Page 251 - Theory and Problems of BEGINNING CHEMISTRY

240 RATES AND EQUILIBRIUM [CHAP. 16

16.14. Write equilibrium constant expressions for the following equations. Tell how these expressions are related.

−→ −→
(a)N 2 O 4 ←− 2NO 2 (b)2 NO 2 ←− N 2 O 4 −→ 1
(c)NO 2 ←− 2 N 2 O 4
[NO 2 ] 2 [N 2 O 4 ] [N 2 O 4 ] 1/2
Ans. (a) K = (b) K = 2 (c) K =
[N 2 O 4 ] [NO 2 ] [NO 2 ]
The K in part b is the reciprocal of that in part a. The K in part c is the square root of that in part b.

16.15. Write an equilibrium constant expression for each of the following:
1 −→ 1 −→
(a)SO 2 + O 2 ←− SO 3 (b)2 NO + O 2 ←− N 2 O 3
2 2
−→
(c)CH 3 C(OH) CHCOCH 3 ←− CH 3 COCH 2 COCH 3
[SO 3 ] [N 2 O 3 ] [CH 3 COCH 2 COCH 3 ]
Ans. (a) K = (b) K = (c) K =
[SO 2 ][O 2 ] 1/2 [NO] [O 2 ] 1/2 [CH 3 C(OH) CHCOCH 3 ]
2
16.16. Write an equilibrium constant expression for each of the following:
−→ −→
(a)N 2 + 3H 2 ←− 2NH 3 (d) PCl 5 (g) ←− PCl 3 (g) + Cl 2 (g)
−→
(b)H 2 + Br 2 (g) ←− 2 HBr(g) (e)2 NO 2 ←− 2NO + O 2
−→
−→
(c)SO 2 Cl 2 ←− SO 2 + Cl 2
2
[NH 3 ] 2 [SO 2 ][Cl 2 ] [NO] [O 2 ]
Ans. (a) K = (c) K = (e) K =
[N 2 ][H 2 ] 3 [SO 2 Cl 2 ] [NO 2 ] 2
[HBr] 2 [PCl 3 ][Cl 2 ]
(b) K = (d) K = [PCl 5 ]
[H 2 ][Br 2 ]
−→
16.17. X + Y ←− Z
A mixture of 1.20 mol of X, 2.10 mol of Y, and 0.950 mol of Z is found at equilibrium in a 1.00-L vessel.
(a) Calculate K.(b) If the same mixture had been found in a 2.00-L reaction mixture, would the value of
K have been the same? Explain.
[Z] 0.950 0.475
Ans. (a) K = = = 0.377 (b) K = = 0.754
[X][Y] (1.20)(2.10) (0.600)(1.05)
The value is not the same. The value of K is related to the concentrations of the reagents, not to their numbers
of moles.

−→
16.18. A + B ←− C + D
After 1.00 mol of A and 1.80 mol of B were placed in a 1.00-L vessel and allowed to achieve equilibrium,
0.30 mol of C was found. In order to determine the value of the equilibrium constant, determine (a) the
quantity of C produced, (b) the quantity of D produced, (c) the quantities of A and B used up, (d) the
quantity of A remaining at equilibrium, by considering the initial quantity and the quantity used up, (e)
the quantity of B remaining at equilibrium, and ( f ) the value of the equilibrium constant.
Ans. (a), (b), and (c) 0.30 mol each. (d)1.00 − 0.30 = 0.70 mol (e)1.80 − 0.30 = 1.50 mol
[C][D] (0.30) 2
( f ) K = = = 8.6 × 10 −2
[A][B] (0.70)(1.50)
−→
16.19. 2A + B ←− C + 2D
If 0.90 mol of A and 1.60 mol of B are placed in a 1.00-L vessel and allowed to achieve equilibrium,
0.35 mol of C is found. Use a table such as shown in Example 16.10 to determine the value of the
equilibrium constant.