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Equilibrium expression for heterogeneous

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. [Pg.335]

For a heterogeneous equilibrium (a chemical equilibrium with components in different phases), reactants or products may be pure liquids or solids. The concentration of a pure liquid or solid in moles/liter cannot change. It is a constant property of the material, and these constants are incorporated into the equilibrium constant. Therefore the concentrations of pure liquids and solids are absent from equilibrium expressions for heterogeneous equilibria. [Pg.159]

Therefore the concentrations of pure liquids and solids are absent from equilibrium expressions for heterogeneous equilibria. [Pg.56]

Writing equilibrium expressions for heterogeneous equilibria is also straightforward, but it is slightly different from what we have done so far for homogeneous equilibria. For example, carbon dioxide can combine with elemental carbon to produce carbon monoxide ... [Pg.599]

Think About It Like writing equilibrium expressions for homogeneous equilibria, writing equilibrium expressions for heterogeneous equilibria becomes second nature if you practice. The importance of developing this skill now cannot be overstated. Your ability to tmderstand the principles and to solve many of the problems in this and Chapters 16 to 19 depends on your ability to write equilibrium expressions correctly and easily. [Pg.600]

Example 16.3 Writing Equilibrium Expressions for Heterogeneous Equilibria... [Pg.495]

The equilibrium constants for heterogeneous reactions are also given by the general expression in Eq. 2 all we have to remember is that the activity of a pure solid or liquid is 1. For instance, for the calcium hydroxide equilibrium (reaction H),... [Pg.482]

The equilibrium constants for heterogeneous reactions are given by the general expression in Eq. 8a all we have to remember is that the... [Pg.563]

This equilibrium is known as a heterogeneous equilibrium, which is to say that it consists of substances that are in different states (phases). If you recall from Chapter 13, the equilibrium constant expressions for such equilibria do not contain the concentrations of liquids or solids. The equilibrium expression for this reaction will describe the degree to which the solid dissolves in solution, which is another way to say the degree to which it is soluble. And since the solid is not shown in the equilibrium expression, the equilibrium constant will express the product of the concentrations of the dissolved solute ions. For this reason, the equilibrium constant is referred to as the solubility-product constant. For this equilibrium, the solubility-product constant, Ksp is ... [Pg.354]

The quantitative relations involved in heterogeneous solid-solution formation were considered by Doerner and Hoskins, who derived the equilibrium expression. For silver bromide and chloride the two extreme situations are represented by... [Pg.173]

It is believed that the Kp data for the three homogeneous reactions (A, B, and C) are more reliable than those for the heterogeneous reaction (D) due to the need for a machine calibration constant which appears in the equilibrium expression for the latter. Ehlert et al. (2) determined this constant from vaporization experiments performed with silver contained in their Knudsen cell. Further support for this belief is provided by the large positive drift that arises in the 3rd law analysis of these Kp s. Also, it Is felt that the results otained from the flame-spectrophotometric studies (4, are somewhat more... [Pg.331]

Explain why pure solids and liquids are left out of equilibrium constant expressions for heterogeneous equilibria. [Pg.644]

Write the equilibrium expression for each of the following heterogeneous equilibria. [Pg.629]

However, experimentai resuits show that the position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present (see Fig. 13.6). The fundamental reason for this behavior is that the concentrations of pure solids and liquids cannot change. Thus the equilibrium expression for the decomposition of solid calcium carbonate might be represented as... [Pg.604]

Expressions for heterogeneous equilibria You have learned to write Keq expressions for homogeneous equilibria, those in which all reactants and products are in the same physical state. When the reactants and products are present in more than one physical state, the equilibrium is called a heterogeneous equilibrium. When ethanol is placed in a closed flask, a liquid-vapor equilibrium is established, as illustrated in Figure 17.8. [Pg.602]

Strategy Because they are constant quantities, the concentrations of solids and liquids do not appear in the equihbrium constant expressions for heterogeneous systems. The total pressure at equilibrium that is given is due to both NH3 and CO2. Note that for every 1 atm of CO2 produced, 2 atm of NH3 will be produced due to the stoichiometry of the balanced equation. Using this ratio, we can calcrrlate the partial presstrres of NH3 and CO2 at eqtrilibritrm. [Pg.389]

A simplification occurs with heterogeneous equilibria because the concentrations of pure solids and pure liquids are constant at a fixed temperature (see page 137). For example, the equilibrium expression for the production of carbon monoxide from coke and carbon dioxide,... [Pg.282]

We then learn how to write equilibrium-constant expressions for heterogeneous reactions. [Pg.610]

Writing Equilibrium-Constant Expressions for Heterogeneous Reactions... [Pg.623]

The equilibrium-constant expression for the equilibrium between a solid and an aqueous solution of its component ions is written according to the rules that apply to any other equilibrium-constant expression. Remember, however, that solids do not appear in the equilibrium-constant expressions for heterogeneous equilibrium, j- (Section 15.4) Thus, the solubility-product expression for BaS04, which is based on Equation 17.15, is... [Pg.723]


See other pages where Equilibrium expression for heterogeneous is mentioned: [Pg.610]    [Pg.605]    [Pg.557]    [Pg.487]    [Pg.459]    [Pg.618]    [Pg.590]    [Pg.610]    [Pg.605]    [Pg.557]    [Pg.487]    [Pg.459]    [Pg.618]    [Pg.590]    [Pg.179]    [Pg.566]    [Pg.603]   


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