Equilibrium constant numerical examples

Being able to write an expression relating equilibrium concentrations in chemical reactions provides a powerful tool for understanding chemical equilibrium. We will soon see how to evaluate these expressions for several classes of reactions. But first, let s look at what we can learn from the numerical values of equilibrium constants. For example, we could ask the question, Which reaction is more likely to produce hydrogen gas at relatively low temperature—the BMA process or the Fluohmic process How can we use the equilibrium constants for these reactions to find the answer ... 

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. 

Now look at the numerical values of the equilibrium constants. The K s listed range from 10+1 to 10 16, so we see there is a wide variation. We want to acquire a sense of the relation between the size of the equilibrium constant and the state of equilibrium. A large value of K must mean that at equilibrium there are much larger concentrations present of products than of reactants. Remember that the numerator of our equilibrium expression contains the concentrations of the products of the reaction. The value of 2 X 10,s for the K for reaction (19) certainly indicates that if a reaction is initiated by placing metallic copper in a solution containing Ag+ (for example, in silver nitrate solution), when equilibrium is finally reached, the concentration of Cu+2 ion, [Cu+2], is very much greater than the square of the silver ion concentration, [Ag+]2. 

The direction chosen for the equilibrium reaction Is determined by convenience. A scientist interested in producing ammonia from N2 and H2 would use f. On the other hand, someone studying the decomposition of ammonia on a metal surface would use eq,r Either choice works as long as the products of the net reaction appear in the numerator of the equilibrium constant expression and the reactants appear in the denominator. Example applies this reasoning to the iodine-triiodide reaction. 

Another problem which can appear in the search for the minimum is intercorrelation of some model parameters. For example, such a correlation usually exists between the frequency factor (pre-exponential factor) and the activation energy (argument in the exponent) in the Arrhenius equation or between rate constant (appears in the numerator) and adsorption equilibrium constants (appear in the denominator) in Langmuir-Hinshelwood kinetic expressions. 

The method of representing the equilibrium state The value of the equilibrium constant changes if the reversible reaction is considered to proceed in the reverse direction. For example, the reaction A + B C + D can also be written asC + D A+Bso that [A] [B]/[C] [D] = kr/kf = K. In this case the value of the equilibrium constant for the reverse reaction is given by K = 1/K. To avoid such confusion while applying the law of mass action, the concentrations of the products are always placed in the numerator and those of the reactants in the denominator. 

As seen in equations (32)-(34), the forward adsorptive flux depends upon the concentration of free cell surface carriers. Unfortunately, there is only limited information in the literature on determinations of carrier concentrations for the uptake of trace metals. In principle, graphical and numerical methods can be used to determine carrier numbers and the equilibrium constant, As, corresponding to the formation of M — Rcen following measurement of [M] and (M —Rceii. For example, a (Scatchard) plot of (M — RCeii /[M] versus (M — RCeii should yield a straight line with a slope equal to the reciprocal of the dissociation constant and abscissa-intercept equal to the total carrier numbers (e.g. [186]). 

In Cleland nomenclature, the initial velocity and individual rate constants are designated by lower case italicized letters (e.g., v, k, k2, etc.). Dissociation, Michaelis, and equilibrium constants utilize an upper case italicized K with the appropriate unitalicized lower case subscript. For example, the equilibrium constant would be symbolized by whereas the Michaelis constant for substrate B would be designated by K. Dissociation constants for a Michaelis complex contain a subscript i and a letter for the dissociating ligand (e.g., for the EA binary complex, the dissociation constant would be Ki ). Maximum velocities are designated by a capital italicized V, usually with a subscript 1 or 2 depending on whether the forward or reverse reaction is referred to. (If the numerical subscript is not provided, the forward reaction is assumed. In most cases, the unitalicized subscript max is also provided.)... 

Values for equilibrium constants have been tabulated for numerous reactions occurring at various temperatures. Because many of these constants are associated with industrial processes, the temperature at which a is reported varies considerably. That is, it might seem odd that a value for a reaction is given at what seems like an arbitrary temperature, but it should be remembered that these constants have probably been determined with regard to a specific industrial process, for example, 1,000°C for the production of hydrogen gas from steam and coke (carbon). 

The forward reaction is a three-body reaction with values of k ) = 2.2 X 10-3() cm6 molecule-2 s-1 and kx = 1.5 X 10-12 cm3 molecule-1 s-1 at 300 K with Fc = 0.6 (see Chapter 5.A.2 for discussion of this factor) recommended by DeMore et al. (1997). Atkinson et al. (1997b) recommend kf) = 2.8 X 10-3(1 cm6 molecule-2 s-1 and k = 2.0 X 10—12 cm3 molecule-1 s-1 with Fc = 0.45. The recommended value of the equilibrium constant Kt) t) is 2.9 X 10-11 cm3 molecule-1 at 298 K, with an uncertainty of 30% (DeMore et al., 1997). This equilibrium constant has been the subject of numerous experimental studies, which have yielded results that ranged over a factor of two at room temperature. For example, a study by Wangberg et al. (1997) subsequent to the NASA and IUPAC recommendations reports a value of 2.34 X 10" " cm3 molecule-1, about 20% smaller but within the relatively large uncertainty of the recommended value. 

Note that reactions 2.14, 2.15, and 2.23 involve fractional stoichiometric coefficients on the left-hand sides. This is because we wanted to define conventional enthalpies of formation (etc.) of one mole of each of the respective products. However, if we are not concerned about the conventional thermodynamic quantities of formation, we can get rid of fractional coefficients by multiplying throughout by the appropriate factor. For example, reaction 2.14 could be doubled, whereupon AG° becomes 2AG, AH° = 2AH , and AS° = 2ASf, and the right-hand sides of Eqs. 2.21 and 2.22 must be squared so that the new equilibrium constant K = K2 = 1.23 x 1083 bar-3. Thus, whenever we give a numerical value for an equilibrium constant or an associated thermodynamic quantity, we must make clear how we chose to define the equilibrium. The concentrations we calculate from an equilibrium constant will, of course, be the same, no matter how it was defined. Sometimes, as in Eq. 2.22, the units given for K will imply the definition, but in certain cases such as reaction 2.23 K is dimensionless. 

Ion exchange materials have equilibrium exchange capacities of about 5 meq/g or 2.27geq/lb. The percentage of equilibrium exchange that can be achieved practically depends on contact time, the concentration of the solution, and the selectivity or equilibrium constant of the particular system. The latter factor is discussed in Section 15.2 with a numerical example. 

The numerical value of the equilibrium constant for a reaction indicates the extent to which reactants are converted to products that is, it measures how far the reaction proceeds before the equilibrium state is reached. Consider, for example, the reaction of H2 with 02, which has a very large equilibrium constant (Kc = 2.4 X 1047 at 500 K) ... 

Because products appear in the numerator of the equilibrium constant expression and reactants are in the denominator, a very large value of Kc means that the equilibrium ratio of products to reactants is very large. In other words, the reaction proceeds nearly to completion. For example, if stoichiometric amounts of H2 and 02 are allowed to react and [H20] = 1.0 M at equilibrium, then the concentrations of H2 and 02 that remain at equilibrium are negligibly small [H2] = 2.0 X 10-16M and [02] = 1.0 X 10-16M. (Try substituting these concentrations into the equilibrium equation to show that they satisfy the equation.)... 

The search for relationships among the dynamic and equilibrium properties of related series of compounds has been a paradigm of chemists for many years. The discovery of such unifying principles and predictive relationships has practical benefits. Numerous relationships exist among the structural characteristics, physicochemical properties, and/or biological qualities of classes of related compounds. Perhaps the best-known attribute relationships are the correlations between reaction rate constants and equilibrium constants for related reactions commonly known as linear tree-energy relationships (LFERs). The LFER concept led to the broader concepts of QSARs, which seek to predict the environmental fate of related compounds based on correlations between their bioactivity or physicochemical properties and structural features. For example, therapeutic response, environmental fate, and toxicity of organic compounds have been correlated with... 

With an equilibrium constant expression and a numerical value for K, we can compute the composition of an equilibrium mixture. Problems are often stated in terms of initial amounts of products and/or reactants. It is usually convenient to set up a table with a column for each reactant and product species, entering (i) the initial concentrations or partial pressures (ii) the change in each concentration or pressure in terms of an unknown x and (iii) the equilibrium concentrations or partial pressures in terms of x. Substitution of the expressions in row (iii) into the equilibrium constant expression leads to an equation in x, the root of which allows calculations of the equilibrium concentrations or partial pressures. The procedure is best illustrated by examples. 

A system is in a state of equilibrium when there is a balance between reactants and products. This balance is defined by thermodynamic parameters, namely bond strengths and the intermolecular forces between all the molecules in the system. The equilibrium constant (K) is the numerical description of that balance. K is equal to the product of all of the molar concentrations of the products, each raised to the power of their stoichiometric coefficients, divided by the product of the molar concentration of the reactants, each raised to the power of their stoichiometric coefficients. This sounds a lot worse than it is. For example, with this model reaction ... 

The equilibrium constant at 333 K in the numerical example considered above is... 

The Law of Chemical Equilibrium is based on the constancy of the equilibrium constant. This means that if one disturbs the equilibrium, for example by adding more reactant molecules, there will be an increase in the number of product molecules in order to maintain the producl/reactant ratio unchanged and thus preserving the numerical value of the equilibrium constant. The Le Chatelier Principle expresses this as follows If an external stress is applied to a system in equilibrium, the system reacts in such a way as to partially relieve the stress. In our present experiment, we demonstrate the Le Chatelier Principle in two manners (a) disturbing the equilibrium by changing the concentration of a product or reactant (b) changing the temperature. 

It should also be noted however that, under certain conditions no division by c° is necessary, for example because the equilibrium constant quotient contains as many concentration terms in its numerator as in its denominator. In such cases we could say that ... 

Chemical speciation in soil solutions and other natural waters can be calculated routinely with a number of software products offered in a variety of computational media.27 30 Five examples of these products are listed in Table 2.5. They differ principally in the method of solving the chemical equilibrium problem numerically, or in the chemical species and equilibrium constants considered, or in the model used to estimate single-species activity coefficients. Irrespective of these differences, all the examples follow a similar algorithm ... 

What is the numerical value of the equilibrium constant for the formation of HI (Example 2, this chapter) with partial pressures expressed in millimeters of Hg ifistead of in atmospheres What is it in terms of concentrations in moles per liter ... 

As already mentioned, the second system to be studied [116] was the 5-norbornenyl system. The radical clock-5-norbornenyl 149 [118] is known to rearrange to the 3-nortricyclenyl radical with a rate constant, determined by EPR, on the order of 6 x 10 sec at -30 C and 10 -10 see" at 25 C [123, 124]. It has been firmly established from numerous examples of free radical reductions involving this system that the equilibrium in solution is in favor of radical 150. For example, a report by Davies [125] demonstrated that whether one started from 5-bromo-2-norbornene 148 Br or from 3-bromonortricyclene 153 Br when carrying out a cobaltous chloride-catalyzed reaction with methylmagnesium bromide, one obtains a mixture of 30% 2-norbornene 151 and 70% norticyclene 152 (Scheme 45a). 

The double arrow indicates that this is an equilibrium. Recall that the equilibrium constant expression is written by placing the concentrations of the products in the numerator and the concentrations of the reactants in the denominator. In this example, all terms are to the first power because all the coefficients are 1. 

Numerous applications of standard electrode potentials have been made in various aspects of electrochemistry and analytical chemistry, as well as in thermodynamics. Some of these applications will be considered here, and others will be mentioned later. Just as standard potentials which cannot be determined directly can be calculated from equilibrium constant and free energy data, so the procedure can be reversed and electrode potentials used for the evaluation, for example, of equilibrium constants which do not permit of direct experimental study. Some of the results are of analjrtical interest, as may be shown by the following illustration. Stannous salts have been employed for the reduction of ferric ions to ferrous ions in acid solution, and it is of interest to know how far this process goes toward completion. Although the solutions undoubtedly contain complex ions, particularly those involving tin, the reaction may be represented, approximately, by... 

Note from Example 14-3 that the analytical concentrations of acid and conjugate base are identical when an acid has been half neutralized (after the addition of exactly 25.00 mL of base in this case). Thus, these terms cancel in the equilibrium-constant expression, and the hydronium ion concentration is numerically equal to the dissociation constant. Likewise, in the titration of a weak base, the hydroxide ion concentration is numerically equal to the dissociation constant of the base at the midpoint in the titration curve. In addition, the buffer capacities of each of the solutions are at a maximum at this point. These points are often called the half-titration points. 

A numerical example illustrates this discussion. Assuming the following values for the parameters, with equilibrium constants U = 5, fl2 = 10/ and a = 2.0, safety margin jS = 1.05, and phase ratio F = 0.5, we obtain Mj, = = 5 periods... 

---