Equilibrium constant first example

We will use two useful relationships when working with equilibrium constants. First, if we reverse a reaction s direction, the equilibrium constant for the new reaction is simply the inverse of that for the original reaction. For example, the equilibrium constant for the reaction... 

While the pE values of the electrodes just discussed are usually found in tables, we should recognize that they can be derived from the combination of the pE°(M" /M ) and the relevant equilibrium constants. For example, the pE for the AgCl/Ag electrode can be obtained from the pE° of AgVAg and the K p for AgCl. First,... 

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 ... 

Prebiotic reactions are reactions that might have occurred under the conditions prevalent on the Earth before the first living creatures emerged and can lead to analogs of molecules necessary for life as we now know it. To qualify, a reaction must proceed with a favorable rate and have a reasonable value for the equilibrium constant. An example of a prebiotic reaction is the formation of 5-hydroxymethyluracil (HMU) from uracil and formaldehyde (HCHO). Amino acid analogs can be formed from HMU under prebiotic conditions by reaction with various nucleophiles, such as HjS, HCN, indole, and imidazole. For the synthesis of HMU at pH = 7, the temperature dependence of the rate constant is given by... 

In Chapter 2 the Diels-Alder reaction between substituted 3-phenyl-l-(2-pyridyl)-2-propene-l-ones (3.8a-g) and cyclopentadiene (3.9) was described. It was demonstrated that Lewis-acid catalysis of this reaction can lead to impressive accelerations, particularly in aqueous media. In this chapter the effects of ligands attached to the catalyst are described. Ligand effects on the kinetics of the Diels-Alder reaction can be separated into influences on the equilibrium constant for binding of the dienoplule to the catalyst (K ) as well as influences on the rate constant for reaction of the complex with cyclopentadiene (kc-ad (Scheme 3.5). Also the influence of ligands on the endo-exo selectivity are examined. Finally, and perhaps most interestingly, studies aimed at enantioselective catalysis are presented, resulting in the first example of enantioselective Lewis-acid catalysis of an organic transformation in water. 

Besides equilibrium constant equations, two other types of equations are used in the systematic approach to solving equilibrium problems. The first of these is a mass balance equation, which is simply a statement of the conservation of matter. In a solution of a monoprotic weak acid, for example, the combined concentrations of the conjugate weak acid, HA, and the conjugate weak base, A , must equal the weak acid s initial concentration, Cha- ... 

A simple example of the analysis of multicomponent systems will suffice for the present consideration, such as the calculation of the components in a gaseous mixture of oxygen, hydrogen and sulphur. As a first step, the Gibbs energy of formation of each potential compound, e.g. S2, H2S, SO, SO2, H2O etc. can be used to calculate the equilibrium constant for the formation of each compound from the atomic species of the elements. The total number of atoms of each element will therefore be distributed in the equilibrium mixture in proportion to these constants. Thus for hydrogen with a starting number of atoms and the final number of each species... 

Example 9.4 deals with a system at equilibrium, but suppose the reaction mixture has arbitrary concentrations. How can we tell whether it will have a tendency to form more products or to decompose into reactants To answer this question, we first need the equilibrium constant. We may have to determine it experimentally or calculate it from standard Gibbs free energy data. Then we calculate the reaction quotient, Q, from the actual composition of the reaction mixture, as described in Section 9.3. To predict whether a particular mixture of reactants and products will rend to produce more products or more reactants, we compare Q with K ... 

Now set up a concentration table. The equilibrium constant for precipitation is very large, so imagine the precipitation in two steps (see Example ). First, take the reaction to completion by applying limiting reactant stoichiomehy. Then switch on the solubility equilibrium ... 

One way to make Example 19.13 harder is by giving numbers of moles and a volume instead of concentrations at equilibrium. Since the equilibrium constant is defined in terms of concentrations, we must first convert the numbers of moles and volume to concentrations. Note especially that the volume of all the reactants is the same, since they are all in the same system. 

The next step in the procedure for calculating the stability constants is to represent the concentrations of the several complexes in terms of the equilibrium constants for their formation. We do this by solving each equilibrium constant expression for the concentration of the complex. For example, the first step has an equilibrium constant... 

Worked Example 8.19 The data below relate to the first-order isomerization of 2-hexene at 340 K, a reaction for which the equilibrium constant is known from other studies to be 10.0. What are the rate constants k and k i ... 

In the o.s. reaction, the ion pair A+ - B is formed in a first step. The corresponding equilibrium constant can usually be obtained from simple electrostatic models. In this "ideal" case specific chemical interactions can be neglected and the rate constant of the E.T. step follows the theory of R.A. Marcus (see for example Marcus, 1975, or Cannon, 1980). In the i.s. reaction each of the three steps in reaction (9.2) may determine the reaction rates. The lability of the coordinated ligands at the... 

The concentration of M = I(T3 molar. Suppose we want to reduce the concentration of M to 10 5. When we calculate the equilibrium constant from the first example and then calculate the required concentration ofLatM= 1(T5 molar we find the following rule of thumb if an excess ofL is needed, keep [L] the same and lower fM] only ... 

The first term in parentheses on the right side of equation 5.213 is the distribution coefficient (K ), and the second groups activity coefficients related to the mixing behavior of components in the two phases. The equilibrium constant is thus related to the interaction parameters of the two phases at equilibrium. For example, the equilibrium between two regular mixtures is defined as... 

The now classical example is lactate dehydrogenase. Sil-verstein and Boyer were the first to determine the rates of exchange between cognate pairs of reactants ie., lactate and pyruvate as well as NAD and NADH). Convenient [NADH]/[NAD ] and [pyruvate]/[lactate] ratios were chosen such that when combined they satisfied the apparent equilibrium constant for the LDH reaction. These investigators first established that each exchange rate was directly proportional to the duration of exchange and likewise directly proportional to enzyme concentration. As an additional control, they also demonstrated the equality of the pyruvate lactate exchange... 

Given measured species concentrations for a homogeneous reaction in a rock, cooling rate at Tae can be found as follows if the equilibrium constant K and the forward reaction rate coefficient k as a function of temperature are known. First, the apparent equilibrium temperature is calculated from the species concentrations. Then kf and kb at Tae are calculated. Then the mean reaction time Xr at Tae is calculated using expressions in Table 2-1. From x, the cooling rate q at Tae can be obtained using Equation 5-125. Two examples are given below. 

Students assigned to the charge transfer experiment must first find out what constitutes a charge transfer complex. Then they find examples of charge transfer studies from the literature. They usually quickly see that spectrophotometry is commonly used for such studies and that the temperature will need to be varied. The procedure chosen is to determine the equilibrium constant at more than one temperature and from these data, to calculate the thermodynamic parameters. Students generally have difficulty in selecting a system that has an equilibrium constant that is not too big or too small and to select a solvent. This means they need to do some calculations to determine if a reasonable quantity of product is... 

The combination of time marching and Newton s method can be illustrated via a very simple model problem [277]. Consider two reactions, R + A B + P and R + B 2P, where in the first a reactant R reacts with a compound A to produce a compound B and a product P. Then, in the second reaction, R further reacts with B to produce two moles of P. If the reaction rates are significantly different, this will lead to a stiff system. For the sake of our example, presume that the mole fraction of R is fixed at a value of 0.1, and that the rate constants for the reactions are k = 1011 and ki = 1012, respectively. Furthermore take the equilibrium constants for the two reactions to be AT] =5 and K.2 = 15. With these parameters set, the mole fractions of A and B (A and B) are governed by the following system of equations. (The value of P is determined from the fact that the mole fractions must sum to unity.)... 

The result in (2.11) that the reaction time is the sum of two other times is fairly common in the general reaction-diffusion literature, in which a steady-state approximation is used and there is a diffusion toward a sink followed by reaction at that sink. For example, in the scheme A B - C, with forward and reverse rate constants k and Ic2 for the first step (equilibrium constant K = k /k ) and rate constant k for the last step, a steady-state approximation for B yields l/fcrale(l/fc3 )+(l/fci), which has the same functional form as (2.11). The more complete expression, which allows for the back reaction (recrossings), has a slightly more complicated structure [44]. 

Measurement of the rate of polymerization of 4-methyl-1-pentene have shown that in polar solvent the rate is first order in monomer. This is evidently an example of Case 2 (K<41). It is postulated that K, the equilibrium constant, would be small in polar solvent since the concentration of the ion-counterion pair,... 

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