Chemical equations equilibrium expression

We learn to interpret the magnitude of an equilibrium constant and how its value depends on the way the corresponding chemical equation is expressed. 

Before doing calculations with equilibrium constants, it is valuable to understand what the magnitude of an equilibrium constant can tell us about the relative concentrations of reactants and products in an equilibrium mixture. It is also useful to consider how the magnitude of any equilibrium constant depends on how the chemical equation is expressed. 

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 ... 

Strategy First write the chemical equation for the equilibrium system. Then write the expression for K, leaving out pure liquids and solids. Remember that gases are represented by their partial pressures. 

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for K. 

The equilibrium constant expression for the dissolving of SrCr04 can be written following the rules in Chapters 12 and 13. In particular, the solid does not appear in the expression the concentration of each ion is raised to a power equal to its coefficient in the chemical equation. 

Clausius-Clapeyron equation An equation expressing the temperature dependence of vapor pressure ln(P2/Pi) = AHvapCl/Tj - 1/T2)/R, 230,303-305 Claussen, Walter, 66 Cobalt, 410-411 Cobalt (II) chloride, 66 Coefficient A number preceding a formula in a chemical equation, 61 Coefficient rule Rule which states that when the coefficients of a chemical equation are multiplied by a number n, the equilibrium constant is raised to the nth power, 327... 

In this generalized equation, (75), we see that again the numerator is the product of the equilibrium concentrations of the substances formed, each raised to the power equal to the number of moles of that substance in the chemical equation. The denominator is again the product of the equilibrium concentrations of the reacting substances, each raised to a power equal to the number of moles of the substance in the chemical equation. The quotient of these two remains constant. The constant K is called the equilibrium constant. This generalization is one of the most useful in all of chemistry. From the equation for any chemical reaction one can immediately write an expression, in terms of the concentrations of reactants and products, that will be constant at any given temperature. If this constant is measured (by measuring all of the concentrations in a particular equilibrium solution), then it can be used in calculations for any other equilibrium solution at that same temperature. 

SOLUTION First convert the units of partial pressures to bars using 1 bar = 105 Pa P i7 = 4.2 X 10 8 bar, Pcli = 8.3 X 10 8 bar. Write the expression for the equilibrium constant from the chemical equation given. 

The powers to which the activities are raised in the expression for an equilibrium constant must match the stoichiometric coefficients in the chemical equation, which is normally written with the smallest whole numbers for coefficients. Therefore, if we change the stoichiometric coefficients in a chemical equation (for instance, by... 

If a chemical equation can be expressed as the sum of two or more chemical equations, the equilibrium constant for the overall reaction is the product of the equilibrium constants for the component reactions. For example, consider the three gas-phase reactions... 

Step 1 Write the balanced chemical equation for the equilibrium and the corresponding expression for the equilibrium constant. Then set up an equilibrium table as shown here, with columns labeled by the species taking part in the reaction. In the first row, show the initial composition (molar concentration or partial pressure) of each species... 

STRATEGY Because NH4+ is a weak acid and Cl- is neutral, we expect pH < 7. We treat the solution as that of a weak acid, using an equilibrium table as in Toolbox 10.1 to calculate the composition and hence the pH. First, write the chemical equation for proton transfer to water and the expression for Ca. Obtain the value of Ka from Kh for the conjugate base by using K, = KxJKh (Eq. 11a). The initial concentration of the acidic cation is equal to the concentration of the cation that the salt would produce if the salt were fully dissociated and the cation retained all its acidic protons. The initial concentrations of its conjugate base and H30+ are assumed to be zero. 

STRATEGY First, we write the chemical equation for the equilibrium and the expression for the solubility product. To evaluate Ksp, we need to know the molarity of each type of ion formed by the salt. We determine the molarities from the molar solubility, the chemical equation for the equilibrium, and the stoichiometric relations between the species. We assume complete dissociation. 

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... 

When the chemical equation is more complex, the equilibrium constant expression is also more complex and the deductions about the equilibrium concentrations of reactants and products are more involved too. 

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). 

First we write the balanced chemical equation for the reaction. Then we write the equilibrium constant expressions, remembering that gases and solutes in aqueous solution appear in the Kc expression, but pure liquids and pure solids do not. 

B We need the balanced chemical equation in order to write the equilibrium constant expression. We start by translating names into formulas. 

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. 

The equilibrium constant expression associated with systems of slightly soluble salts is the solubility product constant, Ksp. It is the product of the ionic concentrations, each one raised to the power of the coefficient in the balanced chemical equation. It contains no denominator since the concentration of a solid is, by convention, 1, and for this reason it does not appear in the equilibrium constant expression. The Ksp expression for the PbS04 system is ... 

When writing equilibrium constant expressions, use products over reactants. Each concentration is raised to the power of the coefficient in the balanced chemical equation. 

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. 

Step 1 Set up an ICE table. Let the change in molar concentrations of the reactants be x. Use the stoichiometry of the chemical equation to write expressions for the equilibrium concentrations. Record these expressions in your ICE table. 

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 must be noted that Eqs. (35) and (36) are for the case in which the crosslinks in the polymer network were introduced in solution as with the Peppas-Merrill equation for neutral hydrogels and also that a Gaussian chain distribution is assumed. The complete equilibrium expressions accounting for the mixing, elastic-retractive, and ionic contributions to the chemical potential for anionic networks in the two cases described above are then... 

Question (a) is in the province of chemical thermodynamics1 and amounts to evaluating the equilibrium constant (K). Unlike the rate equation, the equilibrium expression for a typical reaction... 

The general treatment of acid-base systems begins with charge and mass balances and equilibrium expressions. There should be as many independent equations as chemical species. Substitute a frac-... 

Third, write what you know and do not know. You are asked for the pH of a basic solution, so you will first have to find the [OH-]. Let [OH-] = x. Because NH and OH- ions are formed in equal amounts, [NHJ] also equals x. Of the original 0. lOOmole/liter ofNH3,x moles/liter will dissociate and leave (0.100 — x) mole/liter at equilibrium. Associate these concentrations with the chemical equation, and substitute them into the K, expression. 

STRATEGY Begin by writing the chemical equation and the expression for the equilibrium constant. Then substitute the equilibrium concentrations in Eq. 12 to find Kc. Raise each concentration to a power equal to the stoichiometric coefficient of that species in the chemical equation. Equation L4 can then be used to obtain K. 

SOLUTION The chemical equation and the expression for the equilibrium constant Kc are given in reaction C and Eq. 12, respectively. We substitute the equilibrium molar concentrations (without their units) into the expression for Kc ... 

Begin by writing the balanced chemical equation for the equilibrium and the corresponding expression for the equilibrium constant. Then draw up an equilibrium table by working through the following steps. 

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