Partial Pressures and the Equilibrium Constant

The concentrations [N2O4] and [NO2] both increase because of the large decrease in volume. However, the number of moles of N2O4 increases, while the number of moles of NO2 decreases. We predict this from LeChatelier s Principle. 

There Are Several Ways to Solve Etiulllbrlum Problems 

It is often more convenient to measure pressures rather than concentrations of gases. Solving the ideal gas equation, PV = nRT, for pressure gives 

The pressure of a gas is directly proportional to its concentration (n/V). For reactions in which all substances that appear in the equilibrium constant expression are gases, we sometimes prefer to express the equilibrium constant in terms of partial pressures in atmospheres (Kp) rather than in terms of concentrations (IQ. 

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Partial Pressures and the Equilibrium Constant 17-10 Relationship between Kp and... 

For a gas-phase reaction, the reactants and products can be described in terms of their partial pressures and the equilibrium constant is called A"p ... 

Numerical values of equilibrium constants can be calculated if the partial pressures of products and reactants at equilibrium are known. Sometimes you will be given equilibrium partial pressures directly (Example 12.3). At other times you will be given the original partial pressures and the equilibrium partial pressure of one species (Example 12.4). In that case, the calculation of K is a bit more difficult, because you have to calculate the equilibrium partial pressures of all the species. 

The data that are required for finding the constants of a rate equation are of the rate as a function of all the partial pressures. When the equilibrium constant also is known, y can be calculated and linear analysis suffices for determination of the constants. Otherwise, nonlinear regression or solution of selected sets of nonlinear equations must be used. 

It is sometimes advantageous to express the equilibrium constant for gaseous systems in terms of either mole fractions, or concentrations, q, rather than partial pressures. The partial pressure, pi, the mole fraction, and the total pressure, p, are related by Pi = Xip. Using this relation for each of the partial pressures in the equilibrium constant, we obtain from Eq. (11.41)... 

This is a (hfficult problem. Express the equilibrium number of moles in terms of the initial moles and the ehange in number of moles (x). Next, ealeulate the mole fraetion of each component. Using the mole fraction, you should come up with a relationship between partial pressure and total pressure for each component. Substitute the partial pressures into the equilibrium constant expression to solve for the total pressure, Pj. 

Plan We can determine the starting partial pressures of all species from the information given. We can then substitute the starting partial pressures into the equilibrium-constant expression to calculate the reaction quotient, Q. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed. 

Fifth, we substitute the equilibrium partial pressures into the equilibrium-constant expression and solve for the single unknown, x ... 

Throughout this book, we express concentrations and partial pressures within the equilibrium constant expression in units of molarity and atmospheres, respectively. When expressing the value of the equilibrium constant, however, we have not included the units. Formally, the values of concentration or partial pressure that we substitute into the equilibrium constant expression are ratios of the concentration or pressure to a reference concentration (exactly 1 M) or a reference pressure (exactly 1 atm). For example, within the equilibrium constant expression, a pressure of 1.5 atm becomes... 

Calculating Equilibrium Partial Pressures from the Equilibrium Constant and Initial Partial Pressures (14.8) Example 14.11 For Practice 14.11 Exercises 59, 60... 

What units should be used when expressing concentrations or partial pressures in the equilibrium constant What are the units of ATp and AT,. Explain. 

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. 

Equilibrium constants do not have units because in the strict thermodynamic definition of the equilibrium constant, the activity of a component is used, not its concentration. The activity of a species in an ideal mixture is the ratio of its concentration or partial pressure to a standard concentration (1 M) or pressure (1 atm). Because activity is a ratio, it is unitless and the equilibrium constant involving activities is also unitless. 

We can calculate a more realistic H2(aq) concentration from the partial pressure of H2(g) in the atmosphere (Table 6.3) and the equilibrium constant for the reaction... 

With multiple rate controlling steps, a steady state is postulated, that is, all rates are equated to the overall rate. Equations for the individual steps are formulated in terms of variables such as interfacial concentrations and various coverages of the catalyst surface. Any such variables that are not measurable are eliminated in terms of measurable partial pressures and the rate, as well as various constants to be evaluated from the data. The solved problems deal with several cases for instance, P6.03.04 has two participants not in adsorptive equilibrium and P6.06.17 treats a process with five steps. 

Many different types of reversible reactions exist in chemistry, and for each of these an equilibrium constant can be defined. The basic principles of this chapter apply to all equilibrium constants. The different types of equilibrium are generally denoted using an appropriate subscript. The equilibrium constant for general solution reactions is signified as or K, where the c indicates equilibrium concentrations are used in the law of mass action. When reactions involve gases, partial pressures are often used instead of concentrations, and the equilibrium constant is reported as (p indicates that the constant is based on partial pressures). and are used for equilibria associated with acids and bases, respectively. The equilibrium of water with the hydrogen and hydroxide ions is expressed as K. The equilibrium constant used with the solubility of ionic compounds is K p. Several of these different K expres-... 

Write the equilibrium equation by setting Kp equal to the equilibrium constant expression using partial pressures. Put the partial pressures of products in the numerator and the partial pressures of reactants in the denominator, with the pressure of each substance raised to the power of its coefficient in the balanced chemical equation. Then substitute the partial pressures into the equilibrium equation and solve for Kp. 

The reaction volume may be of interest in itself, and furthermore its determination can provide a route to the volume of activation in the reverse direction if that parameter is not experimentally accessible and when AV for the reaction in the forward direction is known. As indicated above, AV may be determined from the dependence upon pressure of the equilibrium constant. It may also be obtained under certain circumstances from the partial molar volumes of the reactants and products. Density measurements d are made on several solutions of different concentrations of the reactant(s) and the product(s). The following equation is used to obtain the apparent molar volume of each species, tp, at each molar concentration c. 

Thus the partial differential of the logarithm of Kx with respect to pressure p is equal to minus the molar expansion, dVId, divided by RT. We see then that an increase in pressure increases the equilibrium constant Kx if the reaction is accompanied by a decrease in volume CdVIdf < 0), and conversely if dVIdt > 0 the equilibrium constant is decreased. 

We can now use the five equilibrium equations, the two ratios of fictitious partial pressures, and the total pressure equation to solve for the equilibrium among the eight species at any given temperature. The initial flame temperature assumed is 3000°K. At this temperature the values of the equilibrium constants (N2) are ... 

Once again, a set of equations relating the defect concentrations to the partial pressure of X2 and the equilibrium constants Ks, IQ, and K3 are written and plotted. The result is shown in the left hand part of Figure 3. 

If H2 is produced inside the cell, AE0 is negative the Gibbs s free energy change, AG0 =-nFAE0 , becomes positive, 23.9 kJ/mol at pH 8 (well known inside pH of Escherichia coli), and the equilibrium constant, K =[NAD+] [H2]/[NADH], becomes 6.5xl0 5. This means that if H2 is produced inside the cell, only 10 " atm of partial H2 pressure will stop H2 production. But, if H2 is produced outside the cell at pH 6 and NADH is oxidized inside the cell at pH 8, AFV becomes nearly 0 and therefore H2 production is possible even at approx. 1 atm of partial pressure of H2. In fact, bacteria produced H2 continuously under a partial pressure 0.6 atm. 

Ustyugov et al. measured the total unsaturated vapour pressure of selenium-tellurium mixtures with a quartz membrane manometer. The partial pressures of Sc2(g), Tc2(g), and TeSe(g) were calculated from a knowledge of the total pressure, the amounts of substance taken, and the equilibrium constants for the side-reactions with formation of the 864, See, Seg, Te, and Te4 gaseous species. From the partial pressures presented it is found that the minor species contribute, in total, to less than 10% of the total pressure. The results can be expressed by log A °((V.48), T) = 0.6270 - 368.4 7 . A third law evaluation of these data yields A //° ((V.48), 298.15 K) = (0.1 0.2) kJ-moU. ... 

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