Download EQUILIBRIUM - SCH4U1-CCVI

EQUILIBRIUM
1. The graph at the right shows the
concentrations of reactants and
products in a mixture of two
gases A2 and B2 that are reacting
to form another gas, AB.
A2 + B2
 2 AB
AB
Concentration
(mol/L)
B2
A2
0
10
Time (min)
20
(a) What happens to the concentrations of A2 and B2 as the reaction proceeds?
(b) Since rate depends on concentration, what happens to the rate of consumption of A2 and B2 as the
reaction proceeds?
(c) In principle this simple reaction ought to be reversible; that is, two molecules of AB can collide to
form A2 and B2. What is the initial concentration of AB and what will be the rate of the reverse
reaction at the beginning?
(d) What happens to the concentration of AB and therefore the rate of the reverse reaction as time
passes?
(e) The overall rate of reaction is the net result of the forward rate and the reverse rate. According to
the slope of the concentration/time graph what eventually happens to the overall rate of reaction.
(f) Does this mean that the reactions have stopped? Describe an experiment that could be done to
verify your answer.
2.
After 20 minutes this closed system has come to chemical equilibrium. Also, conditions such as
temperature or pressure, which might affect the rates of the reactions, are constant. Under these
circumstances the concentrations (and properties related to concentration such as density and colour)
will be constant. Predict what would happen to the concentrations over the next 20 minutes if an
additional 2 mol/L of B2 is suddenly injected into the reacting mixture. Explain your predictions in
terms of collision theory.
3.
This same reaction could be carried out in a process where A2 and B2 are added and AB is removed
continually at the same rate. This would also result in constant concentrations of reactants and products.
(a) Why should this be called a steady state rather than a dynamic equilibrium?
(b) What would be the advantages of such a process for commercial production of a substance in
industry?
4.
To ensure that a system is at equilibrium one must test its reversibility. If the system can be shown to
change concentrations when some factor such as pressure or temperature is altered, and then return to the
original concentrations when the original conditions are re-established, then it is demonstrated that the
reaction is readily reversible. Otherwise, apparently constant concentrations in a system could be due to
a high activation energy, and a correspondingly slow reaction rate.
(a) In a reaction similar to that above, two other substances may combine as shown in the equation
X2 (g)
+
Y2 (g)

2 XY(g)
When 6 moles of X2 , 4 moles of XY and 2 moles of Y2 were compressed into a one litre steel
cylinder at 35°C, the concentrations were found to remain constant, and no significant amount of
product was detected. What test would you do to determine whether this system is at equilibrium?
(b) When methane gas is mixed with air in a steel cylinder at room temperature the concentration does
not change over a long period of time. What could be done to determine whether the mixture is at
equilibrium?
THE EQUILIBRIUM CONSTANT
1. What happens to the equilibrium constants of the reactions below as the temperature increases? Heat
of reaction may be calculated from the ∆Hf° values given in appendix in the textbook.
 N2
(a) 2 NH3
(b) 2 CO(g)
+
+ 3 H2

O2 (g)
2 CO2 (g)
2. What happens to the equilibrium constants of the following reactions as the temperature decreases?
(a) N2H4
+
(b) 16 H2S
H2
+

2 NH3

8 SO2
16 H2O
+ 3 S8
3. The value of Keq depends on how the chemical equation is written. Reversing the equation gives the
inverse of the equilibrium constant. Changing the molar coefficients in the equation alters the powers
in the equilibrium law expression. If the equilibrium constant, Keq = 40, for the reaction
A2 (g)
+
B2 (g)

2 AB(g)
(a) What is the equilibrium constant for the reaction
2 AB(g)

A2 (g)
+
B2 (g)
(b) What is the equilibrium constant for the reaction
1
2 A2 (g)
+
1
2 B 2 (g)

AB(g)
4. The yield of products at equilibrium is related to the magnitude of Keq. A large Keq corresponds to
a relatively large amount of products at equilibrium. For which reaction below is a greater percentage
of reactant converted to product; that is, which has the greater yield at equilibrium?
+1
-1
(a) HC2H3O2 (aq)  H(aq) + C 2H3O (aq)
Keq = 1.8 x 10-5
3+
(b) Fe(aq)
-1
+ SCN (aq)

2+
FeSCN(aq)
Keq = 300
5. Only substances appearing in the equilibrium law expression will be those in solution. Solids are
omitted. Write equilibrium law expressions for the following reactions:
(a) CaCO3 (s) 
(b) NH3 (g)
CaO(s)
+
+
CO2 (g)

HCl(g)
NH4Cl(s)
6. The equilibrium law as explained in this course applies only to dilute solutions. Since the
concentration of the solvent in a dilute solution is large and nearly constant, the solvent is not usually
included in the equilibrium law expression. Write equilibrium law expressions for the following
reactions:
(a) H2SO4 (aq)
(b) NH3 (aq)
+
+
H2O(l)
H2O(l)


-1
HSO4(aq)
+1
NH4(aq)
+
+
+1
H 3O(aq)
-1
OH (aq)
7. When 0.40 moles of N2 is placed in a 5.0 L container with 0.20 moles of H2 it reaches an equilibrium
in which there is 0.10 mol H2. The reaction equation is
N2 (g)
+
3 H2 (g)

2 NH3 (g)
(a) Calculate the number of moles of H2 consumed in reaching equilibrium.
(b) Calculate the number of moles of N2 consumed in reaching equilibrium.
(c) Calculate the number of moles of NH3 produced in reaching equilibrium.
(d) Calculate the equilibrium concentrations of the reactants and products.
(e) Write the equilibrium law expression for this reaction.
(f) Calculate the equilibrium constant for this reaction.
Keq and ICE Problems Worksheet
1. Calculate the equilibrium constant, Keq, for the following reaction at 25 °C, if [NO]eq = 0.106 M,
[O2]eq = 0.122 M and [NO2]eq = 0.129 M.
(Answer: Keq = 12.1)
2 NO (g)
+
O2 (g)

2 NO2 (g)
2. Given the balanced equation and the value for Keq from #1, calculate new value of Keq for the
following:
(Answer: a. Keq = 1.87, b. Keq = 0.0826, Keq = 0.287)
a. 1/2 NO (g)
+
1/4 O2 (g)
b. 2 NO2 (g)

2 NO (g)
c. NO2 (g)

NO (g)
+

+
1/2 NO2 (g)
O2 (g)
1/2 O2 (g)
3. Find the equilibrium constant, Keq, for the following equilibrium. The initial concentrations of AB and
A2D are 0.30 M before they are mixed and when equilibrium is reached, the equilibrium concentration of
A2D is 0.20 M. Use an ICE table for your calculation. (Answer: Keq = 0.80)
2 AB (g)
+
C2D (s)

A2D (g)
+
2 CB (s)
4. If 0.50 mol of NO2 is placed in a 2.0L flask to create NO and O2, calculate [ ]eq if Keq = 1.2 x 10-5.
(Answer: [O2]eq = 0.0057 M, [NO]eq = 0.0114 M, [NO2] eq = 0.24 M)
5. For the system, if we start with 0.100mol/L of CO2 and H2, what are the concentrations of the reactants
and products at equilibrium given that Keq = 0.64 at 900K?
(Answer: [CO]eq = [H2O]eq = 0.044 M, [CO2]eq = [H2]eq = 0.056 M)
CO2 (g)
+ H2 (g)

CO (g)
+
H2O (g)
6. For the system, if we start with 0.010 mol/L of H2 and I2 and 0.096 mol/L of HI, what are their
concentrations at equilibrium given that Keq = 0.016?
(Answer: [HI]eq = 0.093 M, [H2]eq = [I2]eq = 0.012 M)
2HI(g)

H2 (g)
+
I2 (g)
7. At 650°C, the reaction below has a Keq value of 0.771. If 2.00 mol of both hydrogen and carbon dioxide
are placed in a 4.00 L container and allowed to react, what will be the equilibrium concentrations of all
four gases?
(Answer: [CO]eq = [H2O]eq = 0.234 M, [H2]eq = [CO2]eq = 0.266 M)
H2 (g)
+
CO2 (g)

CO (g)
+
H2O (g)
8. Carbonyl bromide, COBr2, can be formed by reacting CO with Br2. A mixture of 0.400 mol CO, 0.300
mol Br2, and 0.0200 mol COBr2 is sealed in a 5.00L flask. Calculate equilibrium concentrations for all
gases, given that the Keq = 0.680.
(Answer: [CO]eq = 0.0807 M, [Br2]eq = 0.0607 M, [COBr2]eq = 0.0033 M)
CO (g) +
Br2 (g)

COBr2 (g)
DETERMINATION OF Keq
OBJECTIVE:
Determine the equilibrium constant for the reaction
3+
-1
2+
Fe(aq) + SCN (aq)
 FeSCN (aq)
INTRODUCTION:
Iron(III) ion reacts with thiocyanate ion to produce a coloured complex ion.
3+
Fe(aq)
+
-1
SCN (aq)

2+
FeSCN (aq)
The intensity of the colour of the solution depends on two things, the concentration of the coloured species and the
path length through which light travels in the solution. For example, light traveling through a 3 cm depth of a 0.1
mol/L solution of this ion will appear to have the same colour intensity as light traveling through a 1 cm depth of a
0.3 mol/L solution. Therefore, if the depth of two solutions is adjusted until they appear to have the same colour
intensity, the ratio of their concentrations will be inversely proportional to the ratio of their depths:
C1 D2
C2 = D1
If the concentration of one solution is known then concentration of the other may be calculated. The equilibrium
concentrations may be used to calculate the value of the equilibrium constant for this reaction. Then predictions may
be made about the equilibrium concentrations in other solutions.
To prepare a solution with a known concentration of the coloured complex, a dilute solution of thiocyanate ion is
reacted with a concentrated solution of iron (III) ion. It may be assumed that the large excess of iron (III) ion causes
most of the thiocyanate ion to be converted to the coloured complex. Therefore, for this mixture the equilibrium
concentration of the coloured complex will be equal to the initial concentration of the thiocyanate ion.
PROCEDURE:
(a) Line up five clean test tubes all of the same diameter, and label them. Add 5.0 mL of 0.002 mol/L potassium
thiocyanate solution to each of these five test tubes. To test tube (1) add 5.0 mL of 0.2 mol/L iron (III) nitrate
solution. This tube will be used as the standard.
(b) Measure 10.0 mL of 0.2 mol/L iron (III) nitrate solution in a graduated cylinder, and fill to the 25.0 mL mark
with distilled water. Pour the solution into a clean dry beaker to mix it. Measure 5.0 mL of this solution and pour
it into test tube 2. Save the remainder of this solution for Part (c).
(c) Pour 10.0 mL of the diluted iron (III) nitrate solution from the beaker into your graduate. Discard the remainder.
Again fill the graduate to the 25.0 mL mark with distilled water, and pour the solution into a clean dry beaker to
mix. Pour 5.0 mL of this solution into test tube 3. Continue dilution in this manner until you have 5.0 mL of
successively more dilute solution in each test tube.
(d) Now compare the solutions in each of the test tubes with the standard tube (1) in order to determine the
concentration of the coloured complex ion. Wrap a strip of paper around test tubes (1) and (2) to exclude light
from the side. Look vertically down through the solutions toward a diffused light source. If the colour
intensities appear the same, measure the depth of each solution to the nearest millimetre and record this. If the
colour intensities do not appear the same, remove some of the solution form the standard tube with a dropper
until the colour intensities are the same. Put the portion you removed into a clean dry beaker, since you may
have to use some of this solution later. The matching may be accomplished by removing more standard than
seems necessary and then replacing part of it drop by drop. When the colour intensities are the same in each
test tube, measure the depth of both solutions to the nearest millimetre. Repeat the procedure comparing test
tubes (1) and (3), (1) and (4), (1) and (5).
CALCULATIONS
1. Remember that the solution in test tube (1) was taken as the standard. The equilibrium concentration of coloured
complex in this solution is very nearly equal to the concentration of thiocyanate in the mixture before the reaction
began. However, a calculation is necessary since the 0.002 mol/L solution of thiocyanate used was diluted by the
addition of the other solution even before the reaction occurred. The initial concentration of thiocyanate in the
mixture is found by multiplying its concentration before mixing (0.002) by the dilution factor, (initial
volume/final volume), which in this case is (5/10). Therefore for the standard solution (1), the equilibrium
concentration of the coloured complex was
[FeSCN2+]eq = 0.002 x (5/10) mol/L.
2. The initial concentration of thiocyanate ion in each test tube was
[SCN-1]in = 0.002 x (5/10) mol/L.
3. The initial concentration of the iron (III) ion was different in each test tube since this solution was diluted
repeatedly by a factor of (10/25) in preparing the other solutions. In addition, the iron (III) ion was further diluted
by mixing with the thiocyanate solution.
[Fe3+]in = 0.2 x (10/25) n-1 x (5/10) mol/L
where n is the test tube number.
4. The equilibrium concentration of the coloured complex in each test tube is equal to the assumed concentration in
(1) multiplied by the ratio of depths which you measured in your experiment.
5. The amount of thiocyanate ion consumed is equal to the amount of coloured ion produced. Therefore to get the
equilibrium concentration of thiocyanate ion, subtract the equilibrium concentration of the coloured product from
the initial concentration of the thiocyanate ion.
6. Similarly, subtract the equilibrium concentration of the coloured product from the initial concentration of ferric
ion to get the equilibrium concentration of ferric ion.
7. Write the equilibrium expression for this reaction.
8. Calculate the values for the equilibrium constant from the data for the second through fifth test tubes.
9. Summarize your calculations in the data table below.
1
2
3
4
5
Initial [SCN-1]
Initial [Fe3+]
Depth of Standard ( D1)
5.20
4.60
3.90
2.70
1.50
Depth of Sample (D2)
5.20
5.20
5.20
5.20
5.20
Depth Ratio (D1/D2)
Equilibrium [FeSCN2+]
Equilibrium [SCN-1]
Equilibrium [Fe3+]
Keq
CONCLUSION:
What is the average Ke value for the mixtures in test tubes 2 to 5?
LE CHATELIER'S PRINCIPLE
1. The graph at the right shows the
affect of adding more reactant, B2,
(at time = 20 minutes) to a system
initially at equilibrium with
concentrations of 6 mol/L, 2 mol/L
and 4 mol/L respectively for A2, B2
and AB.
6
[A 2 ]
[AB]
4
Concentration
(mol/L)
[B 2 ]
2
0
15
20
The reaction is
A2
+ B2 
2 AB
Use collision theory to explain the change in concentrations.
25
Time
(min)
30
35
2. If pressure on the system is increased, the equilibrium will shift to reduce the number of moles of gas in
the mixture, thus lowering the pressure again. What would be the affect on the yield of products of
increasing the pressure in each of the following systems at equilibrium?
(a) 3 H2 (g)
+
(b) 2 NaCl(s)
+
(c) SO3 (g)
+

N2 (g)
H2SO4 (l)
CaCO3 (s)
2 NH 3 (g)


2 HCl(g)
+
Na2SO4 (s)
CaSO4 (s)
+
CO2 (g)
3. If the temperature is raised, the equilibrium will shift in the endothermic direction to consume some of
the added heat. What would be the affect of raising the temperature on the concentrations of products in
each of the following systems at equilibrium? Explain using Collision Theory.
(a) HCl
(b) 1/2 N2
+
NH3
+

3/2 H2
NH4Cl

NH3
+
heat
∆H = -46.2 kJ/mol NH3
(c) C2H4 (g)
+
heat

C2H2 (g)
+
H2 (g)
4. Of course, lowering the concentration, pressure or temperature will have the opposite affect to the
changes described above. Each question below refers to the reaction:
N2O4 (g) + heat  2 NO2 (g)
(a) What is the affect on the yield of products of decreasing the pressure?
(b) What would be the affect of decreasing the temperature?
(c) What would be the affect of removing the NO2 as it is formed?
5. H2(g)
Cl2(g) 
+
2 HCl(g)
What direction will the equilibrium shift when the partial pressure of hydrogen is increased?
6. 3 H2(g)
+
N2(g)

2 NH3(g)
Given that this reaction is exothermic, what direction will the equilibrium shift when the temperature of
the reaction is decreased?
7. 2 NO2(g)

N2O4(g)
If a large quantity of argon is added to the container in what direction will the equilibrium shift?
8. NH4OH(aq)

NH3(g)
+
H2O(l)
In what direction will the equilibrium shift if ammonia is removed from the container as soon as it is
produced?
9. 2 BH3(g)

B2H6(g)
If this equilibrium is taking place in a piston with a volume of 1 L and I compress it so the final volume is
0.5 L, in what direction will the equilibrium shift?
EQUILIBRIUM
Qualitative Observations of Le Chatelier’s Principle
PART 1 Equilibrium involving thymol blue
1.
Into one 250 mL Erlenmeyer flasks add approximately 50 mL of distilled water and 3 mL of thymol
blue solution to the flask. Add drops of either HCl or NaOH until the solution is blue.
2.
To the flask add a drops of 0.1 M HCl while swirling until a definite colour change exists.
3.
Continue the slow addition of 0.1 M HCl to the flask until a second colour shift occurs
4.
Now add 0.1 M NaOH slowly to the flask woth swirling until a definite colour change is observed.
5.
Continue the slow addition of 0.1 M NaOH until the colour changes again.
6.
Bring the solution to yellow and dispose down the sink with plenty of water.
Tb2- + H3O+ 
HTb 1- +
Blue
yellow
1+
HTb
+ H3O
 H2Tb
+
yellow
red
stress
[ ] change
[ ] 
H2O
H2O
shift direction
[ ] 
visible change
Starting
blue
HCl 1st
HCl 2nd
NaOH 1st
NaOH 2nd
PART B Equilibrium Involving Thiocyanatoiron (III) Ion
1.
Using the amber equilibrium solution, pour approximately 2-4 mL into each of 5 test tubes.
2.
The 1st test tube will serve as a control.
3.
To the 3rd test tube, add 1 mL of 0.2 M Fe(NO3)3
4.
To the 4th test tube, add 1 mL of 0.1 M KSCN
5.
To the 5th test tube E, add 1 mL of 3.0 M NaOH
6.
To the 2nd test tube add 1 mL of 0.1 M HCl
7.
Dispose of all solutions in the Anion waste bucket in the fumehood.
PART 2 Equilibrium involving Thiocyanatoiron (III) Ion
Fe3+(aq)
yellow
stress
+ SCN-(aq) 
colourless
[ ] change
FeSCN2+(aq)
dark red
[ ] 
shift direction
[ ] 
visible change
Fe(NO3)3
KSCN
NaOH
HCl
PART C Equilibrium Involving Copper (II) Complexes
1.
Place 2-5 mL of 0.1 M CuSO4 in a test tube
2.
Add 3 drops of 1M NH3 and observe the result
3.
Continue adding 1 M NH3 until another change occurs
4.
Add 3 M HCl drop by drop until a change occurs.
5.
Dispose of the solutions in a beaker for Ammonia/ammonium waste in the fumehood.
Cu(H2O)42+ (aq) + 4 NH3 (aq)  Cu(NH3)42+ (aq)
blue
colourless
dark blue
Note: another equilibrium exists:
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
stress
NH3 1st
NH3 2nd
HCl
[ ] change
[ ] 
shift direction
[ ] 
+ 4 H2O
visible change
Equilibrium Constant, Keq
1. Once a system has reached equilibrium, are the following true or false?
a. The reaction is finished, no more products are forming. __________
b. The concentrations of the reactants and the products are equal. ______________
c. The concentrations are no longer changing. ___________
d. The reaction is not over, but will continue forever if isolated. _____________
e. The speed at which products are made equals the speed at which reactants form. __________
2. What is equal at equilibrium? ______________________
3. What general information can be gathered by observing the magnitude of the equilibrium constant?
4. Write the expression for Keq for the reaction: 2 NO (g) + Cl2 (g)  2 NOCl (g)
5. Write the Keq for: 2 K3PO4 (aq) + 3 Ca(NO3)2 (aq)  6 KNO3 (aq) + Ca3(PO4)2 (s)
6. Write the expression for Keq for the reaction :
H2 (g) + Br2 (l)  2 HBr (g)
7. Write the expression for Keq for the reaction:
CO2 (g) + CaO (s)  CaCO3 (s)
8. For the reaction: SiH4 (g) + 2 O2 (g)  SiO2 (g) + 2 H2O (l)
a) Write the equilibrium expression for the forward reaction:
b) Write the equilibrium expression for the reverse reaction:
c) What is the equilibrium constant in the forward direction if [SiH4] = 0.45M; [O2] = 0.25M; and
[SiO2] = 0.15M at equilibrium?
d) What is the equilibrium constant in the reverse reaction?
e) If [SiH4] = 0.34M; [O2] = 0.22M; [SiO2] = 0.35M; and [H2O] = 0.20M, what would be the reaction
quotient (Q) in the forward direction and which direction will the reaction go?
Calculating Keq, Q, ICE tables
9. Calculate the equilibrium constant for this reaction: 2 PO2Br (aq)  2 PO2 (aq) + Br2 (aq)
[PO2Br] = 0.0255M, [PO2] = 0.155M, and [Br2] = 0.00351M at equilibrium.
Given:
10. For the combination reaction: H2 (g) + F2 (g)  2 HF (g), calculate all three equilibrium concentrations
when [H2]i= [F2]i = 0.200 M and Keq = 64.0.
11. For the decomposition reaction, COCl2 (g)  CO (g) + Cl2 (g), calculate all three equilibrium
concentrations when Keq = 0.680 with [CO]i = 0.500 and [Cl2]i = 1.00 M.
12. We place 0.064 mol N2O4 (g) in a 4.00 L flask at 200 K. After reaching equilibrium, the concentration of
NO2(g) is 0.0030 M. What is Keq for the reaction N2O4(g)  2 NO2(g)?
13. Carbonyl bromide decomposes to carbon monoxide and bromine: COBr2(g)  CO(g) + Br2(g) Keq is
1.90 x 10-4 at 73oC. If an initial concentration of 0.300 M COBr2 is allowed to equilibrate, what are the
equilibrium concentrations of COBr2, CO, and Br2?
14. Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas. What is the initial
concentration of phosphorus pentachloride if at equilibrium the concentration of chlorine gas is 0.500M?
Given: Keq = 10.00 (Hint: ICE table)
Le Chatelier’s Principle
15. Consider this endothermic reaction: 3 O2(g)  2 O3(g). To shift this reaction to the reactants:
a) You could ________ the pressure. ( increase or decrease)
b) You could ________ the volume. ( increase or decrease)
c) You could ________ oxygen gas. ( add or remove)
d) You could ________ the temperature. ( increase or decrease)
e) Which of the following, if increasing, will change the value of the equilibrium constant?
a) Pressure
b) Volume
c) [Product]
d) Temperature
e) [Reactant]
16. Consider this reaction: 2 SO2(g) + O2(g)  2 SO3(g). To shift this reaction towards the products:
a) You could _________ the pressure. (increase or decrease)
b) You could _________ the volume. (increase or decrease)
c) You could _________ oxygen gas. (add or remove)
17. For each system described below, indicate in which direction the equilibrium will shift when each stress
is added or removed.
a) N2 (g) + 3 H2 (g)  2 NH3 (g): more H2 is added to reaction at equilibrium. ( or  or same)
b) For the same reaction, some NH3 is removed from the reaction at equilibrium. ( or  or same)
c) 2 SO2 (g) + O2 (g)  2 SO3 (g) + heat: heat is added. ( or  or same)
d) Using the same reaction, heat is removed ( or  or same)
e) PCl3 (g) + Cl2 (g)  PCl5 (g): volume is reduced by half. ( or  or same)
f) Using the same system as above, a catalyst is added to the system. ( or  or same)
g) H2 (g) + Cl2 (g)  2 HCl (g): volume is doubled. ( or  or same)
h) Using the same system as above, some neon is added to the system. ( or  or same)
18. Explain how the following changes in reaction conditions will affect the position of the equilibrium:
A (g) + B (aq)

C (s)
ΔHrxn= -453 kJ/mol
a) The pressure of A in the reaction chamber is increased. ( or  or same)
b) The temperature of the reaction is increased by 200 C. ( or  or same)
c) A catalyst is added to the system. ( or  or same)
d) More of compound B is steadily added to the reaction chamber. ( or  or same)
e) More of compound C is steadily added to the reaction chamber. ( or  or same)
f) Argon gas is added to the reaction chamber, doubling the pressure. ( or  or same)
Ksp PROBLEM SET #1
1. Write the solubility expression for the following:
(a) Zn(OH)2
(b) Ca3(PO4)2
(c) Fe2(SO4)3
2. Calculate the solubility of Mg(OH)2 in g/L given that Ksp = 5.6 x 10-12. (6.5 x 10-3
g/L).
3. The solubility of SrF2 is 7.3 x 10-3 g/100 mL. Calculate Ksp. (7.8 x 10-10)
4. The solubility of Cu(IO3)2 is 3.245 x 10-3 mol/L. Calculate Ksp. (1.367 x 10-7)
5. The solubility of PbI2 is 0.058 g/100mL. Calculate the Ksp for PbI2. (8.0 x 10-9)
6. Calculate the concentration in mol/L of Barium ion in a saturated soln of Ba(IO3)2, if the
Ksp = 1.15 x 10-9. (6.6 x 10-4 mol/L)
7. The solubility of Cd(OH)2 is 2.01 x 10-4 g/100mL. Calculate Ksp. (1.04 x 10-14)
8. The Ksp for Silver Phosphate is 2.8 x 10-18. Calculate the solubility in g/100 mL.
(7.49 x 10-4 g/100 mL)
Ksp PROBLEM SET #2
1. If 0.010 mg of NaCl is added to 200. mL of a 2.0 x 10-5 M AgNO3 will a precipitate form? Ksp AgCl =
1.8 x 10-10. (Q = 1.71 x 10-11)
2. If one gram of AgNO3 is added to 50. mL of a 0.050 M HC2H3O2 which is soluble, will a precipitate
form? Ksp for AgC2H3O2 is 2.0 x 10-3. (Q = 5.9 x 10-3)
3. In each of the following cases, show whether a precipitate will form under the given conditions:
(a) 10. mL of 0.10 M AgNO3 is added to 600. mL of a 0.010 M Na2SO4 solution. Ksp for Ag2SO4 is
1.5 x 10-5. (Q = 2.7 x 10-8)
(b) 1.0 g of Pb(NO3)2 is put in 100. mL of 0.010 M HCl. Ksp (PbCl2) = 1.2 x 10-5. (Q = 3.0 x 10-6)
(c) 1.0 milligram of CaCl2 and 1.0 milligram of Na2C2O4 are added to 1.0 litre of water. Ksp for
CaC2O4 is 2.0 x 10-9. (Q = 6.7 x 10-11)
4. Given that the Ksp for CaC2O4 is 2.0 x 10-9, calculate how many grams of CaC2O4 will dissolve in 1.0
litre of:
(a) water (m = 5.7 x 10-3g)
(b) 0.10 M Na2C2O4 (m = 2.6 x 10-6g)
(c) 0.010 M CaCl2 (m = 2.6 x 10-5g)
(d) 0.10 M NaNO3 (m = 5.7 x 10-3g)
5. Given the Ksp for PbI2 = 8.5 x 10-9, calculate how many grams of PbI2 will dissolve in 250. mL of the
following systems:
(a) water (m = 1.5 x 10-1g)
(b) 0.010 M Pb(NO3)2 (m = 5.3 x 10-2g)
(c) 0.010 M CaI2 (m = 2.5 x 10-3g)
6. 75 mL of a 2.0 x 10-3 M AgNO3 solution is mixed with 45 mL of a 1.0 x 10-4 M NaCl soln. Will a ppte
form? Ksp AgCl = 1.8 x 10-10. (Q = 4.7 x 10-8)
7. How many grams of AgCl will dissolve in 100.0 mL of a 0.010 M NaCl soln. Ksp = 1.8 x 10-10. (m =
2.6 x 10-7g)
8. 100. mL of a 0.010 M Mg(NO3)2 is mixed with 50. mL of a 0.010 M Ba(OH)2. Will a ppte form? Ksp
Mg(OH)2 = 5.6 x 10-12. (Q = 3.0 x 10-7)
9. Calculate the number of moles of AgCl that will dissolve in 1.0 litre of a 0.10 M CaCl2 solution. (n =
9.0 x 10-10 mol)
10. 50. mL of a 5.0 x 10-4 M Ca(NO3)2 is mixed with 50. mL of a 2.0 x 10-4 M NaF to give 100. mL of
solution. Will a ppte form? Ksp CaF2 is 3.9 x 10-11. (Q = 2.5 x 10-12)
11. Calculate the solubility of RaSO4 in mol/L in a 0.10 M Na2SO4 sol’n. Ksp RaSO4 = 4.3 x 10-11. (x =
4.3 x 10-10 M)
12. A solution contains 0.010 moles of Cl- and 0.0010 moles of CrO4-2 per litre. Ag+ is added.
Which will ppte first AgCl or Ag2CrO4. Ksp AgCl = 1.8 x 10-10, Ksp Ag2CrO4 = 1.2 x 10-12. ([Ag+]
with Cl- is smaller than CrO42-, therefore AgCl will form a precipitate first)
EQUILIBRIUM REVIEW
1. Two colourless solutions are mixed in a stoppered flask. As the reaction proceeds, the resulting solution
turns red, and a colourless gas is formed. After a few minutes, no more gas is evolved but the red colour
remains. What evidence is there that equilibrium has been established?
2. What evidence is there to indicate that equilibrium is a dynamic state?
3. Using the Ideal gas law and the formula relating Concentration and moles, derive the relationship that
explains why pressure can be considered as a concentration unit for gases.
4. Write the equilibrium expressions (Ke or K) for each of the following reactions:
(a)
H2 (g) + F2 (g)  2 HF (g) ;
(b)
4 NO (g) + 3 O2 (g)  2 N2O5 (g) ;
(c)
BaCO3 (s)  BaO (s) + CO2 (g) ;
(d)
Fe (s) + Cu2+ (aq)
(e)
HSO4- (aq)  H+ (aq) + SO42- (aq) ;
(f)
2 CrO42- (aq) + H+ (aq)  Cr2O72- (aq) + OH- (aq) ;
(g)
Ag2CO3 (s) 
 Fe2+ (aq) + Cu (s) ;
2 Ag+ (aq) + CO32- (aq) ;
5. The equilibrium constants for three different reactions are:
(a) Keq = 1.5 x 1012 ;
In which reaction is there:
(b) Keq = 0.15;
(c) Keq = 4.3 x 10-15
(a) a large ratio of product to reactant?
(b) a small ratio of reactant to product?
6. When 0.035 mole of PCl5 is heated to 250°C in a 1-litre vessel, an equilibrium is established in which
the concentration of Cl2 is 0.025 mol/L. Find the Keq.
PCl5 (g)  PCl3 (g) + Cl2 (g)
7. Assume that the analysis of another equilibrium mixture of the system in Problem 6 shows that the
equilibrium concentration of PCl5 is 0.012 mol/L and that of Cl2 is 0.049 mol/L. What will be the
equilibrium concentration of PCl3? The reaction is carried out at 250°C. Use Keq from question 6.
8. How many moles of PCl5 must be heated in a 1-litre flask at 250°C in order to produce enough chlorine
to give an equilibrium concentration of 0.10 mol/L? Use Keq from question 6.
9. Will there be a net reaction when 2.5 moles of PCl5 0.60 mole of Cl2, and 0.60 mole of PCl3 are placed
in a 1-litre flask and heated to 250°C? If so, will PCl5 decompose, or will Cl2 and PCl3 react to form
more PCl5? Use Keq from question 6.
10. Under a given set of conditions, an equilibrium mixture:
SO2 (g) + NO2 (g)  SO3 (g) + NO (g)
in a 1 litre container was analyzed and found to contain 0.300 mole of SO3, 0.200 mole of NO, 0.050
mole of NO2, and 0.400 mole of SO2. Calculate the Keq.
11. At 55°C, for the reaction
2 NO2 (g) 
N2O4 (g) :
Keq = 1.15
(a) Write the equilibrium expression.
(b) Calculate the concentration of N2O4 present in equilibrium with 0.5 mole of NO2.
12. Calculate the Keq for the following reaction from the data given below.
CO2 (g) + H2 (g)  CO (g) + H2O (g)
[CO] = [H2O] = 1.33 x 10-3 mol/L, [CO2] = [H2] = 1.17 x 10-3 mol/L
13. One mole of pure NH3 was injected into a 1-litre flask at a certain temperature. The equilibrium mixture
below was then analyzed and found to contain 0.300 mol of H2:
2 NH3
 N2
+ 3 H2
(a) Calculate the concentration of N2 at equilibrium.
(b) Calculate the concentration of NH3 at equilibrium.
(c) Calculate the equilibrium constant for this system at this temperature and pressure.
(d) Which way would the equilibrium be shifted if 0.600 mole of H2 were injected into the flask?
(e) How would the injection of hydrogen into the flask affect the equilibrium constant?
(f) How would the equilibrium constant be affected if the pressure of this system were suddenly
increased?
14. When 0.50 mole of CO2 and 0.50 mole of H2 were forced into a 1-litre reaction container, the following
equilibrium was established:
CO2 (g) + H2 (g) 
H2O (g) + CO (g) and Keq = 2.00.
(a) Find the equilibrium concentration of each reactant and product.
(b) How would the equilibrium concentrations differ if 0.50 mole of H2O and 0.50 mole of CO had
been introduced into the reaction vessel instead of the CO2 and H2?
15. The following equation represents a gaseous system at equilibrium:
2 H2O (g) + heat
 2 H2 (g) + O2 (g)
Indicate in which direction the equilibrium will shift when the following changes are made:
(a)
The concentration of H2 is increased..
(b)
The partial pressure (concentration) of H2O is increased.
(c)
The concentration of O2 is decreased.
(d)
The temperature is increased.
(e)
The total pressure is increased. To the left,
To the right,
16. Consider the following reaction:
N2O4 (g)
 2 NO2 (g) ; H = +ve ; Keq = 0.87 at 55 C
What will be the effect of each of the following changes on the concentration of N2O4 at equilibrium:
(a)
increasing the pressure.
(b)
increasing the temperature.
(c)
increasing the volume.
(d)
adding more NO2 (g) to the system without changing P or T.
(e)
adding a catalyst?
17. Answer the same questions (a,b,c,e) for the following reaction as you did for the reaction given in
Problem 16 for the concentration of water.
H2 (g) + 1/2 O2 (g)
 H2O (g) ; H = -ve ; Keq = 1.0 x 1040 at 25 C
18. How can you increase the concentration of the product in each of the following reactions by varying the
temperature and pressure?
(a)
4 NH3 (g) + 5 O2 (g)
(b)
Br2 (g) + Cl2 (g) 
 4 NO (g) + 6 H2O (g) ;
2 BrCl (g) ;
H = +ve
H = -ve
(c)
BaSO4 (s)  Ba2+ (aq) + SO42- (aq) ;
H = +ve
19. Write the equation for the chemical equilibrium that exists between undissolved and dissolved solute in a
saturated solution for each of the following slightly soluble sulfides.
T12S
Ksp = 1 x 10-24
CuS
Ksp = 1 x 10-35
Cu2S
Ksp = 1 x 10-48
HgS
Ksp = 1 x 10-52
Rank each of these sulfides in order of decreasing molar solubility in their saturated solutions.
20. 500 mL of a saturated solution of silver carbonate, Ag2CO3, is evaporated to dryness leaving 0.0698 g of
Ag2CO3. What is the Ksp of silver carbonate?
21. What mass of barium fluoride, BaF2, will dissolve in 500. mL of a 0.100 M NaF solution? The Ksp of
barium fluoride is 1.7 x 10-6. Fluoride ion is the common ion.
22. Will a precipitate of BaF2 form when 0.035 mol of Ba2+ and 0.015 mol of F- are combined in 1.000 L of
solution?
Here is the same question, but asked in an alternate way:
Will a precipitate of BaF2 form when the following solutions are mixed?
700. mL of 0.050 M BaC12
+
300. mL of 0.050 M NaF
Ksp(BaF2) = 1.7 x 10-6.
Extra SCH 4U Review questions
1. What is the solubility of PbCl2 in 0.10 M NaCl? (Ksp PbCl2 is 1.2 x 10-5).
2. What is the Keq for
A
+ 2B
 AB2
if [AB2] i = 0.20 M and [A]eq = 0.050 M.
3. If 1.00 g of NaCl(s) is added to 500.0 mL of a 0.010 M Pb(NO3)2 will a precipitate form? (Ksp PbCl2 =
1.2 x 10-5)
4. What is the [AB]eq in
A
+ B
 AB
if [A]i = [B]i = 0.10 M (Keq = 101).
5. What is the solubility in g/100 mL of Ag2CO3 if the Ksp for Ag2CO3 is 8.5 x 10-12?