Page 248 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 16] RATES AND EQUILIBRIUM 237

−→
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 + D
Initial concentrations 1.25 1.50 0.00 0.00
Changes produced by reaction
Equilibrium concentrations x
The changes due to the chemical reaction are as easy to determine as before, and therefore so are the equilibrium
concentrations:

−→
A + B ←− C + D
Initial concentrations 1.25 1.50 0.00 0.00
Changes produced by reaction −x −x x x
Equilibrium concentrations 1.25 − x 1.50 − x x x

We could substitute these equilibrium concentrations in the equilibrium constant expression and solve by using the
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) x 2
K = = = = 0.0010
[A][B] (1.25)(1.50) 1.88
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.

(x) 2
K = = 0.100
(1.25 − x)(1.50 − x)
2
2
x = (0.100)(1.25 − x)(1.50 − x) = 0.100(x − 2.75x + 1.88)
2
0.900x + 0.275x − 0.188 = 0

2
2
−b ± (b − 4ac) −0.275 ± (0.275) − 4(0.900)(−0.188)
x = =
2a 2(0.900)