Using Equilibrium Expressions to Solve Problems

SECTION 15.4 Using Equilibrium Expressions to Solve Problems... 

The use of the extended Debye-Hlickel equation with the appropriate equilibrium constants for mass action expressions to solve a complex chemical equilibrium problem is known as the ion-association (lA) method. 

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. 

Many problems do not involve perfect squares. You may need to use a quadratic equation to solve the equilibrium expression. 

Using this expression for [OH ] in the Kw equilibrium enables us to solve the problem ... 

The second variation is to determine either the pH or the hydrogen ion concentration of a solution when given the hydroxide ion concentration, [OH-], for the solution. To solve these problems you need to utilize the equilibrium constant expression for the self-ionization of water (Kw). This expression will allow you to convert from the hydroxide ion concentration, [OH-], to the hydrogen ion concentration, [H+], The [H+] can then be used to calculate pH if necessary. One of the free-response questions on the 1999 test required this calculation. 

However, N202 is a reaction intermediate, which is not allowed in the rate law. In order to solve the problem, we will have to use the first step in the mechanism and find a suitable substitution for N2Oz. In the first step, which is an equilibrium step, the rates of the forward and reverse reactions are identical. Therefore, we can set up an expression such that kx = k x. In this expression, we obtain the following equality ... 

This expression is consistent with the experimental observations. For this example, the reaction equilibrium represented by step 3 is never used to solve the problem since the most abundant reaction intermediate is assumed to be H (accounted for in the equilibrated step 2.) Thus, a complex set of elementary steps is reduced to two kinetically significant reactions. 

If we wish to calculate the solubility of barium sulfate in a system containing hydronium and acetate ions, we must take into account not only the solubility equilibrium but also the other three equilibria. We find, however, that using four equilibrium-constant expressions to calculate solubility is much more difficult and complex than the simple procedure illustrated in Examples 9-4, 9-5, and 9-6. To solve this type of problem, the systematic approach described in Section llA is helpful. We then use this approach to illustrate the effect of pH and complex fonna-tion on the solubility of typical analytical precipitates. In later chapters, we use this same systematic method for solution of problems involving multiple equilibria of several types. 

If we do not know the pH, the logarithmic concentration diagram can also be used to give an approximate pH value. For example, find the pH of a 0.1 M maleic acid solution. Since the log concentration diagram expresses mass balance and the equilibrium constants, we need only one additional equation such as charge balance to solve the problem exactly. The charge-balance equation for this system is... 

Take a mixture of two or more chemicals in a temperature regime where both have a significant vapor pressure. The composition of the mixture in the vapor is different from that in the liquid. By harnessing this difference, you can separate two chemicals, which is the basis of distillation. To calculate this phenomenon, though, you need to predict thermodynamic quantities such as fugacity, and then perform mass and energy balances over the system. This chapter explains how to predict the thermodynamic properties and then how to solve equations for a phase separation. While phase separation is only one part of the distillation process, it is the basis for the entire process. In this chapter you will learn to solve vapor-liquid equilibrium problems, and these principles are employed in calculations for distillation towers in Chapters 6 and 7. Vapor-liquid equilibria problems are expressed as algebraic equations, and the methods used are the same ones as introduced in Chapter 2. 

Chemists frequently need to calculate the amounts of reactants and products present at equilibrium in a reaction for which they know the equilibrium constant. The approach in solving problems of this type is similar to the one we used for evaluating equilibrium constants We tabulate initial concentrations or partial pressures, changes in those concentrations or pressures, and final equilibrium concentrations or partial pressures. Usually we end up using the equilibrium-constant expression to derive an equation that must be solved for an unknown quantity, as demonstrated in Sample Exercise 15.11. 

This expression leads to a quadratic equation in x, which we can solve by using either an equation-solving calculator or the quadratic formula. We can simplify the problem, however, by noting that the value of K is quite small. As a result, we anticipate that the equilibrium lies far to the left and that x is much smaller than the initial concentration of acetic acid. Thus, we assume that x is negligible relative to 0.30, so that 0.30 — x is essentially equal to 0.30. We can (and should ) check the validity of this assumption when we finish the problem. By using this assumption. Equation 16.30 becomes... 

In the first example, chosen to show that it is possible to solve a whole family of problems in little more time than it takes to do a single one when one doesn t use the spreadsheet, we will learn how to label columns needed in the exercise, dealing with expressions of concentrations of aqueous hydrochloric acid, enter data of interest, and use formulas so that the data can be transformed into the desired answers to the question(s) asked. We will also learn how to prepare a graph comparing the variation of molarity and molality as a function of %HC1. Also, by determining the percent difference between M and m at various concentrations, we will be able to judge when it is reasonable to substitute the more convenient M for the more rigorous m in equilibrium expressions. 

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