In the previous section, you learned about reactions that can reach a state of equilibrium, in which the concentration of reactants and products aren't changing. If these amounts are changing, we should be able to make a relationship between the amount of product and reactant when a reaction reaches equilibrium.
Equilibrium reactions are those that do not go to completion, but are in a state where the reactants are reacting to yield products and the products are reacting to produce reactants. In a reaction at equilibrium, the equilibrium concentrations of all reactants and products can be measured.
These are all equilibrium constants and are subscripted to indicate special types of equilibrium reactions. Also, note that the concentrations of products in the numerator are multiplied. The same is true of the reactants in the denominator. That leaves just 1 on top in the numerator.
Dividing by 1 does not change the value of K. The equilibrium constant value is the ratio of the concentrations of the products over the reactants. We do this by setting up a chart that shows the initial concentrations of all species, how the concentrations of each will change, and what the final concentrations of each species will be. For this process, the chart is given next don't panic, we'll explain how we got all these values in a minute :.
To figure out what the concentrations of each species are, we plug these values into the expression for finding the equilibrium constant. Since we know that the equilibrium constant is 1. Now, solving for x won't be a lot of fun because we'll need to use the quadratic equation. Let's face it, we don't really want to memorize the quadratic equation, much less use it. Fortunately, there's a shortcut we can use to get around using the quadratic equation when the K eq values are very small.
If K eq is small, then we can safely guess that the amount of product formed x will be very, very small when compared to the initial quantity of the reactant. This is because a very small K eq value means that very little product has been formed. As a result, we can omit the "x" in the denominator to simplify the [1. Our new and easier expression to solve is transformed into:. Let's see if this was a good assumption. If the amount of product formed was 1. As a result, our assumption was valid and we saved ourselves a lot of mathematical heartache and toil.
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Consequently, the numerical values of K and K p are usually different. They are, however, related by the ideal gas constant R and the temperature T :. The temperature is expressed as the absolute temperature in kelvins. According to Equation The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0. The balanced equilibrium equation is as follows:. What is K p for this reaction at the same temperature? Given: equilibrium equation, equilibrium constant, and temperature.
Asked for: K p. Then use Equation Thus, from Equation When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium An equilibrium in which the reactants and products of an equilibrium reaction form a single phase, whether gas or liquid.
In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium An equilibrium in which the reactants of an equilibrium reaction, the products, or both are in more than one phase. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids.
The reference states for pure solids and liquids are those forms stable at 1 bar approximately 1 atm , which are assigned an activity of 1. Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:.
The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:. Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value:. We can rearrange Equation Incorporating the constant value of [C] into the equilibrium equation for the reaction in Equation The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:.
Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. As shown in Figure In the system, the equilibrium composition of the gas phase at a given temperature, K in this case, is the same whether a small amount of solid carbon left or a large amount right is present.
Write each expression for K , incorporating all constants, and K p for the following equilibrium reactions. Given: balanced equilibrium equations. Asked for: expressions for K and K p. Find K by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation.
Then express K p as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation. This reaction contains a pure solid PCl 5 and a pure liquid PCl 3. Their concentrations do not appear in the equilibrium constant expression because they do not change significantly. This reaction contains two pure solids Fe 3 O 4 and Fe , which do not appear in the equilibrium constant expressions.
The two gases do, however, appear in the expressions:. Write the expressions for K and K p for the following reactions. For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products recall that pure water is about Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid.
The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction. The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium.
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions. As we stated in Section The reaction normally occurs in two distinct steps.
In the first reaction 1 , N 2 reacts with O 2 at the high temperatures inside an internal combustion engine to give NO. Summing reactions 1 and 2 gives the overall reaction of N 2 with O 2 :. The equilibrium constant expressions for the reactions are as follows:.
Multiplying K 1 by K 2 and canceling the [NO] 2 terms,. Thus the product of the equilibrium constant expressions for K 1 and K 2 is the same as the equilibrium constant expression for K 3 :. The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. To determine K for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Calculate the equilibrium constant for the following reaction at the same temperature. Given: two balanced equilibrium equations, values of K , and an equilibrium equation for the overall reaction.
Asked for: equilibrium constant for the overall reaction. Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of K for that equation. Calculate K for the overall equation by multiplying the equilibrium constants for the individual equations. The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and The values for K 1 and K 2 are given, so it is straightforward to calculate K 3 :.
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide.
Calculate the equilibrium constant for the overall reaction at this same temperature. The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant K , a unitless quantity.
The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same K.
For a system at equilibrium, the law of mass action relates K to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation.
The ratio is called the equilibrium constant expression.
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