13 Chapter Homework Solutions
Textbook Homework
H2O O2
Kc
2
H2O2
2
T11.
(a)
(b) Kc
PCl5
PCl3 Cl2
(c) Kc = [CO]2
(d) Kc
H2S
H2
T13.
To obtain the second equation, one must multiply the first equation by 2 and reverse reactants
and products. Therefore, one squares Kc and inverts it:
1
(e) K c 2 2
K c1
T14.
To obtain the second equation, one must multiply the first equation by 3. Therefore,
the second equilibrium constant is the cube of the first one.
3
(d) K P 2 K P1
T18.
Kc
PCl5 PCl3 Cl2 [PCl5] = 4.2x10-5 M , [PCl3] = 1.3x10-2 M , [Cl2] = 3.9x10-3 M
Kc
T22.
[ PCl3 ][Cl2 ] (1.3 x102 )(3.9 x103 )
1.21 1.2
[ PCl5 ]
(4.2 x105 )
If 11.4% of the HI reacts, then 88.6% of the HI remains; i.e. 0.886 mol HI remains
Initial Concentration: [HI]o = 1.00 mol/1.00 L = 1.00 M
Concentration at equilibrium: [HI] = 0.886 mol/1.00 L = 0.886 M
(a) Lets prepare ICE Table for: 2 HI F H2 + I2
HI
Initial
1.00
Change
-2x
Equilibrium
1.00 - 2x
Equilibrium (num.) 0.886 M
H2
0
+x
x
0.057 M
I2
0
+x
x
0.057 M
We have computed x (used in the last row above) from:
1.00 - 2x = 0.886 2x = 0.114 x = 0.057 M
(b) K c
T25.
[ H 2 ][ I 2 ] (0.057)(0.057)
0.0041
[ HI ]2
(0.886) 2
Butane F 2-Methylpropane BUT F 2MP (for convenience)
Kc = 2.5 [BUT]o = 0.100 M , [2MP]o = 0.100 M
(a) Lets first determine which way the reqaction will proceed from initial concentrations
[2MP ] 0.100
1 2.5( K c )
[ BUT ] 0.100
Because Q < Kc, the reaction will proceed towards the right.
Q
Lets prepare an ICE table
Initial
Change
Equilibrium
BUT
0.100
-x
1.00 - x
2MP
0.100
+x
0.100 + x
(b) Equilibrium Concentrations
At equilibrium: K c
[2MP] 0.100 x
2.5
[ BUT ] 0.100 x
0.100 + x = 2.5(0.100 x) = 0.25 2.5x
x + 2.5x = 0.25 0.100
3.5x = 0.15
x = 0.043
Therefore, [BUT] = 0.100 - x = 0.0057 M and [2MP] = 0.100 + x = 0.143 M
Check: K c
[2MP]
0.143
2.5 Therefore, the concentrations are correct
[ BUT ] 0.0.057
(c) [BUT]o = 0.017 mol/0.50 L = 0.034 M , [2MP]o = 0.00
Initial
Change
Equilibrium
BUT
0.034
-x
0.034 - x
2MP
0.00
+x
x
At Equilibrium: K c
x
0.034 x
[2MP]
x
2.5
[ BUT ] 0.034 x
= 2.5
x = 2.5(0.034 x) = 0.085 2.5x
x + 2.5x = 0.085
3.5x = 0.085
x = 0.024
[BUT] = 0.034 - x = 0.010 M , [2MP] = x = 0.024 M
Check: K c
T27.
[2MP] 0.024
2.4 2.5 K c Agrees to within roundoff error
[ BUT ] 0.01
H2 + I2 F 2 HI
[HI]o = 0.05 mol/1.0 L = 0.05 M , [H2] = [I2] = 0
Initial
Change
Equilibrium
H2
0
+x
x
At Equilibrium: K c
I2
0
+x
x
HI
0.05
-2x
0.05 - 2x
[ HI ]2
(0.05 2 x) 2
76
[ H 2 ][ I 2 ]
( x)( x)
To avoid the need to solve the quadratic equation, take the square root of both sides
(0.05 2 x)
76 8.72
x
0.05 - 2x = 8.72 x
10.72x = 0.05
x = 0.0047
Therefore, [H2] = [I2] = x = 0.0047 M , [HI] = 0.05 - 2x = 0.0406 0.041
Check: K c
T28.
[ HI ]2
(0.041) 2
76.1 76 ( K c ) Concentrations are correct
[ H 2 ][ I 2 ] (0.0047)(0.0047)
N2 + O2 F 2 NO
[NO]o = 0.15 mol/10.0 L = 0.015 M , [N2] = [O2] = 0
Initial
Change
Equilibrium
N2
0
+x
x
O2
0
+x
x
NO
0.015
-2x
0.015 - 2x
[ NO]2
(0.015 2 x) 2
1.7 x103
[ N 2 ][O2 ]
( x)( x)
To avoid the need to solve the quadratic equation, take the square root of both sides
(0.015 2 x)
1.7 x103 0.041
x
0.015 - 2x = 0.041x
2.041x = 0.015
x = 0.0073
Therefore, [N2] = [O2] = x = 0.0073 M , [NO] = 0.015 - 2x = 0.0040
At Equilibrium: K c
T31.
2 SO3 F 2 SO2 + O2
Kc = 3.58x10-3
Init. Concs.: [SO3]o = 0.15 mol/10 L = 0.015 M
[SO2]o = 0.015 mol/10 L = 0.0015 M
[O2]o = 0.0075 mol/10 L = 0.00075 M
[ SO2 ]2 [O2 ] (0.0015) 2 (0.00075)
1.13x107 3.58 x103 ( K c )
[ SO3 ]
(0.015)
No, the reaction is not at equilibrium because Q Kc
(a) Q
(b) The reaction will proceed towards the right (products) because Q < Kc
T33.
N2 + O2 F 2 NO
Kc = 1.7x10-3
Init. Concs.: [N2]o = 0.25 mol/10 L = 0.025 M
[O2]o = 0.25 mol/10 L = 0.025 M
[NO]o = 0.015 mol/10 L = 0.0015 M
[ NO]2
(0.0015) 2
(a) Q
3.6 x103 1.7 x103 ( K c )
[ N 2 ][O2 ] (0.025)(0.025)
No, the reaction is not at equilibrium because Q Kc
(b) The reaction will proceed towards the left (reactants) because Q > Kc
(c) Lets make an ICE Table
Initial
Change
Equilibrium
N2
0.025
+x
0.025 + x
At Equilibrium: K c
O2
0.025
+x
0.025 + x
NO
0.0015
-2x
0.0015 - 2x
[ NO]2
(0.0015 2 x) 2
1.7 x103
[ N 2 ][O2 (0.025 ] x)(0.025 x)
To avoid the need to solve the quadratic equation, take the square root of both sides
(0.0015 2 x)
1.7 x103 0.041
(0.025 x)
0.0015 2x = 0.041(0.025 + x)
0.0015 2x = 0.001025 + 0.041x
2.041x = 0.0015 - 0.001025 = 4.75x10-4
x = 2.33 10-4 M
[N2] = [O2] = 0.025 + x = 0.025 + 2.33x10-4 = 0.0252 0.025 M (very little change in concs)
[NO] = 0.0015 - 2x = 0.0015 - 2(2.33x10-4) = 0.00103 0.001 M
T37.
H2(g)+ Br2(g) F 2 HBr(g)
Hrxn= -103.7 kJ
Notes in preparing table:
1. Only a change in temperature will change the value of Kc or KP.
2. Because Hrxn < 0 (exothermic), an increase in temperature will move the
reaction to the left (more reactants), and decrease Kc or KP
Change
[Br2]
[HBr]
Kc
Kp
Some H2 (a reactant) is
added to the container
The temperature of the
gases in the container is
increased. (increase in
energy, energy released in
the reaction)
The pressure of HBr
(a product) is increased.
decrease
increase
no change
no change
increase
decrease
decrease
decrease
increase
increase
(since extra is
added)
no change
no change
T39.
2 NOBr(g) F 2 NO(g) + Br2(l)
H = +16.1 kJ
(a) Adding more Br: No effect - Adding a pure liquid or solid has no effect on equilibrium
(b) Removing some NOBr(g): The equilibrium will shift to Left.
(c) Lowering the Temperature: The equilibrium will shift to the Left (and Kc will decrease)
T70.
(a) N2(g) + O2(g)
2 NO(g) H = + 180.0 kJ
(i) Increasing the temperature adds more energy. The reaction shifts toward the products
(right), to absorb the energy.
(ii) Decreasing the pressure, affects the products and the reactants equally (2 mol gas on
each side). That results in no shift at all.
(iii) Increasing the amount of O2(g) increases the concentration of a reactant. The reaction
shifts toward the products (right), to decrease the [O2].
(b) CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
H = 802.3 kJ
(i) Increasing the temperature adds more energy. The reaction shifts toward the reactants (left),
to absorb the energy.
(ii) Decreasing the pressure affects the products and the reactants equally (3 mol gas on
each side). That results in no shift at all.
(iii) Increasing the amount of CO2(g) increases the concentration of a product.
The reaction shifts toward the reactants (left), to decrease the [CO2].
(c) CaCO3(s)
2 CaO(s) + CO2(g)
H = + 177.9 kJ
(i) Increasing the temperature adds more energy. The reaction shifts toward the products
(right), to absorb the energy.
(ii) Decreasing the pressure affects the products (1 mol of gas) more than the reactants (0 mol
of gas). That results in a shift of the reaction toward the products (right), to increase the
pressure.
(iii) Increasing the amount of CO2(g) increases the concentration of a product.
The reaction shifts toward the reactants (left), to decrease the [CO2].
Supplementary Homework
S1.
Ans: d. the same; less
S2.
Ans: c. shift to the left; higher pressure favors fewer moles of gas
S3.
Ans: d. low; high
S4.
Ans: e. be unchanged; shift to the right
S5.
Ans: c. shift to the right; decreased temperature favors an exothermic reaction
S6.
(a)
(b)
(c)
(d)
(e)
Increase
Remain the same
Decrease
Increase
Increase. In addition, Kc will increase
S7.
In the second reaction, the reactants and products are interchanged, and the stoichiometric \
coefficients are doubled. Therefore, one has
Kc2 = (1/Kc1)2 = (1/1.23)2 = 0.66
S8.
In the second reaction, the reactants and products are interchanged and the stoichiometric
coefficients are halved. Therefore, one has
1/2
1
Kc2
K c1
1/2
1
39.
0.16
S9.
N2(g) + 3 H2(g) 2 NH3(g)
Kc = 0.060
[NH3] = 0.24 M , [H2] = 1.05 M , [N2] = ??
[ NH 3 ]2
[ NH 3 ]2
(0.24) 2
Kc
Solving for [N2] gives: [ N 2 ]
0.83 M
[ N 2 ][ H 2 ]3
K c [ H 2 ]3 (0.060)(1.05)3
S10.
Br2(g) + 2 NO(g) 2 NOBr(g)
Lets make an ICE Table
Initial
Change
Equilibrium
Br2
0
+x
x
NO
0
+2x
2x
[NOBr]o = 0.200 M
[NOBr] = 0.176 M at equilibrium
NOBr
0.200
-2x
0.200 - 2x
We can determine the value of x by using the equilibrium concentration, [NOBr].
[NOBr] = 0.176 = 0.200 - 2x 2x = 0.200 - 0.176 = 0.024 x = 0.012
Therefore, at equilibrium, [NOBr] = 0.176 M, [Br2] = x = 0.012 M , [NO] = 2x = 0.024 M
[ NOBr ]2
(0.176) 2
Kc
4.48 x103 4500.
2
2
[ Br2 ][ NO]
(0.012)(0.024)