Partial pressure equilibrium expressions

S is the amount of substance (gas) per unit volume of solvent (polymer) in equilibrium with a unit partial pressure, as expressed in the equation ... 

Heterogeneous reaction rates on particles in polar stratospheric clouds (PSCs) are more difficult to estimate because of the uncertainties in the type of PSC particles present in the lower stratosphere. The particle composition, volume density and radius must be derived from a thermodynamic model (e.g., Carslaw cl at, 1994 1997). Condensation of HNO3 occurs when the partial pressure of nitric acid in air exceeds the equilibrium vapor pressure Pjjnq F°r example, in the case of nitric acid trihydrates (NAT), if partial pressures are expressed in Torr and the temperature T in Kelvin, we have (Hanson and Mauersberger, 1988)... 

Kp, the partial pressure equilibrium constant, is the relationship that exists among the partial pressures of gaseous reactants and products in a reversible reaction at equilibrium. Partial pressures are expressed in bar (after 1982) or atm (before 1982). 

Experiments on sufficiently dilute solutions of non-electrolytes yield Henry s laM>, that the vapour pressure of a volatile solute, i.e. its partial pressure in a gas mixture in equilibrium with the solution, is directly proportional to its concentration, expressed in any units (molar concentrations, molality, mole fraction, weight fraction, etc.) because in sufficiently dilute solution these are all proportional to each other. 

Other conventions for treating equiUbrium exist and, in fact, a rigorous thermodynamic treatment differs in important ways. Eor reactions in the gas phase, partial pressures of components are related to molar concentrations, and an equilibrium constant i, expressed directiy in terms of pressures, is convenient. If the ideal gas law appHes, the partial pressure is related to the molar concentration by a factor of RT, the gas constant times temperature, raised to the power of the reaction coefficients. 

The equilibrium eonstant for ammonia synthesis is expressed as a funetion of the partial pressure as... 

It turns out that there is a relatively simple relationship between the partial pressures of different gases present at equilibrium in a reaction system. This relationship is expressed in terms of a quantity called the equilibrium constant, symbol K In this chapter, you will learn how to—... 

This equilibrium constant is often given the symbol Kp to emphasize that it involves partial pressures. Other equilibrium expressions for gases are sometimes used, including Kc ... 

Strategy First write the chemical equation for the equilibrium system. Then write the expression for K, leaving out pure liquids and solids. Remember that gases are represented by their partial pressures. 

Strategy First write the expression for K. Then substitute equilibrium partial pressures into that expression and solve. 

The form of the expression for Q, known as the reaction quotient, is the same as that for the equilibrium constant, K. The difference is that the partial pressures that appear in Q are those that apply at a particular moment, not necessarily when the system is at equilibrium. By comparing the numerical value of Q with that of K, it is possible to decide in which direction the system will move to achieve equilibrium. 

In air, the partial pressures of N2 and 02 are about 0.78 and 0.21 atm, respectively. The expression for K can be used to calculate the equilibrium partial pressure of NO under these conditions ... 

Express the equilibrium partial pressures of all species in terms of a single unknown, x. To do this, apply the principle mentioned earlier The changes in partial pressures of reactants and products are related through tite coefficients of the balanced equation. To keep track of these values, make an equilibrium table, like the one illustrated in Example 12.4. 

You must express all equilibrium partial pressures in terms of a single unknown, x. A reasonable choice for x is... 

The equation just written is generally applicable to any system. The equilibrium constant may be the K referred to in our discussion of gaseous equilibrium (Chapter 12), or any of the solution equilibrium constants (Rw Ra, Rj, K, . . . ) discussed in subsequent chapters. Notice that AG° is the standard free energy change (gases at 1 atm, species in solution at 1M). That is why, in the expression for K, gases enter as their partial pressures in atmospheres and ions or molecules in solution as their molarities. 

Reaction quotient (Q) An expression with the same form as Kbut involving arbitrary rather than equilibrium partial pressures, 333-334 Reaction rate The ratio of the change in concentration of a species divided by the time interval over which the change occurs, 285 catalysis for, 305-307 collision model, 298-300 concentration and, 287-292,314q constant, 288 enzymes, 306-307 egression, 288... 

NOTE The operational efficiency of a deaerator is governed by its potential to reduce the partial pressure of oxygen in the FW to a minimum. The equilibrium concentration of any gas (such as Of) dissolved in a liquid (i.e., the gas s solubility) is proportional to the partial pressure of that gas in contact with the liquid. This relationship is expressed by Henry s Law ... 

In a packed column, operating at approximately atmospheric pressure and 295 K, a 10% ammonia-air mixture is scrubbed with water and the concentration of ammonia is reduced to 0.1%. If the whole of the resistance to mass transfer may be regarded as lying within a thin laminar film on the gas side of the gas-liquid interface, derive from first principles an expression for the rate of absorption at any position in the column. At some intermediate point where the ammonia concentration in the gas phase has been reduced to 5%. the partial pressure of ammonia in equilibrium with the aqueous solution is 660 N/nr and the transfer rate is ]0 3 kmol/m2s. What is the thickness of the hypothetical gas film if the diffusivity of ammonia in air is 0.24 cm2/s ... 

To express the composition of the vapor in equilibrium with the liquid phase of a binary liquid mixture, we first note that the definition of partial pressure (PA = xAP for component A) and Dalton s law (P = PA + PB) allow us to express the composition of the vapor of a mixture of liquids A and B in terms of the partial pressures of the components ... 

We use a different measure of concentration when writing expressions for the equilibrium constants of reactions that involve species other than gases. Thus, for a species J that forms an ideal solution in a liquid solvent, the partial pressure in the expression for K is replaced by the molarity fjl relative to the standard molarity c° = 1 mol-L 1. Although K should be written in terms of the dimensionless ratio UJ/c°, it is common practice to write K in terms of [J] alone and to interpret each [JJ as the molarity with the units struck out. It has been found empirically, and is justified by thermodynamics, that pure liquids or solids should not appear in K. So, even though CaC03(s) and CaO(s) occur in the equilibrium... 

A note on good practice In some cases you will see an equilibrium constant denoted KP to remind you that it is expressed in terms of partial pressures. However, the subscript P is unnecessary. 

Now we come to the most important point in this chapter. At equilibrium, the activities (the partial pressures or molarities) of all the substances taking part in the reaction have their equilibrium values. At this point the expression for Q (in which the activities have their equilibrium values) has become the equilibrium constant, K, of the reaction. That is,... 

STRATEGY At equilibrium, the partial pressures of the reactants and products (in bar, l bar = 10s Pa) satisfy the expression for K. Therefore, rearrange the expression to give the one unknown concentration, and substitute the data. 

SOLUTION First convert the units of partial pressures to bars using 1 bar = 105 Pa P i7 = 4.2 X 10 8 bar, Pcli = 8.3 X 10 8 bar. Write the expression for the equilibrium constant from the chemical equation given. 

We have seen that the value of an equilibrium constant tells us whether we can expect a high or low concentration of product at equilibrium. The constant also allows us to predict the spontaneous direction of reaction in a reaction mixture of any composition. In the following three sections, we see how to express the equilibrium constant in terms of molar concentrations of gases as well as partial pressures and how to predict the equilibrium composition of a reaction mixture, given the value of the equilibrium constant for the reaction. Such information is critical to the success of many industrial processes and is fundamental to the discussion of acids and bases in the following chapters. 

Step 1 Write the balanced chemical equation for the equilibrium and the corresponding expression for the equilibrium constant. Then set up an equilibrium table as shown here, with columns labeled by the species taking part in the reaction. In the first row, show the initial composition (molar concentration or partial pressure) of each species... 

STRATEGY The general procedure is like that set out in Toolbox 9.1. Write the expression for the equilibrium constant, and then set up an equilibrium table. In this case, use as the initial partial pressures those found immediately after addition of the reagent, but before the reaction has responded. Because a product has been added, we expect the reaction to respond by producing more reactants. Obtain the initial equilibrium concentrations and the value of K from Example 9.8. 

Although Kp for a gas phase reaction is defined in terms of the partial pressures of reactants and products, at times it is more convenient to express the equilibrium constant for... 

We will list the elementary steps and decide which is rate-limiting and which are in quasi-equilibrium. For ammonia synthesis a consensus exists that the dissociation of N2 is the rate-limiting step, and we shall make this assumption here. With quasi-equilibrium steps the differential equation, together with equilibrium condition, leads to an expression for the coverage of species involved in terms of the partial pressures of reactants, equilibrium constants and the coverage of other intermediates. 

All coverages of adsorbed species (which, of course, can not necessarily be measured experimentally) can now be expressed as equilibrium constants and partial pressures. The reader may verify that this leads to... 

For different acceptor particle adsorption isotherms expressions (1.85) - (1.89) provide various dependencies of equilibrium values of <7s for a partial pressure P (ranging from power indexes up to exponential). Thus, in case when the logarithmic isotherm Nt InP is valid the expression (1.85 ) leads to dependence <75 P" often observed in experiments [20, 83, 155]. In case of the Freundlich isotherm we arrive to the same type of dependence of - P" observed in the limit case described by expression (1.87). 

Therefore, the activation energy of quasi-equilibrium conductivity changes as a logarithm of concentration of adsorption particles which, when the linear dependence between Nt and P is available, corresponds to situation observed in experiment [155]. We should note that due to small value m function (1.91) satisfactorily approximates the kinetics oit) A - B n(i + t/t>) observed in experiments [51, 167, 168]. Moreover, substantially high partial pressures of acceptor gas, i.e. at high concentrations of Nt expression (1.81) acquires the shape ait) Oait/toc) it,Nty " when t>toc>. This suggests that for... 

Alternatively, by making use of the partial pressures of the species involved instead of their concentrations, the expression for the equilibrium constant may be obtained as... 

The expression for K involving the concentrations of the species involved is found to be independent of volume. This implies that any change of pressure is not going to change the final state of equilibrium. The same result can be obtained by taking into consideration the alternative expression involving the partial pressures. If the pressure on the system is increased to n times its original value then all the partial pressures will be increased in the same proportion. This obviously implies that the equilibrium is independent of the pressure. The effect of some other factors on this reaction may now be considered. One such factor can be the addition of substances. For example, on addition of more A2, the partial pressure of A2 in the reactor would increase momentarily from pAl to some value, p A/. It has already been seen that... 

Similar relationships can be written for the dissolution of hydrogen and oxygen. These relationships are expressions of Sievert s law which can be stated thus the solubility of a diatomic gas in a liquid metal is proportional to the square root of its partial pressure in the gas in equilibrium with the metal. The Sievert s law behaviour of nitrogen in niobium is illustrated in Figure 3.8. The law predicts that the amount of a gas dissolved in a metal can be reduced merely by reducing the partial pressure of that gas, as for example, by evacuation. In practice, however, degassing is not as simple as this. Usually, Sievert s law is obeyed in pure liquid metals only when the solute gas is present in very low concentrations. At higher concentrations deviations from the law occur. 

In contrast to kinetic models reported previously in the literature (18,19) where MO was assumed to adsorb at a single site, our preliminary data based on DRIFT results suggest that MO exists as a diadsorbed species with both the carbonyl and olefin groups being coordinated to the catalyst. This diadsorption mode for a-p unsaturated ketones and aldehydes on palladium have been previously suggested based on quantum chemical predictions (20). A two parameter empirical model (equation 4) where - rA refers to the rate of hydrogenation of MO, CA and PH refer to the concentration of MO and the hydrogen partial pressure respectively was developed. This rate expression will be incorporated in our rate-based three-phase non-equilibrium model to predict the yield and selectivity for the production of MIBK from acetone via CD. 

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