Page 252 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 16] RATES AND EQUILIBRIUM 241

Ans. 2A + B −→ C + 2D
←−
Initial 0.90 1.60 0 0
Change −0.70 −0.35 +0.35 +0.70
Equilibrium 0.20 1.25 0.35 0.70

[C][D] 2 (0.35)(0.70) 2
K = = = 3.4
2
[A] [B] (0.20) (1.25)
2
16.20. In which line of the table used to calculate the equilibrium concentrations are the values in the ratio of
the coefficients of the balanced chemical equation? Is this true in other lines?
Ans. The terms in the second line are in that ratio, since it describes the changes made by the reaction. The terms
in the other lines are not generally in that ratio.

−7
16.21. For the following reaction, K = 1.0 × 10 :
−→
W + Q ←− R + Z
If 1.0 mol of W and 2.5 mol of Q are placed in a 1.0-L vessel and allowed to come to equilibrium, calculate
the equilibrium concentration of Z, using the following steps: (a) If the equilibrium concentration of Z
is equal to x, how much Z was produced by the chemical reaction? (b) How much R was produced by
the chemical reaction? (c) How much W and Q were used up by the reaction? (d) How much W is left
at equilibrium? (e) How much Q is left at equilibrium? ( f ) With the value of the equilibrium constant
given, will x (equal to the Z concentration at equilibrium) be significant when subtracted from 1.0?
(g) Approximately what concentrations of W and Q will be present at equilibrium? (h) What is the value
of x?(i) What is the concentration of R at equilibrium? ( j) Is the answer to part f justified?
Ans. (a) x. (There was none originally, so all that is there must have come from the reaction.) (b) x. (The Z and
R are produced in equal mole quantities, and they are in the same volume.) (c) x each. (The reacting ratio
ofWtoQtoRto Z is 1: 1: 1: 1.) (d)1.0 − x (the original concentration minus the concentration used up).
(e)2.5 − x (the original concentration minus the concentration used up). ( f ) No. (We will try it first and
later see that this is true.) (g) 1.0 mol/L and 2.5 mol/L, respectively.
[R][Z] x 2 −7 2 −7 −4
(h) K = = = 1.0 × 10 x = 2.5 × 10 x = 5.0 × 10 = [Z]
[W][Q] (1.0)(2.5)
(i) [R] = x = 5.0 × 10 −4 mol/L
−4
( j) Yes. 1.0 − (5.0 × 10 ) = 1.0

16.22. For the following reaction of gases, K = 1.0 × 10 −11 :

−→
W + 2Q ←− 3R + Z
Determine the concentration of Z at equilibrium by repeating each of the steps in the prior problem.
Calculate the equilibrium concentration of R if 1.0 mol of W and 2.5 mol of Q are placed in a 1.0-L
vessel and allowed to attain equilibrium.
Ans. (a) x (b)3x (c)W: x,Q:2x (d)1.0−x (e)2.5−2x ( f )no (g) 1.0 mol/L and 2.5 mol/L, respectively
3
3
[R] [Z] (3x) (x)
4
(h) K = = = 4.3x = 1.0 × 10 −11 x = 1.2 × 10 −3 = [Z]
[W][Q] 2 (1.0)(2.5) 2
(i) [R] = 3x = 3.6 × 10 −3 mol/L
(j)Yes