EQUILIBRIUM CONSTANTS

As its name suggests, “equilibrium” is a balance – a balance between the forward and reverse rates of a chemical reaction. But the balance exists also between the concentrations of the reactants and products. We expect that for any given reaction, the concentrations of the reactants and the products at equilibrium will be related to one another, and that relationship will be determined by the intrinsic rates of the forward and reverse reactions. For a reaction with a fast forward rate and a slow reverse rate, for example, we expect that at equilibrium, the concentrations of the products will be high and that of the reactants will be low. It’s much easier to measure concentrations than rates, and, as it turns out, it’s much easier to characterize the situation at equilibrium using concentrations. In fact, all you need is a balanced equation and measured values for the concentrations of reactants and products at equilibrium.

EQUILIBRIUM CONSTANTS Given the balanced equation, all you need to know to determine the concentrations of the reactants and products at equilibrium is: How much of each reactant and product you started with; and The concentration of just one of the reactants or products at equilibrium. Although that calculation is not difficult, we will always give all of the equilibrium concentrations.

EQUILIBRIUM CONSTANTS The position of equilibrium is described by a constant, K eq, called the equilibrium constant. It is equal to the ratio of product over reactant concentrations. The equation is called the equilibrium constant expression. The ratio is constructed in this way (products divided by reactants) so that the constant will be a large number if the concentrations of products are large (that is, the equilibrium favors the formation of products) and the constant will be small if the concentrations of products are small. In a sense, the equilibrium constant is a measure of the extent to which the reaction goes to completion.

EQUILIBRIUM CONSTANTS The exact expression is constructed by taking the concentration of each product and reactant, raising it to a power equal to its coefficient in the balanced equation, then placing the product concentrations on the top of the ratio and the reactant concentration on the bottom of the ratio. The brackets mean “concentration of,” so the notation [N 2 ] means “concentration of N 2 ” – usually given in moles per liter (molarity).

EQUILIBRIUM CONSTANTS The equilibrium constant, K eq, has a value that is influenced by the temperature and by the total pressure. In this lesson, we will ignore the variation of K eq with T and P and assume in the calculations and problems that we do that these variables don’t change. You already know that changing the temperature can shift the position of equilibrium and that raising or lowering the total pressure by decreasing or increasing the volume of the container can also shift the position of equilibrium, so it shouldn’t surprise you that K eq changes when these two parameters change.

EQUILIBRIUM CONSTANTS Here’s another example of an equilibrium constant expression. Notice again that the exponents in the expression are just the coefficients in the balanced equation.

EQUILIBRIUM CONSTANTS If we know the values of the concentrations at equilibrium, it’s easy to calculate the value of K eq. Suppose, for example, that for the hypothetical reaction shown, we measure the equilibrium concentrations of A, B, and A 2 B and find them to be 0.50 M, 1.0 M, and 2.0 M, respectively. Try doing this problem on paper before you go on.

EQUILIBRIUM CONSTANTS The equilibrium constant expression is [A 2 B] divided by the product of [A] 2 times [B]. Substituting the values for the concentrations into this expression, we find that the value of K eq is 8.0. In most equilibrium constant calculations we ignore the units. In this example, units of K eq would be mol -2 L 2. But the units of K eq vary, depending on the relative numbers of moles of reactants and products in the balanced equation. Since there is no single, universally applicable set of units for the equilibrium constant, we normally do not include them at all.

EQUILIBRIUM CONSTANTS Here is a slightly different kind of problem. Suppose that we know the concentrations of all of the reactants and products at equilibrium but one, and we also know the value of the equilibrium constant. We can use the equilibrium constant expression to calculate the concentration of the unknown reactant or product. Again, we will ignore the units of the equilibrium constant and assume them to be appropriate to the values of the concentrations expressed in moles per liter.

EQUILIBRIUM CONSTANTS Again, we start with the equilibrium constant expression. We then solve it for the variable whose value we want: in this case, the concentration of A. One way to check our calculations is to substitute the values of the concentrations back into the equilibrium constant expression to see if they give us the correct value for K eq : 1/(0.352 x 1) = 8.2. Pretty close. The discrepancy is due to our having rounded the answer to 2 digits – had we rounded [A] to three digits, we would have gotten 8.0.

EQUILIBRIUM CONSTANTS Notice that there are many, in fact an infinite number, of combinations of concentrations that satisfy the equilibrium constant expression. The expression for a given reaction tells us at equilibrium not what the concentrations are, but what the relationship between them must be. That relationship is determined by the relative rates of the forward and reverse reaction. In fact, it is equal to their ratio.

EQUILIBRIUM CONSTANTS Write the Equilibrium constant expression for the reaction: N 2 (g) + 3 H 2 (g) 2 NH 3 (g).

EQUILIBRIUM CONSTANTS N 2 (g) + 3 H 2 (g) 2 NH 3 (g). Where [ ] represents equilibrium concentrations. Notice that substances on right of equation go on top of the expression and coefficients in equations become exponents. Concentrations are multiplied together.

EQUILIBRIUM CONSTANTS How about this one? 2 A + B 2 A 2 B + B

EQUILIBRIUM CONSTANTS 2 A + B 2 A 2 B + B

EQUILIBRIUM CONSTANTS K eq === 8.0 Calculation of K eq for reaction 2 A + B  A 2 B if at equilibrium: [A] = 0.50 M [B] = 1.0 M [A 2 B] = 2.0 M

EQUILIBRIUM CONSTANTS Calculate the numerical value of K eq for the reaction X + 3 Y XY 3 if at equilibrium [X]=0.1 M, [Y]=2.0 M, and [XY 3 ]=0.4 M.

EQUILIBRIUM CONSTANTS K eq === 0.5

EQUILIBRIUM CONSTANTS Write the equilibrium constant equation and calculate the value of the equilibrium constant for this reaction: Ag NH 3 Ag(NH 3 ) 2 + [Ag + ] = 3.0 x M [NH 3 ] = 1.0 M [Ag(NH 3 ) 2 + ] = 5.0 x M

EQUILIBRIUM CONSTANTS K eq === 1.7x10 7

EQUILIBRIUM CONSTANTS For the reaction A + B C, which of the following cases is not at equilibrium (given that only one is not)? Case I: [A] = 5 [B] = 2 [C] = 1 Case II: [A] = 2 [B] = 5 [C] = 1 Case III: [A] = 4 [B] = 5 [C] = 2 Case IV: [A] = 1 [B] = 5 [C] = 2