Equations equilibrium expression

The equilibrium expressed by the preceding equation lies overwhelmingly to the side of the zwittenon... 

Thermodynamics of Liquid—Liquid Equilibrium. Phase splitting of a Hquid mixture into two Hquid phases (I and II) occurs when a single hquid phase is thermodynamically unstable. The equiUbrium condition of equal fugacities (and chemical potentials) for each component in the two phases allows the fugacitiesy andy in phases I and II to be equated and expressed as ... 

From this equation, we can write the equilibrium expression ... 

Barrer s discussion4 of his analog of Eq. 28 merits some comment. Equation 28 expresses the equilibrium condition between ice and hydrate. As such it is valid for all equilibria in which the two phases coexist and not only for univariant equilibria corresponding with a P—7" line in the phase diagram. (It holds, for instance, in the entire ice-hydratell-gas region of the ternary system water-methane-propane considered in Section III.C.(2).) In addition to Eq. 28 one has Clapeyron s equation... 

The above rate equation is in agreement with that reported by Madhav and Ching [3]. Tliis rapid equilibrium treatment is a simple approach that allows the transformations of all complexes in terms of [E, [5], Kls and Kjp, which only deal with equilibrium expressions for the binding of the substrate to the enzyme. In the absence of inhibition, the enzyme kinetics are reduced to the simplest Michaelis-Menten model, as shown in Figure 5.21. The rate equation for the Michaelis-Menten model is given in ordinary textbooks and is as follows 11... 

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. 

Examining the equilibrium expressions in Table 16-2, the equations can be put into a logarithmic form, e.g., for the dissociation of a simple acid, HA,... 

If we multiply one of these equations by the other, the concentration of the intermediate, [NO3 ]g, cancels to produce an equilibrium expression entirely in terms of concentrations of reactants and products ... 

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... 

If the mechanism for this equilibrium were given, we could derive an equilibrium expression using the principle of reversibility, but Equation shows that we can write the equilibrium expression directly without knowing details of the mechanism. 

To determine percent ionization, we need to know the equilibrium concentration of hydronium ions. This requires an equilibrium calculation, for which we follow the seven-step method. We need to set up the appropriate equilibrium expression and solve for [H3 O, after which we can use Equation to... 

Equation (31) is true only when standard chemical potentials, i.e., chemical solvation energies, of cations and anions are identical in both phases. Indeed, this occurs when two solutions in the same solvent are separated by a membrane. Hence, the Donnan equilibrium expressed in the form of Eq. (32) can be considered as a particular case of the Nernst distribution equilibrium. The distribution coefficients or distribution constants of the ions, 5 (M+) and B X ), are related to the extraction constant the... 

By using the mass law equilibrium expressions to substitute for pR,pR, andpRR in the last equation, we obtain ... 

The equilibrium constant expressions arc the same because the chemical equations arc the same. It is easy to see why the coefficients of the chemical equation are used as exponents in the equilibrium constant expression by writing out the equation and expression as in part (a). 

Equation 1 expresses a state of equilibrium between an alcohol A. on a molecule whose degree of polymerization is j, the catalyst C and the alkoxide anion A.C. In Relation 2 this activated intermediate reacts with monomeric anhydride A, forming an acid adduct A.AC, which dissociates, forming an unassociable carboxylic acid A.A. Reactions 3-5 depict the union of a carboxylic intermediate with a monomeric epoxide E, or with pendant oxiranes on macromole- ... 

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

Equations (18) and (16) define a temperature where Gaussian behavior is observed (the phase separation temperature) where % — 1/2 and thermal energy is just sufficient to break apart PP and SS interactions to form PS interactions. Equation (12) using (17) for Vc is called the Flory-Krigbaum equation. This expression indicates that only three states are possible for a polymer coil at thermal equilibrium ... 

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. 

Equation (2) expresses the Gibbs free energy in terms of the mole numbers nt, which appear both explicitly and implicitly (in the /x,) on the right-hand side. The Gibbs free energy is a minimum when the system is at equilibrium. The basic problem, then, becomes that of finding that set of nt that makes G a minimum. 

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... 

Where competitive inhibition is observed between two solutes (i.e. binding to a single, identical carrier), it is also possible to estimate carrier concentrations using a steady-state treatment [193-195], In that case, data from the competing solutes are used to generate a sufficient number of equilibrium expressions (e.g. equations (38) and (39)) and corresponding mass balance equations (e.g. equations (40) and (41)) to resolve for the total carrier concentration. 

Constant D is probably a good approximation in so far as the degree of melting is significantly smaller than the proportion of the least abundant mineral phase. Integrating the differential equation gives expressions for the solid and the instantaneous liquid in equilibrium with it... 

If the equilibrium expressed in Equation (7) is attained only between the crystal surface and the aqueous solution, the equilibrium is described by ... 

Although one can probably find exceptions, most equilibrium calculations involving flue gas slurries are performed with temperature as a known variable. With temperature known, the numerical values of the appropriate equilibrium constants can be immediately calculated. The remaining unknown variables to be determined are the activities, activity coefficients, molalities, and the gas phase partial pressures. The equations used to determine these variables are formulated from among the equilibrium expressions presented in Table 1, the expressions for the activity coefficients, ionic strength, material balance expressions, and the electroneutrality balance. Although there are occasionally exceptions, the solution sequence generally is an iterative or cyclic sequence. 

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°,... 

Equation 27 is similar to the solid-liquid equilibrium relation used for non-electrolytes. As in the case of the vapor-liquid equilibrium relation for HC1, the solid-liquid equilibrium expression for NaCl is simple since the electrolyte is treated thermodynamically the same in both phases. 

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 equilibrium expression has the product terms in the numerator and the reactant terms in the denominator. The exponents in the equilibrium expression match the corresponding coefficients in the chemical equation. The molar concentrations at equilibrium were substituted into the expression. 

Generally, it is best to let xrepresent the substance with the smallest coefficient in the chemical equation. This helps to avoid fractional values of x in the equilibrium expression. Fractional values make solving the expression more difficult. 

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

In this problem, the right-hand side of the equilibrium expression is a perfect square. Noticing perfect squares, then taking the square root of both sides, makes solving the equation easier. It avoids solving a quadratic equation. 

Solving an Equilibrium Expression Using a Quadratic Equation... 

Step 2 Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the equilibrium expression. Rearrange the equilibrium expression into the form of a quadratic equation. Solve the quadratic equation for x. 

The coefficients in the chemical equation match the exponents in the equilibrium expression. To check your concentrations, substitute them back into the equilibrium expression. 

We can apply the criterion of equilibrium expressed in Equation (9.17) to chemically reacting systems. Consider the reaction... 

For a system undergoing R independent chemical reactions among N chemical species, R equilibrium expressions are to be added to the relationships among the intensive variables. From Equation (13.1), the total number of intensive variables in terms of N becomes... 

It should also be noted that when the rate of change in the protein-ligand complex concentration is zero (by definition, when the system is at equilibrium), this equation reduces to the equilibrium expression below, with the binding affinity constant defined as the ratio of the dissociation rate koff to tho association rate kon-... 

Equation 9e expresses the assumption of local equilibrium of the partitioning process at the stationary phase - mobile phase Interface. 

The exchange-rate equation is obtained by substituting the expressions for ki2, k-i, and the equilibrium expressions for the enzyme intermediates into Eq. (4a). 

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