynamic calculations based on the compositional dependence of the equilibrium constant are applied to solubility data in the KCl-KBr-H20 system at 25°C. The experimental distribution coefficient and activity ratio of Br /Cl in solution is within a factor of two of the calculated equilibrium values for compositions containing 19 to 73 mole percent KBr, but based on an assessment of uncertainties in the data, the solid solution system is clearly not at equilibrium after 3-4 weeks of recrystallization. Solid solutions containing less than 19 and more than 73 mole percent KBr are significantly farther from equilibrium. As the highly soluble salts are expected to reach equilibrium most easily, considerable caution should be exercised before reaching the conclusion that equilibrium is established in other low-temperature solid solution-aqueous solution systems. [Pg.561]

This figure shows that the equilibrium values Xf, yf depend upon the solid/water ratio. There is no thermodynamic inconsistency in this because, in terms of Fig. 1 all that Fig. 2 implies is that X-f for two solid/water ratios will yield two values of Xj, each with its thermodynamically corresponding Yf (see Figure... [Pg.646]

Although classical thermodynamics can treat only limiting cases, such a restriction is not nearly as severe as it may seem at first glance. In many cases, it is possible to approach equilibrium very closely, and the thermodynamic quantities coincide with actual values, within experimental error. In other simations, thermodynamic analysis may rule out certain reactions under any conditions, and a great deal of time and effort can be saved. Even in their most constrained applications, such as limiting solutions within certain boundary values, thermodynamic methods can reduce materially the amount of experimental work necessary to yield a definitive answer to a particular problem. [Pg.6]

In the above example e is equal to 0.830. The AG of the overall reaction can be expressed in terms Of the corresponding battery voltage and for the hydrogen-oxygen reaction at 25 C, its value is 1.229 V. As the temperature increases this thermodynamic equilibrium value will decrease by a factor of 0.84 mV per C. If the water produced remains in the gas phase, the ratio of AG /AH increases to 0.911. So as we see, these values are much higher than what can be obtained by a heat engine where the efficiency is defined by the ratio of the temperature difference of the hot... [Pg.304]

In contrast to the kinetic approach, deviations from the terminal model have also been treated from a thermodynamic viewpoint [Kruger et al., 1987 Lowry, 1960 Palmer et al., 2000, 2001]. Altered copolymer compositions in certain copolymerizations are accounted for in this treatment in terms of the tendency of one of the monomers (M2) to depropagate. An essential difference between the kinetic and thermodynamic treatments is that the latter implies that the copolymer composition can vary with the concentrations of the monomers. If the concentration of monomer M2 falls below its equilibrium value [M]c at the particular reaction temperature, terminal M2 units will be prone to depropagate. The result would be a... [Pg.515]

The formation of an oxide layer is thermodynamically favourable and kinetically rapid at room temperature, but as the temperature rises, the free energy of oxide formation (originally negative) increases to the point where the metal, oxide and oxygen are in equilibrium. At temperatures above this equilibrium value, and if the oxygen partial pressure is low enough, the oxide can decompose. [Pg.495]

Within non-equilibrium thermodynamics, the driving force for relaxation is provided by deviations in the local chemical potential from it s equilibrium value. The rate at which such deviations relax is determined by the dominant kinetics in the physical system of interest. In addition, the thermal noise in the system randomly generates fluctuations. We thus describe the dynamics of a step edge by the equation. [Pg.17]

It is apparent that CMC values can be expressed in a variety of different concentration units. The measured value of cCMC and hence of AG c for a particular system depends on the units chosen, so some uniformity must be established. The issue is ultimately a question of defining the standard state to which the superscript on AG C refers. When mole fractions are used for concentrations, AG c directly measures the free energy difference per mole between surfactant molecules in micelles and in water. To see how this comes about, it is instructive to examine Reaction (A) —this focuses attention on the surfactant and ignores bound counterions — from the point of view of a phase equilibrium. The thermodynamic criterion for a phase equilibrium is that the chemical potential of the surfactant (subscript 5) be the same in the micelle (superscript mic) and in water (superscript W) n = n. In general, pt, = + RTIn ah in which... [Pg.372]

The fundamental question in transport theory is Can one describe processes in nonequilibrium systems with the help of (local) thermodynamic functions of state (thermodynamic variables) This question can only be checked experimentally. On an atomic level, statistical mechanics is the appropriate theory. Since the entropy, 5, is the characteristic function for the formulation of equilibria (in a closed system), the deviation, SS, from the equilibrium value, S0, is the function which we need to use for the description of non-equilibria. Since we are interested in processes (i.e., changes in a system over time), the entropy production rate a = SS is the relevant function in irreversible thermodynamics. Irreversible processes involve linear reactions (rates 55) as well as nonlinear ones. We will be mainly concerned with processes that occur near equilibrium and so we can linearize the kinetic equations. The early development of this theory was mainly due to the Norwegian Lars Onsager. Let us regard the entropy S(a,/3,. ..) as a function of the (extensive) state variables a,/ ,. .. .which are either constant (fi,.. .) or can be controlled and measured (a). In terms of the entropy production rate, we have (9a/0f=a)... [Pg.63]

Introduced in this manner, the order parameter, , is a hidden thermodynamic variable its equilibrium values, (T), are not independent but are fixed by Eq. 17.3. Therefore, an order parameter is characteristic of the transformation process because it cannot be fixed by an experimental condition. [Pg.421]

How could we have predicted which product would be favored The first step is to decide whether the prediction is to be based on (1) which of the two products is the more stable, or (2) which of the two products is formed more rapidly. If we make a decision on the basis of product stabilities, we take into account AH° values, entropy effects, and so on, to estimate the equilibrium constants Keq for the reactants and each product. When the ratio of the products is determined by the ratio of their equilibrium constants, we say the overall reaction is subject to equilibrium (or thermodynamic) control. Equilibrium control requires that the reaction be reversible. [Pg.374]

Corresponding to these three problems our paper is divided into three parts. In the first (Sections 3-6) is set forth the experimental material on the amount of nitric oxide formed in explosions of mixtures of various combustibles with oxygen and nitrogen. It is shown that heating the mixture is equivalent to adding a corresponding amount of combustible. Thermodynamic computations show that in all cases the observed amount of nitric oxide is less than the equilibrium value at the maximum temperature of the gas. Direct experiments on the combustion of a gas in a stream have proved that the nitric oxide is not formed during but only after combustion. Thus, in Part I the thermal nature of the reaction is proved. [Pg.365]

Equations (11) and (13) are two of the most important thermodynamic relationships for biochemists to remember. If the concentrations of reactants and products are at their equilibrium values, there is no change in free energy for the reactions going in either direction. Living cells, however, maintain some compounds at concentrations far from the equilibrium values, so that their reactions are associated with large changes in free energy. We expand on this point in chapter 11. [Pg.39]

For Plant (3) the exit conversions and yields of methane and carbon dioxide obtained by all three models are much lower than the equilibrium values. Therefore, Plant (3) is run far from its thermodynamic equilibrium. Large differences between the predictions of the three models exist in the data the exit conversion simulations of methane differ by 16 to 23% that of the carbon dioxide yield by 12 to 18%. Since the dusty gas model is the more rigorous one, we can use its simulation output as a base for comparison in place of experimental or industrial data which is unavailable in this case. [Pg.498]

If the reaction includes a mechanism for equilibration of the products, e.g. by direct interconversion following their formation (B = C), the ratio [B]/[C] will not remain constant but drifts towards the equilibrium value. This is easily detected by product analysis during the course of the reaction. Scheme 4.2 with no product interconversion corresponds to kinetic control and the alternative corresponds to thermodynamic ox equilibrium) control, see Chapter 2. [Pg.86]

A more detailed picture of hydroisomerization of n-octane and n-nonane is given in Table II in which product distributions are listed for different degrees of conversion along with thermodynamic equilibrium values. The latter have been calculated from Gibbs free energy data available in literature (F7) the accuracy of which, however, is not known. From Table II the following conclusions may be drawn ... [Pg.13]

Hydroisomerization proceeds towards thermodynamic equilibrium which is approximately reached between the normal, mono-branched and di-branched structures at high degrees of overall conversion. Hydrocracking, however, is severe under these conditions. It is evident from Table II that monomethyl isomers are primary products the same is apparently true for monoethyl isomers although due to thermodynamic reasons lower concentrations are obtained. Dimethyl isomers including those containing a quarternary carbon atom are formed as secondary products. However, trimethyl isomers are formed very slowly so that their concentrations do not reach equilibrium values. It follows from this that the number of ramifications is deciding as to whether a branched isomer is a primary, secondary or tertiary product in hydroisomerization of n-octane and n-nonane. [Pg.13]


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