Page 246 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 16] RATES AND EQUILIBRIUM 235

Ans. For this equation, the terms involving the concentrations of the elements are in the numerator, because they are the
products:
3
[H 2 ] [N 2 ]
K =
[NH 3 ] 2
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 = = 0.143
(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:
←− C + D
A + B −→
Initial concentrations 1.50 2.50 0.00 0.00
Changes produced by reaction
Equilibrium concentrations 0.45
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 1.50 2.50 0.00 0.00
Changes produced by reaction −0.45 −0.45 +0.45 +0.45
Equilibrium concentrations 0.45