EQUILIBRIUM CONSTANTS
16.4. EQUILIBRIUM CONSTANTS
Although Le Chˆatelier’s principle does not tell us how much an equilibrium will be shifted, there is a way to determine the position of an equilibrium once data have been determined for the equilibrium experimentally. At a given temperature the ratio of concentrations of products to reactants, each raised to a suitable power, is constant for a given equilibrium reaction. The letters A, B, C, and D are used here to stand for general chemical species. Thus, for a chemical reaction in general,
a A + bB −→ ←− c C + dD
The following ratio always has the same value at a given temperature:
[C] c [D] d K=
[A] a [B] b
Here the square brackets indicate the concentration of the chemical species within the bracket. That is, [A] means the concentration of A, and so forth. [A] a means the concentration of A raised to the power a, where a is the value of the coefficient of A in the balanced equation for the chemical equilibrium. The value of the ratio of concentration terms is symbolized by the letter K, called the equilibrium constant. For example, for the reaction of hydrogen and nitrogen referred to in Sec. 16.3,
3H 2 + N −→ 2 ←− 2 NH 3
the ratio is
The exponents 2 and 3 are the coefficients of ammonia and hydrogen, respectively, in the balanced equation. Please note carefully the following points about the equilibrium constant expression:
1. This is a mathematical equation. Its values are numbers, involving the concentrations of the chemicals and their coefficients.
2. Each concentration is raised to the power given by the coefficient in the chemical equation.
3. The concentrations of the products of the reaction are written in the numerator of the equilibrium constant expression; the concentrations of the reactants are in the denominator.
4. The terms are multiplied together, not added.
5. Each equilibrium constant expression is associated with a particular chemical reaction written in a given direction.
EXAMPLE 16.7. Write the equilibrium constant expression for the reaction
CHAP. 16]
RATES AND EQUILIBRIUM
Ans. For this equation, the terms involving the concentrations of the elements are in the numerator, because they are the products:
There are a great many types of equilibrium problems. We will start with the easiest and work up to harder ones.
EXAMPLE 16.8. Calculate the value of the equilibrium constant for the following reaction if at equilibrium the concen- tration of A is 1.60 M, that of B is 0.800 M, that of C is 0.240 M, and that of D is 0.760 M.
A + B −→ ←− C + D
Ans. Since all the coefficients in the balanced equation are equal to 1, the equilibrium constant expression is
[C][D] K= [A][B]
We merely substitute the equilibrium concentrations into this equation to determine the value of the equilibrium constant:
( 0.240 M)(0.760 M)
K=
1.60 M)(0.800 M)
One way to make Example 16.8 harder is by giving numbers of moles and a volume instead of concentrations at equilibrium. Since the equilibrium constant is defined in terms of concentrations, we must first convert the numbers of moles and volume to concentrations. Note especially that the volume of all the reactants is the same, since they are all in the same system.
EXAMPLE 16.9. Calculate the value of the equilibrium constant if at equilibrium there are 0.400 mol A, 0.200 mol B, 0.0600 mol C, and 0.190 mol D in 0.250 L of solution.
A + B −→ ←− C + D
Ans. Since numbers of moles and a volume are given, it is easy to calculate the equilibrium concentrations of the species. In this case, [A] = 1.60 M, [B] = 0.800 M, [C] = 0.240 M, and [D] = 0.760 M, which are exactly the concentrations given in the prior example. Thus, this problem has exactly the same answer.
It is somewhat more difficult to determine the value of the equilibrium constant if some initial concentrations instead of equilibrium concentrations are given.
EXAMPLE 16.10. Calculate the value of the equilibrium constant in the following reaction if 1.50 mol of A and 2.50 mol of B are placed in 1.00 L of solution and allowed to come to equilibrium. The equilibrium concentration of C is found to be
0.45 M.
A + B −→ ←− C + D
Ans. To determine the equilibrium concentrations of all the reactants and products, we must deduce the changes that have occurred. We can assume by the wording of the problem that no C or D was added by the chemist, that some
A and B have been used up, and that some D has also been produced. It is perhaps easiest to tabulate the various concentrations. We use the chemical equation as our table headings, and we enter the values that we know now:
A+B
−→ ←− C+D
Initial concentrations
Changes produced by reaction Equilibrium concentrations
We deduce that to produce 0.45 M C it takes 0.45 M A and 0.45 M B. Moreover, we know that 0.45 M D was also produced. The magnitudes of the values in the second row of the table—the changes produced by the reaction—are always in the same ratio as the coefficients in the balanced chemical equation .
A + B ←− −→ C + D
Initial concentrations
Changes produced by reaction
RATES AND EQUILIBRIUM
[ CHAP. 16
Note that the concentrations of reactants are reduced. We then merely add the columns to find the rest of the equilibrium concentrations:
A + B ←− −→ C + D
Initial concentrations
Changes produced by reaction
1.05 2.05 0.45 0.45 Now that we have calculated the equilibrium concentrations, we can substitute these values into the equilibrium
Equilibrium concentrations
constant expression:
[C][D]
[A][B]
When the chemical equation is more complex, the equilibrium constant expression is also more complex and the deductions about the equilibrium concentrations of reactants and products are more involved, too.
EXAMPLE 16.11. Calculate the value of the equilibrium constant for the following reaction if 1.50 mol of A and 2.50 mol of B are placed in 1.00 L of solution and allowed to come to equilibrium. The equilibrium concentration of C is found to be
0.45 M.
2 A + B −→ ←− C + D
Ans. The equilibrium constant expression for this equation is
[C][D] K= [A] 2 [B]
2A+B −→ ←− C
Initial concentrations
Changes produced by reaction Equilibrium concentrations
The changes brought about by the chemical reaction are a little different in this case. Twice as many moles per liter of A are used up as moles per liter of C are produced. Note that the magnitudes in the middle row of this table and the coefficients in the balanced chemical equation are in the same ratio.
2A+ B ←− −→ C + D
Initial concentrations
Changes produced by reaction
Equilibrium concentrations
Adding the columns gives 2A+ B ←− −→ C + D
Initial concentrations
Changes produced by reaction
0.60 2.05 0.45 0.45 The equilibrium values are substituted into the equilibrium constant expression:
Equilibrium concentrations
[C][D]
K=
[A] 2 [B] = ( 0.60) 2 = ( 0.27 2.05)
The next type of problem gives the initial concentrations of the reactants (and products) plus the value of the equilibrium constant and requires calculation of one or more equilibrium concentrations. We use algebraic quantities (such as x) to represent at least one equilibrium concentration, and we solve for the others if necessary in terms of that quantity.
CHAP. 16]
RATES AND EQUILIBRIUM
EXAMPLE 16.12. For the reaction
A + B −→ ←− C + D
1.25 mol of A and 1.50 mol of B are placed in a 1.00-L vessel and allowed to come to equilibrium. The value of the equilibrium constant is 0.0010. Calculate the concentration of C at equilibrium.
Ans.
A + B ←− C −→
Initial concentrations
Changes produced by reaction Equilibrium concentrations
The changes due to the chemical reaction are as easy to determine as before, and therefore so are the equilibrium concentrations:
+ D Initial concentrations
A + B ←− C −→
1.25 1.50 0.00 0.00 Changes produced by reaction − x
x Equilibrium concentrations
x We could substitute these equilibrium concentrations in the equilibrium constant expression and solve by using the
1.25 − x
1.50 − x
quadratic equation (Appendix). However, it is more convenient to attempt to approximate the equilibrium concentra- tions by neglecting a small quantity (x) when added to or subtracted from a larger quantity (such as 1.25 or 1.50). We do not neglect small quantities unless they are added to or subtracted from larger quantities! Thus, we approximate the equilibrium concentrations as [A] ∼ =
1.25 M and [B] ∼ =
1.50 M. The equilibrium constant expression is thus
[C][D]
( x)(x)
[A][B]
x= 0.043 M
Next we must check whether our approximation was valid. Is it true that 1.25 − 0.043 ∼ = 1.25 and 1.50 − 0.043 ∼ = 1.50? Considering the limits of accuracy of the equilibrium constant expression, results within 5% accuracy are considered valid for the general chemistry course. These results are thus valid. The equilibrium concentrations are
[A] = 1.25 − 0.043 = 1.21 M [B] = 1.50 − 0.043 = 1.46 M [C] = [D] = 0.043 M
If the approximation had caused an error of 10% or more, you would not be able to use it. You would have to solve by a more rigorous method, such as the quadratic equation or using an electronic calculator to solve for the unknown value. (Ask your instructor if you are responsible for the quadratic equation method.)
EXAMPLE 16.13. Repeat the prior example, but with an equilibrium constant value of 0.100. Ans.
The problem is done in exactly the same manner until you find that the value of x is 0.434 M.
x 2 = 0.100 1.88
x= 0.434
When that value is subtracted from 1.25 or 1.50, the answer is not nearly the original 1.25 or 1.50. The approximation is not valid. Thus, we must use the quadratic equation.
1.25 − x)(1.50 − x)
x 2 = ( 0.100)(1.25 − x)(1.50 − x) = 0.100(x 2 − 2.75x + 1.88)
0.900x 2 + 0.275x − 0.188 = 0
− b± ( b 2 − 4ac)
x=
2a = 2(0.900)
RATES AND EQUILIBRIUM
[ CHAP. 16
Two solutions for x, one positive and one negative, can be obtained from this equation, but only one will have physical meaning. There cannot be any negative concentrations.
x= 0.329 M = [C] = [D]
The concentrations of A and B are then
[A] = 1.25 − 0.329 = 0.92 M [B] = 1.50 − 0.329 = 1.17 M
The value from the equilibrium constant expression is therefore equal to the value of K given in the problem: