The Partial Pressure Equilibrium Constant

For reactions with more than one pathway, and for reactions with more than one step, the principle for detailed balance states that, at equilibrium, the forward and reverse reaction rates for each step must be equal, and any two or more single reactions or series of reactions resulting in the same products from identical reactants must have the same equilibrium constant for a given temperature. The equilibrium constant does not depend upon whether or not other substances are present. 

For reactions involving gases, the equilibrium constant can be written in terms of partial pressures. The concentration equilibrium constant and the partial pressure equilibrium constant do not have the same value, but they are related by the equation  

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The partial pressure equilibrium constant for the decomposition of CaCO, is K. If Beaker II is removed, under which of the following conditions would equilibrium NOT be achieved ... 

Similarly from Appendix C, Equation (C-9). if we know the partial pressure equilibrium constant at a temperature, 7, and the heat of reaction, we can find the equilibrium constant at any other temperature... 

Kp, the partial pressure equilibrium constant, is the relationship that exists among the partial pressures of gaseous reactants and products in a reversible reaction at equilibrium. Partial pressures are expressed in bar (after 1982) or atm (before 1982). 

Observe that by setting = 0, the term KsKq/Kq) is simply the overall partial pressure equilibrium constant, Kp, for the reaction... 

In the open system, water is in contact with an unlimited amount of a gas that is, the partial pressure is constant and is not changed by the quantity of the gas that is absorbed in the water phase (cf. Figure 4.2c). This system represents an appropriate model for equilibrium of a surface water with the atmosphere or for rainwater in contact with large quantities of air. 

TABLE 9.6 Partial Pressure Equilibrium Constants for the Oxidation of Sulfur Dioxide ... 

Other conventions for treating equiUbrium exist and, in fact, a rigorous thermodynamic treatment differs in important ways. Eor reactions in the gas phase, partial pressures of components are related to molar concentrations, and an equilibrium constant i, expressed directiy in terms of pressures, is convenient. If the ideal gas law appHes, the partial pressure is related to the molar concentration by a factor of RT, the gas constant times temperature, raised to the power of the reaction coefficients. 

When one of the elements is solid, as in tire case of carbon in the calculation of the partial pressures of tire gaseous species in the reaction between methane and air, CO(g) can be used as a basic element together widr hydrogen and oxygen molecules, and thus the calculation of the final partial pressure of methane must be evaluated using the equilibrium constant for CH4 formation... 

It turns out that there is a relatively simple relationship between the partial pressures of different gases present at equilibrium in a reaction system. This relationship is expressed in terms of a quantity called the equilibrium constant, symbol K In this chapter, you will learn how to—... 

At equilibrium, appreciable amounts of both gases are present From that point on, the amounts of both N02 and N204 and their partial pressures remain constant, so long as the volume of the container and the temperature remain unchanged. 

The amount and hence the partial pressure of N204 drop, rapidly at first, then more slowly. The partial pressure of N02 increases. Finally, both partial pressures level off and become constant At equilibrium, at 100°C,... 

Numerical values of equilibrium constants can be calculated if the partial pressures of products and reactants at equilibrium are known. Sometimes you will be given equilibrium partial pressures directly (Example 12.3). At other times you will be given the original partial pressures and the equilibrium partial pressure of one species (Example 12.4). In that case, the calculation of K is a bit more difficult, because you have to calculate the equilibrium partial pressures of all the species. 

The form of the expression for Q, known as the reaction quotient, is the same as that for the equilibrium constant, K. The difference is that the partial pressures that appear in Q are those that apply at a particular moment, not necessarily when the system is at equilibrium. By comparing the numerical value of Q with that of K, it is possible to decide in which direction the system will move to achieve equilibrium. 

The equilibrium constant for a chemical system can be used to calculate the partial pressures of the species present at equilibrium. In the simplest case, one equilibrium pressure can be calculated, knowing all the others. Consider, for example, the system... 

At a certain temperature, the equilibrium constant for the reaction is 0.0255. What are the partial pressures of all gases at equilibrium if the initial partial pressure of the gases (both reactants and products) is 0.300 atm ... 

The equilibrium constant for the decomposition at a certain temperature is 25. Calculate the partial pressures of all the gases at equilibrium if ammonia with a pressure of 1.00 atm is sealed in a 3.0-L flask. 

We have already noted that if we place liquid water in a flask at 20°C and seal the flask, some water molecules leave the liquid and enter the gas phase. The partial pressure rises as more and more water molecules become part of the gas. Finally, however, the pressure stops rising and the partial pressure of water becomes constant. This partial pressure is the vapor pressure and equilibrium now exists. 

By integration the dependence of the equilibrium cell voltage on the partial pressure of the dissolved gas (with the integration constant K equivalent to A%, [10]) is obtained. 

The vapor pressure of ethanol at 25°C is 58.9 Torr. A sample of ethanol vapor at 25°C and 58.9 Torr partial pressure is in equilibrium with a very small amount of liquid ethanol in a 10.0-L container also containing dry air, at a total pressure of 750.0 Torr. The volume of the container is then reduced at constant temperature to 5.0 1,. (a) What is the partial pressure of ethanol in the smaller volume Explain your reasoning. 

SOLUTION We write the equilibrium constant with the partial pressure of the product, NH3, in the numerator, raised to a power equal to its coefficient in the balanced equation. We do the same for the reactants, but place their partial pressures in the denominator ... 

We use a different measure of concentration when writing expressions for the equilibrium constants of reactions that involve species other than gases. Thus, for a species J that forms an ideal solution in a liquid solvent, the partial pressure in the expression for K is replaced by the molarity fjl relative to the standard molarity c° = 1 mol-L 1. Although K should be written in terms of the dimensionless ratio UJ/c°, it is common practice to write K in terms of [J] alone and to interpret each [JJ as the molarity with the units struck out. It has been found empirically, and is justified by thermodynamics, that pure liquids or solids should not appear in K. So, even though CaC03(s) and CaO(s) occur in the equilibrium... 

Now we come to the most important point in this chapter. At equilibrium, the activities (the partial pressures or molarities) of all the substances taking part in the reaction have their equilibrium values. At this point the expression for Q (in which the activities have their equilibrium values) has become the equilibrium constant, K, of the reaction. That is,... 

The equilibrium constant in Eq. 2 is defined in terms of activities, and the activities are interpreted in terms of the partial pressures or concentrations. Gases always appear in K as the numerical values of their partial pressures and solutes always appear as the numerical values of their molarities. Often, however, we want to discuss gas-phase equilibria in terms of molar concentrations (the amount of gas molecules in moles divided by the volume of the container, [I] = j/V), not partial pressures. To do so, we introduce the equilibrium constant Kt., which for reaction E is defined as... 

H20(g). (a) If the partial pressure of CO, is increased, what happens to the partial pressure of CH4 (h) If the partial pressure of CH4 is decreased, what happens to the partial pressure of CO, (c) If the concentration of CH4 is increased, what happens to the equilibrium constant for the reaction ... 

The following plot shows how the partial pressures of reactant and products vary with time for the decomposition of compound A into compounds B and C. All three compounds are gases. Use this plot to do the following (a) Write a balanced chemical equation for the reaction, (h) Calculate the equilibrium constant for the reaction, (c) Calculate the value of Kc for the reaction at 25°C. 

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