Equilibrium expressions, reactions

Many organic reactions involve acid concentrations considerably higher than can be accurately measured on the pH scale, which applies to relatively dilute aqueous solutions. It is not difficult to prepare solutions in which the formal proton concentration is 10 M or more, but these formal concentrations are not a suitable measure of the activity of protons in such solutions. For this reason, it has been necessaiy to develop acidity functions to measure the proton-donating strength of concentrated acidic solutions. The activity of the hydrogen ion (solvated proton) can be related to the extent of protonation of a series of bases by the equilibrium expression for the protonation reaction. 

Some chemical reactions are reversible and, no matter how fast a reaction takes place, it cannot proceed beyond the point of chemical equilibrium in the reaction mixture at the specified temperature and pressure. Thus, for any given conditions, the principle of chemical equilibrium expressed as the equilibrium constant, K, determines how far the reaction can proceed if adequate time is allowed for equilibrium to be attained. Alternatively, the principle of chemical kinetics determines at what rate the reaction will proceed towards attaining the maximum. If the equilibrium constant K is very large, for all practical purposes the reaction is irreversible. In the case where a reaction is irreversible, it is unnecessary to calculate the equilibrium constant and check the position of equilibrium when high conversions are needed. 

Derive an equilibrium expression for the reactive absorption of HiS in diethanolamine DEA . The molarity of DEA is 2 kmol/m. The following reaction takes place ... 

The problem can be solved using the equilibrium expression for the adenylate kinase reaction ... 

In bulb B we approached equilibrium from a higher temperature (which favors NOj) then redaction (5) predominated at first. Using the same sort of argument we applied to bulb A, we see jthat as time progresses, reaction (4) becomes more and more rapid (as N204 is produced) and reaction (5) becomes slower (as N02 is used up). Finally, when the rates become equal, equilibrium is reached and the equilibrium expression (6) is applicable in bulb B. 

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. 

Again Le Chatelier s Principle tells us qualitatively what will occur and the equilibrium expression tells us quantitatively. If we add OH-(ag), a change will take place that tends to counteract partially the resulting increase in [OH-]. This occurs through the reaction between OY (aq) and H+(aq), consuming both ions and reducing the value of [H+] X [OH-]. Reaction continues until this product reaches the equilibrium value, K = 1.00 X 10-14. 

Write the equilibrium expression relating the concentrations of reactants and products in reaction (26). Notice that the concentration of water must be included because it is not necessarily large enough to be considered constant. 

Substituting Eqs. (35) and (36) into Eq. (34), the electrochemical potential fluctuation of dissolved metal ions at OHP is deduced. Then, disregarding the fluctuation of the chemical potential due to surface deformation, the local equilibrium of reaction is expressed as fi% = 0. With the approximation cm x, y, 0, if cm(x, y, (a, tf, we can thus derive the following equation,... 

Write the equilibrium expression Kc for each of the following reactions ... 

Because so many chemical reactions in our bodies take place in buffered environments, biochemists commonly need to make quick estimates of pH by employing a form of the expression for Ka that gives the pH directly. For the equilibrium in reaction A, we can rearrange the expression for Ka into... 

A proponent of "reverse weathering" suggested that gibbsite, kaolinite, and quartz exist in equilibrium according to the following equation. In equilibrium expressions for these reactions, water will appear as the activity, rather than concentration. The activity can be approximated by the mole fraction of water. What is the activity of water if this equilibrium is maintained Could this equilibrium exist in seawater, where the mole fraction of water is about 0.98 AG values (kj/mol) gibbsite — 2320.4 kaolinite — 3700.7 quartz —805.0 water —228.4. 

To illustrate the generality of reversibility and the equilibrium expression, we extend our kinetic analysis to a chemical reaction that has a two-step mechanism. At elevated temperature NO2 decomposes into NO and O2 instead of forming N2 O4. The mechanism for the decomposition reaction, which appears in Chapter 15. 

Each equilibrium expression described so far contains a ratio of concentrations of products and reactants. Moreover, each concentration is raised to a power equal to its stoichiometric coefficient in the balanced equation for the overall reaction. Concentration ratios always have products in the numerator and reactants in the... 

It is possible to carry out this type of kinetic analysis whether a mechanism is simple or elaborate. That is, we can always derive the equilibrium expression for a reaction by applying reversibility and setting forward and reverse rates equal to one another at equilibrium. It is unnecessary to go through this procedure for every chemical equilibrium. As our two examples suggest, inspection of the overall stoichiometry always gives the correct expression for the equilibrium constant. That is, a reaction of the form tjA + iBf ofD + eE has an... 

Both relationships include a constant and both involve concentrations raised to exponential powers. However, a rate law and an equilibrium expression describe fundamentally different aspects of a chemical reaction. A rate law describes how the rate of a reaction changes with concentration. As we describe in this chapter, an equilibrium expression describes the concentrations of reactants and products when the net rate of the reaction is zero. 

In this reaction, Fe is a pure solid and H2 O is the solvent, so they do not appear in the equilibrium expression. The other reagents have exponents equal to their stoichiometric coefficients ... 

After completing our analysis of the effects of the dominant equilibrium, we may need to consider the effects of other equilibria. The calculation of [H3 O ] in a solution of weak base illustrates circumstances where this secondary consideration is necessary. Here, the dominant equilibrium does not include the species, H3 O, whose concentration we wish to know. In such cases, we must turn to an equilibrium expression that has the species of interest as a product. The reactants should be species that are involved in the dominant equilibrium, because the concentrations of these species are determined by the dominant equilibrium. We can use these concentrations as the initial concentrations for our calculations based on secondary equilibria. Look again at Example for another application of this idea. In that example, the dominant equilibrium is the reaction between hypochlorite anions and water molecules H2 0 l) + OCr(c2 q) HOCl((2 q) + OH ((2 q) Working with this equilibrium, we can determine the concentrations of OCl, HOCl, and OH. To find the concentration of hydronium ions, however, we must invoke a second equilibrium, the water equilibrium 2 H2 0(/) H3 O (a q) + OH (a q)... 

The protonation reactions for ionizable molecules have been defined in Section 3.1. When a solute molecule, HA (or B), is in equilibrium with its precipitated form, HA(s) (or B(s)), the process is denoted by the equilibrium expression... 

With the use of the equilibrium expressions for the first three reactions, equation 4.1.22 may be converted to the following form,... 

If you want to prove to yourself that free energies sum, you can write the equilibrium expressions for the first and second reactions and multiply them together, and you ll get the equilibrium expression for the hydrolysis of ATP. Multiplication is equivalent to the addition of logarithms, so that when you multiply equilibrium constants, you re actually adding free energies (or vice versa). 

Siderophore binding sites for iron(III) are for the most part negatively charged and therefore, in aqueous solution there is a competition between H+ and Fe3+ binding. Consequently, the equilibrium expression for the formation of the iron-siderophore complex must take into account proton participation in the reaction. 

Many systems are not at equilibrium. The mass action expression, also called the reaction quotient, Q, is a measure of how far a system is from equilibrium and in what direction the system must go to get to equilibrium The reaction quotient has the same form as the equilibrium constant, K, but the concentration values put into Q are the actual values found in the system at that given moment. 

Note Even though the reverse reaction is favored, we still write the equilibrium expression for the reaction as originally written. This is a major point ... 

To construct such a diagram, a set of defect reaction equations is formulated and expressions for the equilibrium constants of each are obtained. The assumption that the defects are noninteracting allows the law of mass action in its simplest form, with concentrations instead of activities, to be used for this purpose. To simplify matters, only one defect reaction is considered to be dominant in any particular composition region, this being chosen from knowledge of the chemical attributes of the system under consideration. The simplified equilibrium expressions are then used to construct plots of the logarithm of defect concentration against an experimental variable such as the log (partial pressure) of the components. The procedure is best illustrated by an example. 

Words that can be used as topics in essays 5% rale buffer common ion effect equilibrium expression equivalence point Henderson-Hasselbalch equation heterogeneous equilibria homogeneous equilibria indicator ion product, P Ka Kb Kc Keq KP Ksp Kw law of mass action Le Chatelier s principle limiting reactant method of successive approximation net ionic equation percent dissociation pH P Ka P Kb pOH reaction quotient, Q reciprocal rule rule of multiple equilibria solubility spectator ions strong acid strong base van t Hoff equation weak acid weak base... 

We can write a reactant quotient at any point during the reaction, but the most meaningful point is when the reaction has reached equilibrium. At equilibrium, the reaction quotient becomes the equilibrium constant, Kc (or Kp if gas pressures are being used). We usually express this equilibrium constant simply as a number without units since it is a ratio of concentrations or pressures. In addition,... 

With the search variables listed above, the calculation of the remaining molalities (iterative state variables) can be easily performed sequentially by using the equilibrium expressions for the reaction shown in Table 1 and the calculated activity coefficients. The equations are used in the order in which they appear in Table 1. The sequence of calculation of molalities is OH", SO", H2S03, HSO", CO", H CO, CaOH+, CaS0°,... 

This equilibrium expression depends only on the stoichiometry of the reaction. By convention, chemists always write the concentrations of the products in the numerator and the concentrations of the reactants in the denominator. Each concentration term is raised to the power of the coefficient in the chemical equation. The terms are multiplied, never added. 

The following Sample Problem shows how to find the equilibrium expression for a reaction. In this chapter, you will use equilibrium expressions for homogeneous reactions (mostly reactions between gases). In Chapters 8 and 9, you will learn how to use equilibrium expressions for heterogeneous systems. 

Write the equilibrium expression for each homogeneous reaction. 

The equilibrium constant, Kc, is calculated by substituting equilibrium concentrations into the equilibrium expression. Experimentally, this means that a reaction mixture must come to equilibrium. Then one or more properties are measured. The properties that are measured depend on the reaction. Common examples for gaseous reactions include colour, pH, and pressure. From these measurements, the concentrations of the reacting substances can be determined. Thus, you do not need to measure all the concentrations in the mixture at equilibrium. You can determine some equilibrium concentrations if you know the initial concentrations of the reactants and the concentration of one product at equilibrium. 

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