CHM 1046. Chapter 14 Homework Solutions.
Problems: 4, 8, 14, 21, 28, 32, 38, 46, 52, 56, 62, 66, 80
4)
The equilibrium constant tells you the relationship between product and reactant
concentrations that exists, at equilibrium, in a system at some temperature T. A large
equilibrium constant tells you that at equilibrium the concentrations of products will be
much larger than the concentrations of reactants. A small equilibrium constant means the
opposite, that the concentrations of reactants at equilibrium will be large in comparison to
the concentrations of products.
8)
When calculating KC the units that should be used for concentration are molarity
(mol/L). When calculating Kp the units that should be used for pressure are atmospheres
(atm).
KC and Kp do not themselves have units. This is because what appears in either of
these equilibrium constants are quantities divided by a standard quantity.
For KC
[A]/(1. mol/L), and so forth...
For Kp
pA/(1. atm), and so forth...
14)
a) If Q < K the reaction will shift from left to right (from reactants to products) as
equilibrium is approached.
b) If Q > K the reaction will shift from right to left (from products to reactants) as
equilibrium is approached.
c) If Q = K then the system is at equilibrium, and so no shifting will occur
21)
I will write both KC and Kp. Note that in general Kp can be written only for
processes where the only thing appearing in the equilibrium constant are gases.
a)
KC = [SbCl3] [Cl2]
[SbCl5]
Kp = (pSbCl3) (pCl2)
(pSbCl5)
b)
KC = [NO]2 [Br2]
[BrNO]2
Kp = (pNO)2 (pBr2)
(pBrNO)2
c)
KC = [CS2] [H2]4
[CH4] [H2S]2
Kp = (pCS2) (pH2)4
(pCH4) (pH2S)2
d)
KC =
[CO2]2
[CO]2 [O2]
Kp =
1
(pCO2)2
(pCO)2 (pO2)
28)
For the reaction as written, Kp = (pCO2) (pCF4) = 2.2 x 106
(pCOF2)2
a) Kp = (pCO2)1/2 (pCF4)1/2 = { (pCO2)(pCF4)/(pCOF2)2 }1/2 ; so Kp = (2.2 x 106)1/2
(pCOF2)
= 1.5 x 103 (products favored)
b) Kp = (pCO2)3 (pCF4)3 = { (pCO2)(pCF4)/(pCOF2)2 }3 = ; so Kp = (2.2 x 106)3
(pCOF2)6
= 1.1 x 1019 (products favored)
c) Kp =
(pCOF2)4
(pCO2)2 (pCF4)2
= { (pCO2)(pCF4)/(pCOF2)2 }-2 so Kp = (2.2 x 106)-2
= 2.1 x 10-13 (reactants favored)
32)
For all of these problems we use the following relationship, derived in class
Kp = KC (RT)n , where
n = (moles of gas as product) - (moles of gas as reactant)
Note that T must be in K, and R must be 0.08206 (it normally would have units of
L.atm/mol.K, but KC and Kp have no units for reasons discussed in class).
a) KC = 5.9 x 10-3
T = 298 K
n = 2 - 1 = 1
Kp = (5.9 x 10-3) (0.08206 . 298) = 0.144
b) KC = 3.7 x 108
T = 298 K
n = 2 - 4 = -2
Kp = (3.7 x 108) (0.08206 . 298)-2 = 6.2 x 105
c) KC = 4.10 x 10-31
T = 298 K
n = 2 - 2 = 0
Kp = (4.10 x 10-31) (0.08206 . 298)0 = 4.10 x 10-31
In fact, if n = 0 , then Kp = KC.
38)
For this reaction KC = [HI]2
[H2] [I2]
At 25 C, missing KC, so KC =
(0.922)2
= 617.
(0.0355) (0.0388)
At 340. C, missing [H2]
KC = [HI]2
[H2] [I2]
, so [H2] = [HI]2 = (0.387)2
= 0.343 mol/L
KC [I2] (9.6) (0.0455)
2
At 445. C, missing [HI]
KC = [HI]2 , so [HI] = {KC [H2] [I2] }1/2 = {(50.2) (0.0485) (0.0468)}1/2
[H2] [I2]
= 0.338 mol/L
46)
Qp = (pH2)2 (pS2) = (0.112)2 (0.055) = 3.5 x 10-3.
(pH2S)2
(0.445)2
Since Qp > Kp we need to reduce its value. Therefore the reaction will shift from
right to left (more H2S will form).
52)
KC =
CO
Cl2
COCl2
[COCl2] = 255.
[CO] [Cl2]
Initial
0.1500
0.175
0.000
Change
-x
-x
+x
Equilibrium
0.1500 - x
0.1750 - x
x
(Note we have defined x as the concentration of COCl2 that forms).
(x)
(0.1500 - x) (0.1750 - x)
= 255.
Since it does not look like x is small compared to 0.1500, we will have to solve
the above by the quadratic formula.
x = (0.1500 - x) (0.1750 - x) (255.) = 6.6938 - 82.875 x + 255 x2
255 x2 - 83.875 x + 6.6938 = 0
x = 83.875  [(83.875)2 - 4 (255) (6.6938)]1/2 = 0.1927, 0.1362
2 (255)
The underlined root is correct. The other root (0.1927) would lead to a negative
value for the concentration of both CO and Cl2.
So at equilibrium
[CO] = 0.0138 M
[Cl2] = 0.0388 M
[COCl2] = 0.1362 M
As a check, if we substitute into our expression for KC, we get
KC =
(0.1362)
= 254.  255.
(0.0138) (0.0388)
3
56)
KC =
SO2
Cl2
SO2Cl2
[SO2] [Cl2]
[SO2Cl2]
= 2.99 x 10-7
Initial
0.000
0.000
0.175
Change
+x
+x
-x
Equilibrium
x
x
0.175 - x
(Note we have defined x as the concentration of SO2 that forms).
(x)2
(0.1750 - x)
= 2.99 x 10-7
Since KC << 1, it looks like it is worth assuming that x << 0.1750. If we make
that assumption, then
(x)2

x2
= 2.99 x 10-7
(0.1750 - x)
(0.1750)
So x2 = (2.99 x 10-7) (0.1750) = 5.23 x 10-8
x = (5.23 x 10-8)1/2 = 2.29 x 10-4
Now we go back and check our assumption. Is 2.29 x 10-4 << 0.1750 ? YES, so
our assumption was good.
So at equilibrium
[SO2] = [Cl2] = 2.29 x 10-4 M
[SO2Cl2] = 0.1748 M
62)
We use Le Chatlier's principle here (we could also use the reaction quotient, but
that takes longer and will give the same result).
a) System will shift from right to left (to get rid of some of the added NO).
b) System will shift from left to right (to get rid of some of the added BrNO).
c) System will shift from left to right (to replace some of the Br2 that was
removed).
4
66)
a) Since volume is decreased all of the concentrations will increase. The system
will respond by trying to reduce the total concentration. However, since there are the
same number of moles of reactants and products, there will be no shift in the reaction.
b) Since volume is increased all of the concentrations will decrease. The system
will respond by trying to increase the total concentration. Since there are more moles of
reactants than products, the system will shift from right to left.
c) Since volume is increased the concentration of CO2 will decrease. Note we
don't have to consider the "concentration" of solids, since they don't enter into the
expression for the equilibrium constant.
So the system will respond by increasing the concentration of CO2. therefore the
reaction will shift from left to right.
80)
It is easier to find Kp and then calculate KC.
Kp =
(pSO3)2
(pSO2)2 (pO2)
Initial
SO2
O2
SO3
Change
3.00
1.00
0.00
- 2x
-x
+ 2x
Equilibrium
3.00 - 2x
1.00 - x
2x
(Note we have defined 2x as the pressure of SO3 that forms).
Now (pSO2) + (pO2) + (pSO3) = 3.75 atm at equilibrium, so
(3.00 - 2x) + (1.00 - x) + 2x = 3.75
4.00 - x = 3.75 ; x = 0.25
So at equilibrium
Kp =
pSO2 = 2.50 atm
pO2 = 0.75 atm
pSO3 = 0.50 atm
(0.50)2
= 0.0533 Kp = KC (RT)n ; so KC = Kp/(RT)n ; n = -1
2
(2.50) (0.75)
So KC = Kp
= Kp (RT) = (0.0533) (0.08206) (300.) = 1.31
-1
(RT)
5