The Equilibrium Constant Expression

At equilibrium, the rate of forward and reverse reactions are equal. For the system N204(g) 2N02(g) 

Since the rate of N2O4 decomposition and the rate of NO2 formation are equal at equilibrium. 

Since kj and k, are constants a new constant can be derived from them. It is named as an equilibrium constant in terms of concentration, and is symbolized as K, . 

Solid and liquid substances are not represented in a K, expression, since their concentrations do not vary in the course of reaction. For example, 

Ammonia can be produced from hydrogen and nitrogen, N2(g) + 3H2(g) 2NH3(g) 

For every gaseous chemical system, an equilibrium constant expression can be written stating the condition that must be attained at equihbrium. For the general system involving only gases, 

This equihbrium constant is often given the symbol to emphasize that it involves partial pressures. Other equihbrium expressions for gases are sometimes used, including A,  

It is important to realize that the expression for K depends on the form of the chemical equation written to describe the equilibrium system. To illustrate what this statement means, consider the N2O4-NO2 system  

Many other equations could be written for this system, for example, 

In this case, the expression for the equilibrium constant would be 

Comparing the expressions for K and K, it is clear that K is the square root of K. This illustrates a general rule, sometimes referred to as the coefficient rale, which states, 

Consider a hypothetical reaction in which a molecule. A, is converted to its isomer, B, that is, the reversible reaction A B. Start with a flask containing 54 molecules of A, represented by open circles. Convert the appropriate number of open circles to filled circles to represent the isomer B and portray the equilibrium condition if /C = 0.02. Repeat the process for K = 0.5 and then for K = 1. 

Before proceeding to other matters, we must emphasize that the reaction quotient expression for reaction (15.1) is just a specific example of a more general case. Let s focus on the following hypothetical, generalized reaction. 

In this equation, the uppercase letters A, B, C, D, etc., refer to chemical substances, and the lowercase letters a, b, c, d, etc., represent the coefficients required to balance the equation. By applying the method described on pages 690 and 691, we obtain the following expression for the reaction quotient. 

At equilibrium, we may substitute the equilibrium values of the activities into equation (15.5) and set the resulting expression equal to K. 

Although the notation used above is clear, few chemists use it. The numerous subscripts make the expression appear rather cluttered. Many chemists would eliminate the subscript eq and write the expression above in the following abbreviated form. 

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The equilibrium position for any reaction is defined by a fixed equilibrium constant, not by a fixed combination of concentrations for the reactants and products. This is easily appreciated by examining the equilibrium constant expression for the dissociation of acetic acid. 

Count the number of species whose concentrations appear in the equilibrium constant expressions these are your unknowns. If the number of unknowns equals the number of equilibrium constant expressions, then you have enough information to solve the problem. If not, additional equations based on the conservation of mass and charge must be written. Continue to add equations until you have the same number of equations as you have unknowns. 

Substituting equations 6.39 and 6.40 into the equilibrium constant expression for the dissociation of HE (equation 6.35) and solving for the concentration of H3O4 gives us... 

Counting unknowns, we find that there are seven—[Ag+], [I ], [Ag(NH3)2 ], [NH3], [NH4+], [OH ], and [H3O+]. Four of the equations needed to solve this problem are given by the equilibrium constant expressions... 

Since the concentrations of Na+, A-, HA, H3O+, and OH- are unknown, five equations are needed to uniquely define the solution s composition. Two of these equations are given by the equilibrium constant expressions... 

Substituting the equilibrium constant expressions for reactions 8.3-8.S into equation 8.6 defines the solubility of AgCl in terms of the equilibrium concentration of Ch. 

Starting with the equilibrium constant expressions for reactions 8.1, and 8.3-8.5, verify that equation 8.7 is correct. 

The change in the concentration of H3O+ is monitored with a pH ion-selective electrode, for which the cell potential is given by equation 11.9. The relationship between the concentration of H3O+ and CO2 is given by rearranging the equilibrium constant expression for reaction 11.10 thus... 

At a given temperature, the pressure of iodine vapor is constant, independent of the amount of solid iodine or any other factor. The equilibrium constant expression is... 

Applying the reciprocal rule (page 327), we can deduce the equilibrium constant expression for the reverse reaction... 

Given the following descriptions of reversible reactions, write a balanced equation (simplest whole-number coefficients) and the equilibrium constant expression (X) for each. 

Recall (page 331, Chapter 12) that in the equilibrium constant expression for reactions in solution—... 

Hence for the ionization of water, the equilibrium constant expression can be written ... 

Substituting into the equilibrium-constant expression, we find... 

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. 

A corollary of this rule has us note that the orders in the rate law [Eq. (6-26)] are lower than those in the equilibrium constant expression for Eq. (6-25). This signals that an intermediate is produced but rapidly consumed. 

As mentioned after Equation (10), the equilibrium constant may be expressed when the reactants are in several phases. As an example, the equilibrium between ammonia in a large cloud droplet and in the gas phase, NH3(aq) and NH3(g), is described by the equilibrium constant expression... 

Solubility equilibria are described quantitatively by the equilibrium constant for solid dissolution, Ksp (the solubility product). Formally, this equilibrium constant should be written as the activity of the products divided by that of the reactants, including the solid. However, since the activity of any pure solid is defined as 1.0, the solid is commonly left out of the equilibrium constant expression. The activity of the solid is important in natural systems where the solids are frequently not pure, but are mixtures. In such a case, the activity of a solid component that forms part of an "ideal" solid solution is defined as its mole fraction in the solid phase. Empirically, it appears that most solid solutions are far from ideal, with the dilute component having an activity considerably greater than its mole fraction. Nevertheless, the point remains that not all solid components found in an aquatic system have unit activity, and thus their solubility will be less than that defined by the solubility constant in its conventional form. 

The equilibrium constant expresses the relationship between the concentrations of products and reactants at equilibrium. 

The equilibrium constant expression is determined from the rates for the forward and reverse reactions ... 

Compare the equilibrium constant expression with the equilibrium reaction to verify that products and reactants are in the right places, with the correct powers. 

C16-0002. Write the equilibrium constant expression for the conversion of molecular oxygen into ozone, O2 O3 (unbalanced). 

Equilibrium constant expressions have some similarities with the rate laws described in Chapter 15. For example, compare the equilibrium constant expression and the rate law for the NO2 / NO reaction ... 

The equilibrium constant expression given by Equation is completely general and can be applied to any chemical reaction. Three features of equilibrium constants are especially important ... 

The direction chosen for the equilibrium reaction Is determined by convenience. A scientist interested in producing ammonia from N2 and H2 would use f. On the other hand, someone studying the decomposition of ammonia on a metal surface would use eq,r Either choice works as long as the products of the net reaction appear in the numerator of the equilibrium constant expression and the reactants appear in the denominator. Example applies this reasoning to the iodine-triiodide reaction. 

We can write the equilibrium constant expression by inspection of the decomposition equilibrium [ 2]eq[I]eq... 

Even in relatively concentrated solutions, the mole fraction of water remains close to 1.00. At a solute concentration of 0.50 M, for example, H2 O = 0.99, only 1% different from 1.00. Equilibrium calculations are seldom accurate to better than 5%, so this small deviation from 1.00 can be neglected. Consequently, we treat solvent water just like a pure substance its concentration is essentially invariant, so it is omitted from the equilibrium constant expression. 

Write the equilibrium constant expression for the reaction of iron metal with strong aqueous acid, and indicate the concentration units for each reagent ... 

The stoichiometry of the reaction determines the form of the equilibrium constant expression. Pure solids, liquids, or solvents do not appear in the expression, since their concentrations are constant. 

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