Page 254 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 16] RATES AND EQUILIBRIUM 243

−→
(a) In the reaction B ←− C + D, 0.16 mol of B, 0.10 mol of C, and 0.10 mol of D are found in a 1.0-L reaction
mixture at equilibrium. Calculate the value of K. If the same mixture were found in a 2.0-L reaction mixture
at equilibrium, would K be different?
−→
(b) In the reaction 2 B ←− C + D, 0.16 mol of B, 0.10 mol of C, and 0.10 mol of D are found in a 1.0-L reaction
mixture at equilibrium. Calculate the value of K. If the same mixture were found in a 2.0-L reaction mixture
at equilibrium, would K be different? Compare this answer to that in part a, and explain.
Ans. (a) In the two reactions we have
(0.10)(0.10) (0.050)(0.050)
K 1 = = 0.063 K 2 = = 0.031
0.16 0.080
and the values are different.
(b) In the two reactions we have
(0.10)(0.10) (0.050)(0.050)
K 1 = = 0.39 K 2 = = 0.39
(0.16) 2 (0.080) 2
The values are the same. In part b, 2.0 L appears twice in the numerator and twice in the denominator,
so that it cancels out. In part a, it appears only once in the denominator and twice in the numerator, and
so the constant in the 2.0-L volume would be half as large as that in the 1.0-L volume. Note that in part
a, the same mixture would not appear in the 2.0-L mixture at the same temperature, because the value
of K does not change with volume.

16.29. Calculate the units of K in Problem 16.14b and c. Can you tell from the units which of the equations is referred to?
Ans. TheunitforpartbisL/mol;thatforpartcis(L/mol) 1/2 .IfavalueisgiveninL/molforthereaction,theequation
in part b is the one to be used. If the units are the square root of those, the equation of part c is to be used.

16.30. What is the sequence of putting values in the table in the solution of Problem 16.19?
Ans. The superscript numbers before the table entries give the sequence.

2A + B −→ C + 2D
←−
Initial 1 0.90 1 1.60 1 0 1 0
Change 3 −0.70 3 −0.35 2 +0.35 3 +0.70
Equilibrium 4 0.20 4 1.25 1 0.35 4 0.70
−→
−4
16.31. For the reaction A + B ←− X + heat, K = 1.0 × 10 , would an increase in temperature raise or lower the value
of K?
Ans. According to Le Chˆatelier’s principle, raising the temperature would shift this equilibrium to the left. That
means that there would be less X and more A and B present at the new equilibrium temperature. The value of
the equilibrium constant at that temperature would therefore be lower than the one at the original temperature.

[X]
K =
[A][B]
16.32. Calculate the value of the equilibrium constant at a certain temperature for the following reaction if there are present
at equilibrium 0.10 mol of N 2 , 0.070 mol of O 2 , and 1.4 × 10 −3 mol of NO 2 in 2.0 L.
−→
N 2 + 2O 2 ←− 2NO 2
Ans. [N 2 ] = 0.10 mol/2.0L = 0.050 mol/L
[O 2 ] = 0.070 mol/2.0L = 0.035 mol/L
[NO 2 ] = 0.0014 mol/2.0L = 0.000 70 mol/L
[NO 2 ] 2 (0.000 70) 2
K = = = 8.0 × 10 −3
[N 2 ][O 2 ] 2 (0.050)(0.035) 2