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

Concentration (mol/L)

B2

A2

AB

Time (min)

0 10 20

(a) What happens to the concentrations of A2 and B2 as the reaction proceeds?

The concentrations both drop. (b) Since rate depends on concentration, what happens to the rate of consumption of A2 and B2 as the

reaction proceeds?

The rate decreases. (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?

The initial concentration is 0 mol/L.

The initial rate of reactin is 0 mol/L·s. (d) What happens to the concentration of AB and therefore the rate of the reverse reaction as time

passes?

[AB] increases and therefore the reverse rate increases. (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.

The overall rate of reaction falls to zero. (f) Does this mean that the reactions have stopped? Describe an experiment that could be done to

verify your answer.

No, the 0 slope does not mean that the reaction has stopped and to test this, add more

reactant or product and observe if the reaction continues (equilibrium verified) or it does

nothing (complete reaction verified).

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.

The sudden increase in B2 concentration will result in additional A2 - B2 collisions which

will produce additional AB. As a result there will be a decrease in A2 concentrations, a

sudden increase followed by a decrease in B2 concentrations and an increase in AB

concentrations. The rise in [AB] would result in increased reverse reaction slowing the

[A2] and [B2] decrease until rate forward = rate reverse and the new equilibrium is

reached.

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?

Since the forward rate does not match the reverse rate, the constancy of concentrations is a

result of removal of products and the addition of reactants rather than an equilibrium. (b) What would be the advantages of such a process for commercial production of a substance in

industry?

This process allows the continuous production of product rather than a batch mode in which

the process must be started and stopped to remove product.

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?

If it were at equilibrium under these conditions, what would you observe?

Change the temperature and measure the new reactant and product concentrations. If the

system is an equilibrium one, the concentrations will have changed.

If it was at equilibrium there would be no visible change of macroscopic properties. (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?

Change the pressure or temperature. If the system is an equilibrium one, the concentrations

will have changed. What do you expect to observe? Unless the temperature and/or the

pressure is raised a great deal, expect no change since the activation energy for this reaction is

quite high and the reaction is very exothermic. Once the rection is triggered it will be

explosive as the released enthalphy sustains the reaction. Since the reverse activation energy

is even larger than the forward one, little reverse reaction is expected. The reaction essentially

goes to completion.

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.

(a) 2 NH3 → N2 + 3 H2

∆H° = (1 mol(0) + 3 mol(0)) - (2 mol(-45.9 kJ/mol) = +91.8 kJ (endothermic)

Ea is largest in the → direction and the → reaction is most affected by a temperature increase

Keq which equals the ratio kf /kr increases.

(b) 2 CO(g) + O2 (g) → 2 CO2 (g)

∆H° = (2 mol(-393.5 kJ/mol)) – (2 mol (-110.5 kJ/mol)) + = -566.0 kJ (exothermic)

Ea is largest in the ← direction and the ← reaction is most affected by a temperature increase,

Keq which equals the ratio kf /kr decreases.

2. What happens to the equilibrium constants of the following reactions as the temperature decreases?

(a) N2H4 + H2 → 2 NH3

∆H° = (2 mol(-45.9 kJ/mol)) - (1 mol(50.6 kJ/mol + 1 mol(0)) = -142.4 kJ (exothermic)

Ea is largest in the ← direction and the ← reaction is most affected by a temperature decreaaes,

Reverse of #1 - Keq increase.

(b) 16 H2S + 8 SO2 → 16 H2O + 3 S8

∆H° = (16 mol(-285.8 kJ/mol)) - (16 mol(-20.6kJ/mol) + 8 mol(-296.8 kJ/mol)) = -1868.8 kJ (exo),

Ea is largest in the ← direction and the ← reaction is most affected by a temperature decreaaes,

Reverse of #1 - Keq increase.

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)

0.025 40

1

Keq

1 Keq' then 40 Keq If

(b) What is the equilibrium constant for the reaction

1

2 A2 (g) +

1

2 B 2 (g) AB(g)

6.32 = 40 = Keq Keq" then 40 = Keq If

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?

(a) HC2H3O2 (aq) H+1

(aq) + C 2H3O -1

(aq) Keq = 1.8 x 10-5

(b) Fe3+

(aq) + SCN -1

(aq) FeSCN2+

(aq) Keq = 300

Reaction (b) will have the greater yield

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) CaO(s) + CO2 (g)

Keq = [CO2]

(b) NH3 (g) + HCl(g) NH4Cl(s)

][HCl]3[NH

1 eqK

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) + H2O(l) HSO-1

4(aq) + H 3O+1

(aq)

]4SO2[H

]O3][H-14[HSO

eqK

(b) NH3 (aq) + H2O(l) NH+1

4(aq) + OH -1

(aq)

]3[NH

]][OH4[NH eqK

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

(a) Calculate the number of moles of H2

consumed in reaching equilibrium.

mol 0.10 mol )10.020.0( Hn2

2x 0.10 x - 0.40 E

2x 3x - x - C

0 0.20 0.40 I

(g) 3NH 2 (g) 2H 3 (g) 2N

(b) Calculate the number of moles of N2 consumed in reaching equilibrium.

mol 0.033 2H mol 3

2N mol 1 x 2H mol 0.10 Nn

2

(c) Calculate the number of moles of NH3 produced in reaching equilibrium.

mol 0.067 2H mol 3

3NH mol 2 x 2H mol 0.10 NHn

3

(d) Calculate the equilibrium concentrations of the reactants and products.

mol/L 0.013 L 5.0

mol 067.0 ]3[NH

mol/L 0.020 L 5.0

mol )10.020.0( ]2[H

mol/L 0.073 L 5.0

mol )033.040.0( ]2[N

(e) Write the equilibrium law expression for this reaction.

3]2][H2[N

2]3[NH eqK

(f) Calculate the equilibrium constant for this reaction.

290 )020.0)(073.0(

(0.013)

]][H[N

][NH K

3

2

3

22

2

3eq

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) ; ][ONO][

][NO K

2

2

2

2eq

E 0.106 0.122 0.129

12.1 ][0.122]106.0[

[0.129] K

2

2

eq

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. ½ NO (g) + ¼ O2 (g) ½ NO2 (g)

Multiply original reaction by ¼, then 1.87 1.12 K K' 44/1

eqeq

b. 2 NO2 (g) 2 NO (g) + O2 (g)

Multiply original reaction by -1 then 0.0826 (12.1) K K' -11

eqeq

c. NO2 (g) NO (g) + ½ O2 (g)

Multiply original reaction by - ½ then 0.287 12.1

1 K K'

2/1

eqeq

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. Be sure to show an ICE table for your calculation.

(Answer: Keq = 0.80)

2 AB (g) + C2D (s) A2D (g) + 2 CB (s)

I 0.30 n/a 0.30 n/a

C +2x -x

E 0.30 + 2x 0.30 –x = 0.20

0.80 0.50

0.20

[AB]

D][AK

0.500.10)(20.30[AB]

0.10 x 0.20x-0.30D][A

22

2eq

2

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.

100 833 20 = 10 x 1.2

0.25 0.25 2 - 0.25

applies Rule 100 x + x 2+ x 2 - 0.25 E

K 0 Q x + x 2+ x 2- C

10 x 1.2 = 0 0 0.25 I

][NO

]O[][NO = K ; O NO 2 NO 2

5-

eq

5-

2

2

2

2

eq22

x

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?

x x x -0.100x - 0.100 E

applyt doesn' - Rule 100 x x x - x - C

Keq 0 Q 0 0 0.100 0.100 I

64.0]][CO[H

O][CO][H K OH CO H CO

22

2eq222

0.62 )056.0(

(0.044) Q :Check

M 0.056 0.044- 0.100 ][H ][CO and M 0.044 O][H [CO]

0.044 x and 0.080 1.80x

0.80x - 0.080 x

0.80 x-0.100

x

0.64 x)- (0.100

x then 0.64

x)- x)(0.100- (0.100

(x)(x) K

2

2

eq2eq2eq2eq

2

2

eq

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?

applyt doesn' 0.625 0.016

0.010 x 0.010x 0.0102x - 0.096 E

Rule 100 x x 2x - C

Keq 0.010 096.0

0.010 Q 0.010 0.010 0.096 I

016.0[HI]

]][I[H K I H HI 2

2

2

2

22eq22

0.017 )093.0(

(0.012) Q :Check

M 0.012 0.0016 0.010 ][I ][H and M 0.093 2(0.0016) - 0.096 [HI]

0.0016 x

0.002 1.252x

0.252x - 0.012 x 0.010

0.126 2x-0.096

x 0.010

0.016 2x) - (0.096

x) (0.010 then 0.016

2x) - (0.096

x) x)(0.010 (0.010 K

2

2

eq2eq2eq

2

2

2eq

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?

0.234 x hen 0.878x t - 0.439 x

0.878 x-0.500

x

0.771 x)- (0.500

x then 0.771

x)- x)(0.500- (0.500

(x)(x) K

applyt doesn' x x x -0.500x - 0.500 E

rule 100 x x x - x - C

Keq 0 Q 0 0 0.500 0.500 I

771.0]][H[CO

O][CO][H K OH CO CO H

2

2

eq

22

2eq222

0.774 )266.0(

(0.234) Q :Check

M 0.266 0.234- 0.500 ][CO ][H and M 0.234 O][H , M260.0 [CO]

2

2

eq2eq2eq2eq

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.

0.674 )0607.0)(0807.0(

(0.0033) Q :Check

0.0033M 0007.000400.0 ][COBr and M 0.0607 ][Br M, 0.0807 [CO]

negatives) (no 1.611- or x 0.0007 x

1.36

1.096 1.095- x

1.36

1.20 1.095- x

0.0007- c 1.095 b 680.0 a

0 0.0007 -1.095x 0.680x

0.680x 0.095x 0.0033 x - 0.00400

) x0.14x (0.0048 0.680 x - 0.00400

0.680 x) x)(0.0600(0.0800

x- 0.00400 K

x -0.00400x -0.0600 x 0.0800 E

applyt doesn' - Rule 100 x - x x C

Keq 0.83 Q 0.00400 0.0600 0.0800 I

680.0][CO][Br

][COBr K COBr Br CO

eq2eq2eq

2

2

2

eq

2

2eq22

DETERMINATION OF Keq OBJECTIVE: Determine the equilibrium constant for the reaction

Fe3+(aq) + SCN

-1(aq) FeSCN

2+(aq)

INTRODUCTION: Iron(III) ion reacts with thiocyanate ion to produce a coloured complex ion.

Fe3+(aq) + SCN

-1(aq) FeSCN

2+(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

C2 =

D2

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] 0.00100 0.00100 0.00100 0.00100 0.00100

Initial [Fe3+] 0.100 0.0400 0.0160 0.00640 0.00256

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) N/A 0.8846 0.75 0.5192 0.2885

Equilibrium [FeSCN2+] 0.001 8.85x10-4 7.50x10-4 5.19x10-4 2.88x10-4

Equilibrium [SCN-1] ~0 1.15x10-4 2.50x10-4 4.81x10-4 7.12x10-4

Equilibrium [Fe3+] N/A 3.91x10-2 1.53x10-2 5.92x1--3 2.27x10-3

Keq N/A 197 196 182 178 CONCLUSION: What is the average Ke value for the mixtures in test tubes 2 to 5?

1884

178182196197

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.

Concentration

(mol/L)

0

2

4

6

15 20 25 30 35

Time

(min)

[AB]

[A2]

[B2 ]

The reaction is

A2 + B2 2 AB

Use collision theory to explain the change in concentrations.

The addition of B2 results in an increased number of collisions between the reactants which causes a

decrease in the concentrations of A2 and B2 and an increase in the concentration of the product AB. As.

the reactants are consumed the forward reaction rate decreases and as the product is produced, the reverse

reaction rate increases until the forward and reverse rates become equal and equilibrium is reestablished. 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) + N2 (g) 2 NH 3 (g)

Increased pressure results in a greater increase in reactant particle collisions than products and a greater

increase in the forward reaction than the reverse and requires a reduction in gaseous moles which is the

product side therefore increased products

(b) 2 NaCl(s) + H2SO4 (l) 2 HCl(g) + Na2SO4 (s)

Increased pressure results in an increase in product collisions and an increase in the reverse reaction and

requires a reduction in gaseous moles which is the reactant side therefore decreased products

(c) SO3 (g) + CaCO3 (s) CaSO4 (s) + CO2 (g)

Since the gaseous moles are the same for reactants and products therefore nothing happens 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 + NH3 NH4Cl + heat

To absorb the increased heat the reverse reaction must proceed therefore product concentrations decrease,

OR since exothermic, activation energy in the reverse is larger and therefore the reverse reaction

increases more than the forward and product concentrations decrease.

(b) 1/2 N2 + 3/2 H2 NH3 ∆H = -46.2 kJ/mol NH3

To absorb the increased heat the reverse reaction must proceed therefore product concentrations decrease,

OR since exothermic, activation energy in the reverse is larger and therefore the reverse reaction

increases more than the forward and product concentrations decrease.

(c) C2H4 (g) + heat C2H2 (g) + H2 (g)

To absorb the increased heat the forward reaction must proceed therefore product concentrations

increase, OR since endothermic, activation energy in the forward is larger and therefore the forward

reaction increases more than the reverse and product concentrations increase. 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? Increase

(b) What would be the affect of decreasing the temperature? Decrease

(c) What would be the affect of removing the NO2 as it is formed? Increase

5. H2(g) + Cl2(g) 2 HCl(g)

What direction will the equilibrium shift when the partial pressure of hydrogen is increased?

It will shift to the right to decrease the pressure of hydrogen.

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?

It will shift to the right so that the heat that’s being removed will be replaced.

7. 2 NO2(g) N2O4(g)

If a large quantity of argon is added to the container in what direction will the equilibrium shift?

It won’t shift, because the partial pressures of each gas will be the same.

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?

It will shift to the right in an effort to increase the quantity of ammonia present.

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?

It will shift to the right so the volume of the gases in the equilibrium will also be decreased.

PART 1 Equilibrium Involving Thymol Blue

Tb2- (blue) + H3O+ HTb 1- (yellow) + H2O

HTb 1- (yellow) + H3O+ H2Tb (red) + H2O

stress [ ] [ ] shift direction [ ] visible change

Starting blue

HCl 1st [H3O+] [Tb2-]

[HTb 1-] yellow

HCl 2nd [H3O+] [HTb2-] [H2Tb ] red

NaOH 1st [OH-] [H3O+] &

[H2Tb]

[HTb 1-] yellow

NaOH 2nd [OH-] [H3O+] &

[HTb 1-]

[Tb2-] blue

PART 2 Equilibrium involving Thiocyanatoiron (III) Ion

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

stress [ ] [ ] shift direction [ ] visible change

HCl [Cl-] on both

sides nothing no change nothing

slightly lighter

yellow/orange

due to dilution

Fe(NO3)3 [Fe3+] [SCN-]

[FeSCN2+] Dark

orange/red

KSCN [SCN-] [Fe3+]

[FeSCN2+] Dark

orange/red

NaOH [OH-] [Fe3+] &

[FeSCN2+]

[Fe3+] yellow

PART 3 Equilibrium Involving Copper Complexes

Cu(H2O)42+ (aq) + 4 NH3 (aq) Cu(NH3)42+ (aq) + 4 H2O

stress [ ] [ ] shift direction [ ] visible change

NH3 1st [NH3] [Cu(H2O)42+] [Cu(NH3)42+] Blue precipitate

NH3 2nd [NH3] [Cu(H2O)42+]

[Cu(NH3)42+] Darker blue

precipitate

HCl [H3O+] [NH3] &

[Cu(NH3)42+]

[Cu(H2O)42+]

Light blue

solution

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).

g/L 310 x 6.5 2Mg(OH) mol

g 58.33 x

L

2Mg(OH) mol 410 x 1.12 2Mg(OH) of solubility then,

solubilitymolar ]2[Mg 410 x 1.12 x

1210 x 5.6 34x ; 1210 x 5.6 2(x)(2x) ; 1210 x 5.62]-OH][2[Mg spK

1210 x 5.6 x 2 x ICE

2]-OH][2[Mg spK ; (aq)-OH 2 (aq)

2Mg 2Mg(OH)

3. The solubility of SrF2 is 7.3 x 10-3 g/100 mL. Calculate Ksp. (7.8 x 10-10)

1010 x 7.8 2)410 x 5.81 x )(2410 x (5.81 2]-F][2[Sr spK then,

mol/L 410 x 5.81 2SrF mol 1

2Sr mol 1 x

L 1

mL 1000 x

g62.125

2SrF mol 1 x

mL 100

g 310 x 7.3 x solubilitymolar

2x x ICE

2]-F][2[Sr spK ; (aq)-F 2 (aq)

2Sr 2SrF

4. The solubility of Cu(IO3)2 is 3.245 x 10-3 mol/L. Calculate Ksp. (1.367 x 10-7)

710 x 1.367 2)310 x 3.245 x )(2310 x (3.245 2]-3IO][2[Cu spK

2x x ICE

2]-3IO][2[Cu spK ; (aq)

-3IO 2 (aq)

2Cu 2)3Cu(IO

5. The solubility of PbI2 is 0.058 g/100mL. Calculate the Ksp for PbI2. (8.0 x 10-9)

910 x 8.0 2)310 x 1.26 x )(2310 x (1.26 2]-F][2[Sr spK then,

mol/L 310 x 1.26 2PbI mol 1

2Pb mol 1 x

L 1

mL 1000 x

g 00.461

2PbI mol 1 x

mL 100

g 0.058 x solubilitymolar

2x x ICE

2]-[I]2[Pb spK ; (aq)-I 2 (aq)

2Pb 2PbI

3]-24[SO2]2[Ca spK ; (aq)

-24SO 3 (aq)

2Fe 2 (s) 3)4(SO2Fe (c)

2]-34[PO3]2[Ca spK ; (aq)

-34PO 2 (aq)

2Ca 3 (s) 2)4(PO3Ca (b)

2]-][OH2[Zn spK ; (aq)-OH 2 (aq)

2 Zn (s) 2 Zn(OH)(a)

6. Calculate the molar concentration of Ba2+ ion in a saturated soln of Ba(IO3)2, if the Ksp = 1.15 x 10-9.

(6.6 x 10-4 mol/L)

]2[Ba 410 x 6.6 x

910 x 1.15 34x ; 910 x 1.15 2(x)(2x) ; 910 x 1.15 2]-3][IO2[Ba spK

910 x 15.1 2x x ICE

2]-3][IO2[Ba spK ; (aq)

-3IO 2 (aq)

2Ba 2)3Ba(IO

7. The solubility of Cd(OH)2 is 2.01 x 10-4 g/100mL. Calculate Ksp. (1.04 x 10-14)

1410 x 1.04 2)510 x 1.373 x )(2510 x (1.373 2]-][OH2[Cd spK then,

mol/L 510 x 1.373 2Cd(OH) mol 1

2Cd mol 1 x

L 1

mL 1000 x

g 43.146

2Cd(OH) mol 1 x

mL 100

g 410 x 2.01 x

2x x ICE

2]-[OH]2[Cd spK ; (aq)-OH 2 (aq)

2Cd 2Cd(OH)

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)

mL g/100 410 x 7.49

g/L 310 x 7.49 4PO3Ag mol

g 418.58 x

L

4PO3Ag mol 510 x 1.79 4PO3Ag of solubility then,

solubilitymolar ]-34[PO 510 x 1.79 x

1810 x 8.2 427x ; 1810 x 8.2 (x)3(3x) ; 1810 x 8.2]-34[PO3][Ag spK

1810 x 2.8 x 3x ICE

]-34[PO3][Ag spK ; (aq)

-34PO (aq)Ag 3 4PO3Ag

9. The solubility of Mg(OH)2 is 9.0 x 10-4 g/100 mL at 18°C. Calculate Ksp. (1.5 x 10-11)

1110 x 1.5 2)410 x 1.54 x )(2410 x (1.54 2]-][OH2[Mg spK then,

mol/L 410 x 1.54 2Mg(OH) mol 1

2Mg mol 1 x

L 1

mL 1000 x

g 33.58

2Mg(OH) mol 1 x

mL 100

g 410 x 9.0 x

2x x ICE

2]-[OH]2[Mg spK ; (aq)-OH 2 (aq)

2Mg 2Mg(OH)

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)

form willppte no Ksp Q

11-10 x 71.1)7-10 x 8.56)(5-10 x (2.0 Q

310 x 2.0]-Cl][[Ag spK ; -Cl Ag AgCl

5-10 x 2.0

-3NO Ag 3AgNO

mol/L 7-10 x 8.56 L 0.200

1 x

g 44.58

NaCl 1mol x g5-10 x 1.0 g/mol 58.44 M

-Cl Na NaCl

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)

form willppte Ksp Q

3-10 x 9.5)050.0)((0.118 Q

310 x 2.0]-2O3H2C][[H spK ; -

2O3H2C H 2O3H2AgC

mol/L 0.118 L 0.050

1 x

g 88.169

3AgNO 1mol x g 1.0 g/mol 169.88 M

-3NO Ag 3AgNO

mol/L 0.050

-2O3H2C H 2O3H2HC

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 = 2.65 x 10-8)

form willppte no Ksp Q -810 x 65.2)3-10 x 84.9(2)3-10 x (1.64 Q

3-10 x 9.84 3-10 x 1.64

L 0.610

mol3-10 x 6.00

L 0.610

mol3-10 x 1.00

510 x 5.1]-24SO[2][Ag spK ; -2

4SO Ag 2 4SO2Ag

mol3-10 x 1.00 n L 0.010 V mol/L, 0.10 C

-3NO Ag 3AgNO

mol3-10 x 6.00 n L 0.600 V mol/L, 0.010 C

-24SO Na 2 4SO2Na

(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 = 3.02 x 10-6)

form willppte no Ksp Q 6-10 x 02.32)010.0)(2-10 x (3.02 Q

510 x 2.12]-][Cl2[Pb spK ; -Cl 2 2Pb 2PbCl

mol/L 0.010 C

-Cl H HCl

M 2-10 x 3.02 g/mol 331.22 M

L 0.100

1 x

g 331.22

2)3Pb(NO mol 1 x g 1.0 C g 1.0 m

-3NO 2 2Pb 2)3Pb(NO

(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 = 6.72 x 10-11)

form willppte no Ksp Q 11-10 x 72.6)6-10 x 46.7)(6-10 x (9.01 Q

910 x 0.2]-4O2][C2[Ca spK ; -2

4O2C 2Ca 4O2CaC

M 6-10 x 7.46 L 1.0

1 x

g 134.00

4O2C2Na mol 1 x g 3-10 x 1.0 C

-24O2C Na 2 4O2C2Na

M 6-10 x 9.01 L 1.0

1 x

g 110.98

2CaCl mol 1 x g 3-10 x 1.0 C

-Cl 2 2Ca 2CaCl

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)

g 610 x 2.6 L 1.0 x 4O2CaC mol 1

g 128.1 x

L

4O2CaC mol -810 x 2.0 OCaCm

solubilitymolar -810 x 2.0 x 910 x 2.0 (x)(0.10)

10.0 x ICE

910 x 0.2]-4O2][C2[Ca spK ; -2

4O2C 2Ca 4O2CaC

0.10

-24O2C Na 2 4O2C2Na (b)

g 310 x 5.7 L 1.0 x 4O2CaC mol 1

g 128.1 x

L

4O2CaC mol 5-10 x 4.47 OCaCm

solubilitymolar 5-10 x 4.47 x 910 x 2.0 (x)(x)

x x ICE

910 x 0.2]-4O2][C2[Ca spK ; -2

4O2C 2Ca 4O2CaC (a)

42

42

(c) 0.010 M CaCl2 (m = 2.6 x 10-5g) (d) 0.10 M NaNO3 (m = 5.7 x 10-3g)

(a).in as g 310 x 5.7 OCaCm theso ion,common no is There

x x ICE

910 x 0.2]-4O2][C2[Ca spK ; -2

4O2C 2Ca 4O2CaC

0.10 0.10

-3NO Na 3NaNO (d)

g 510 x 2.6 L 1.0 x 4O2CaC mol 1

g 128.10 x

L

4O2CaC mol 7-10 x 2.0 OCaCm

solubilitymolar 7-10 x 2.0 x 910 x 2.0 (0.010)(x)

x 0.010 ICE

910 x 0.2]-4O2][C2[Ca spK ; -2

4O2C 2Ca 4O2CaC

0.010

-Cl 2 2Ca 2CaCl (c)

42

42

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)

g 110 x 1.5 L 0.250 x 2PbI mol 1

g 461.00 x

L

2PbI mol 310 x 1.29 PbIm

solubilitymolar 310 x 1.29 x ; 910 x 5.8 34x ; 910 x 5.8 2(x)(2x)

x2 x ICE

910 x 5.82]-][I2[Pb spK ; -I 2 2Pb 2PbI

2

(b) 0.010 M Pb(NO3)2 (m = 5.3 x 10-2g) (c) 0.010 M CaI2 (m = 2.5 x 10-3g)

g 10 x 2.5 m and solubilitymolar 10 x 2.13 x ; 10 x 5.8 (x)(0.020)

0.020 x ICE

10 x 5.8]][I[Pb K ; I 2 Pb PbI

0.020

I 2 Ca CaI (c)

g 10 x 5.3 m and solubilitymolar 10 x 4.6 x ; 10 x 5.8 )(0.010)(2x

2x 0.010 ICE

10 x 5.8]][I[Pb K ; I 2 Pb PbI

0.010

NO 2 Pb )Pb(NO (b)

3

PbI

592

92-2

sp

-2

2

-2

2

2

PbI

492

92-2

sp

-2

2

-

3

2

23

2

2

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)

form willppte a Ksp Q -810 x 7.4)5-10 x 75.3)(3-10 x (1.25 Q

5-10 x 3.75 3-10 x 1.25

L 0.120

mol6-10 x 4.5

L 0.120

mol4-10 x 1.5

1010 x 8.1]-Cl][[Ag spK ; -Cl Ag AgCl

mol4-10 x 1.5 n L 0.075 V mol/L, 3-10 x 2.0 C

-3NO Ag 3AgNO

mol6-10 x 4.5 n L 0.045 V mol/L, 4-10 x 1.0 C

-Cl Na NaCl

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)

g 710 x 2.6 AgClm and solubilitymolar 810 x 1.8 x ; 1010 x 1.8 (x)(0.010)

0.010 x ICE

1010 x 8.1]-Cl][[Ag spK ; -Cl Ag AgCl

0.010

-Cl Na NaCl

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)

form willppte a Ksp Q 10 x 0.3)10 x )(6.6710 x (6.67 Q

10 x 6.67 10 x 6.67

L 0.150

mol10 x 1.00

L 0.150

mol10 x 1.0

10 x 6.5]OH][[Mg K ; OH 2 Mg Mg(OH)

mol10 x 1.00 n L 0.100 V mol/L, 0.010 C

NO 2 Mg )Mg(NO

mol10 x 1.0 n L 0.050 V 2),(x mol/L 0.010 C

OH 2 Ba Ba(OH)

7-23-3-

3-3-

3-3-

122-2

sp

-2

2

3-

-

3

2

23

3-

-2

2

9. Calculate the number of moles of AgCl that will dissolve in 1.0 litre of a 0.10 M CaCl2 solution. (m =

1.3 x 10-7g)

g 710 x 1.3 AgClm and solubilitymolar 1010 x 9.0 x ; 1010 x 1.8 (x)(0.20)

0.20 x ICE

1010 x 8.1]-Cl][[Ag spK ; -Cl Ag AgCl

0.20 0.10

-Cl 2 2Ca 2CaCl

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)

form willppte no Ksp Q 12-10 x 5.22)4-10 x )(1.004-10 x (2.5 Q

5-10 x 1.0 4-10 x 2.5

L 0.100

mol5-10 x 1.0

L 0.100

mol5-10 x 2.5

1110 x 9.32]-F][2[Ca spK ; -F 2 2Ca 2CaF

mol5-10 x 2.5 n L 0.050 V mol/L, 4-10 x 5.0 C

-3NO 2 2Ca 2)3Ca(NO

mol5-10 x 1.0 n L 0.050 V mol/L, 4-10 x 2.0 C

-F Na NaF

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)

solubilitymolar 1010 x 3.4 x ; 1110 x 3.4 (x)(0.10)

0.10 x ICE

1110 x 3.4]-24SO][2[Ra spK ; -2

4SO 2Ra 4RaSO

0.10

-24SO Na 2 4SO2Na

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)

first. eprecipitat it will AgCl,for smaller is ][Ag as

810 x 1.8 x ; 1010 x 1.8 (x)(0.010)

0.010 x ICE

1010 x 8.1]-Cl][[Ag spK ; -Cl Ag AgCl

510 x 3.5 x ; 1210 x 2.1 (0.0010)2(x)

0.0010 ICE

1210 x 2.1]-24CrO[2][Ag spK ; -2

4CrO Ag 2 4CrO2Ag

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?

There are no macroscopic changes.

2. What evidence is there to indicate that equilibrium is a dynamic state?

The reaction can be shifted in either direction. 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.

CRT = Por RTV

n = P

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) Fe2+ (aq) + Cu (s) ;

(e) HSO4- (aq) H+ (aq) + SO4

2- (aq) ;

(f) 2 CrO42- (aq) + H+ (aq) Cr2O7

2- (aq) + OH- (aq) ;

(g) Ag2CO3 (s) 2 Ag+ (aq) + CO32- (aq) ;

5. The equilibrium constants for three different reactions are:

(a) Keq = 1.5 x 1012 ; (b) Keq = 0.15; (c) Keq = 4.3 x 10-15

In which reaction is there: (a) a large ratio of product to reactant? (a)

(b) a small ratio of reactant to product? (a) 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.

0.063 0.025) - (0.035

025)(0.025)(0.

]5[PCl

]2][Cl3[PCl eqK

0.025 x x - 0.035 ICE

]5[PCl

]2][Cl3[PCl eqK ; 2Cl 3PCl 5PCl

]2][F2[H

2[HF] = eqK

3]2[O4[NO]

2]5O2[N = eqK

]2[CO = eqK

]2[Cu

]2[Fe = eqK

]-4[HSO

]-24][SO[H

= eqK

][H2]-24[CrO

][OH]-27O2[Cr

= eqK

3]-23[CO2][Ag = eqK

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.

M 0.015 ]3[PCl x ; 0.063 0.012

(x)(0.049)

]5[PCl

]2][Cl3[PCl eqK

0.049 x 0.012 ICE

0.063 ]5[PCl

]2][Cl3[PCl eqK ; 2Cl 3PCl 5PCl

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.

0.26y ; 0.010 0.0063 -0.063y ; 0.063 0.10 -y

2(0.10)

]5[PCl

]2][Cl3[PCl eqK

0.10 0.10 0.10-y ICE

0.063 ]5[PCl

]2][Cl3[PCl eqK ; 2Cl 3PCl 5PCl

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.

eqK Q until left theshift to rxn will theand eqK Q 0.144 2.5

20.60 Q

x 0.60x 0.60 x - 2.5 ICE

0.063 ]5[PCl

]2][Cl3[PCl eqK ; 2Cl 3PCl 5PCl

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.

3.0 050)(0.400)(0.

200)(0.300)(0.

]2][NO2[SO

][NO]3[SO eqK

11. At 55°C, for the reaction

2 NO2 (g) N2O4 (g) : Keq = 1.15

(a) Write the equilibrium expression.

1.15 2]2[NO

]4O2[N eqK

(b) Calculate the concentration of N2O4 present in equilibrium with 0.5 mole of NO2.

M 0.29 x ; 1.15 20.5

x

2]2[NO

]4O2[N eqK

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

29.1 )3-10 x )(1.173-10 x (1.17

)3-10 x )(1.333-10 x (1.33

]2][H2[CO

O]2[CO][H eqK

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.

]2[N M 0.100 x ; 0.300 3x x 2x - 1.00 ICE

2]3[NH

3]2][H2[N eqK ; 2H 3 2N 3NH 2

(b) Calculate the concentration of NH3 at equilibrium.

M 0.80 0.200 - 1.00 2x - 1.00 ]3[NH

(c) Calculate the equilibrium constant for this system at this temperature and pressure.

3-10 x 4.2 280.0

3300)(0.100)(0.

2]3[NH

3]2][H2[N eqK

(d) Which way would the equilibrium be shifted if 0.600 mole of H2 were injected into the flask?

The equilibrium will shift to the left, .

(e) How would the injection of hydrogen into the flask affect the equilibrium constant?

There would be no change to the equilibrium constant.

(f) How would the equilibrium constant be affected if the pressure of this system were suddenly

increased?

There would be no change to the equilibrium constant. 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.

M 0.21 0.29- 0.50 ]2[H ]2[CO and M 0.29 ][CO O]2[H

0.29 x ; 0.7 x 2.4 ; x x 1.4 - 0.7 ; x x)(1.4)0.50(

4.1 x- 0.50

x ; 2.00

x- 0.50

x ; 2.00

2)x0.50(

2x

]2[H]2[CO

]O][CO2[H eqK

x x x - 0.50 x - 0.50 ICE

2.00 ]2[H]2[CO

]O][CO2[H eqK ; CO O2H 2H 2CO

(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?

M 0.21 ]2[H ]2[CO and M 0.29 0.21 - 0.50 ][CO O]2[H

0.21 x ; 0.5 x 2.4 ; x 1.4 x 0.50

4.1 x

x- 0.50 ; 2.00

x

x- 0.50 ; 2.00

2x

2)x0.50(

]2[H]2[CO

]O][CO2[H eqK

x- 0.50 x - 0.50 x x ICE

2.00 ]2[H]2[CO

]O][CO2[H eqK ; CO O2H 2H 2CO

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. To the left, .

(b) The partial pressure (concentration) of H2O is increased. To the right, .

(c) The concentration of O2 is decreased. To the right, .

(d) The temperature is increased. To the right, .

(e) The total pressure is increased. To the left, . 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. The [N2O4] will until Keq is reached again.

(b) increasing the temperature. The [N2O4] will until Keq is reached again.

(c) increasing the volume. The [N2O4] will until Keq is reached again.

(d) adding more NO2 (g) to the system without changing P or T. The [N2O4] will .

(e) adding a catalyst? no change 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

(a) [H2O] will (b) [H2O] will (c) [H2O] will (e) no change 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) 4 NO (g) + 6 H2O (g) ; H = -ve

(i) T (ii) P

(b) Br2 (g) + Cl2 (g) 2 BrCl (g) ; H = +ve

(i) T (ii) P no change

(c) BaSO4 (s) Ba2+ (aq) + SO42- (aq) ; H = +ve

(i) T (ii) P no change

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

T12S 2 Tl+ + S2- CuS Cu2+ + S2-

Cu2S Ksp = 1 x 10-48 HgS Ksp = 1 x 10-52

Cu2S 2 Cu+ + S2- HgS Hg2+ + S2-

Rank each of these sulfides in order of decreasing molar solubility in their saturated solutions.

T12S > CuS > Cu2S > HgS 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?

10-10 x 5.190 )4-10 x 5.063(2)4-10 x (1.013 ]-23[CO2][Ag spK

4-10 x 5.063 )4-10 x 2(5.063 ICE

]-23[CO2][Ag spK ; -2

3CO Ag 2 3CO2Ag

M 4-10 x 5.063 L 0.500

mol 4-10 x 2.531 Cthen

mol 4-10 x 2.531 g 75.275

3CO2Ag mol 1 x g 0.0698 n ; g 0.0698 COAgm

32

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.

g 10 x 1.5 m and solubilitymolar 10 x 1.7 x ; 10 x 1.7 (x)(0.100)

0.100 x ICE

10 x 7.1]F][[Ba K ; F 2 Ba BaF

0.100

F Na NaF

2

BaF

462

62-2

sp

-2

2

-

2

22. (a) 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?

form willppte a spK Q ; 610 x 7.9 2015)(0.035)(0. Q

610 x 7.12]-F][2[Ba spK ; -F 2 2Ba 2BaF

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.

form willppte a spK Q ; 610 x 7.9 2015)(0.035)(0. Q

L 1.0

mol 0.015

L 1.0

mol 0.035

610 x 7.12]-F][2[Ba spK ; -F 2 2Ba 2BaF

mol 0.035 n then 0.700L V ; M 0.050 C

-Cl 2 2Ba 2BaCl

mol 0.015 n then 0.300L V ; M 0.050 C

-F Na NaF

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).

g/L 0.33 PbClmol

g 278.10x

L

PbClmol 10x 1.2

x M solubilitymolar PbClof solubility then,

solubilitymolar ][Pb 10x 1.2 x

10x 1.2 0.01x ; 10x 1.2 (x)(0.10) ; 10x 2.1]Cl][[Pb K

10x 2.1 0.10 x ICE

]Cl][[Pb K; Cl 2 Pb PbCl

M0.10

Cl Na NaCl

2

2

3

PbCl2

23

55252-2

sp

5

2-2

sp(aq)-

(aq)2

2

-11

2

2. What is the Keq for A + 2B AB2 if [AB2]i = 0.20 M and [A]eq = 0.050 M?

150 0.001

0.15

0.050))(0.050)(2(

0.050) - (0.20

[A][B]

][AB K

0.050

x - 0.20 2x x ICE

[A][B]

][AB K; AB 2B A

22

2eq

2

2eq2

3. If 1.00 g of NaCl(s) is added to 500. mL of a 0.010 M Pb(NO3)2 will a precipitate form? (Ksp PbCl2 =

1.2 x 10-5).

form willppte no Ksp Q

10x 2.1)10x 42.3)((0.010 Q

]Cl][[Pb K; Cl 2 Pb PbCl

0.010

2NO Pb ) Pb(NO

mol/L 10x 3.42 L0.500

1x

g 44.58

NaCl 1molx g 1.00 g/mol 58.44 M

Cl Na NaCl

5-22-

2-2

sp(aq)-

(aq)2

2

-

3

2

23

2-

-

4. What is the [AB]eq in A + B AB if [A]i = [B]i = 0.10 M (Keq = 101).

Keq Q 100 (0.027)

0.073 Q :Check

0.073 [AB]eq and 0.027 0.073-0.10 [B]eq [A]eq

0.073 x or 0.14 x

202

6.43 21.2

202

408.04 - 449.44 21.2 x

0 1.01 21.2x - 101x

101x x 20.2 - 1.01 x

101 x)-(0.10

(x)

[A][B]

[AB] K

applyt doesn' - Rule100 x x - 0.10x - 0.10 ICE

101 [A][B]

[AB] K; AB B A

2

2

2

2eq

eq

5. What is the solubility in g/100 mL of Ag2CO3 if the Ksp for Ag2CO3 is 8.5 x 10-12?

mL g/100 0.0036 mL 100

L0.1x

COAg mol

g 275.75x

L

COAg mol 10x 1.29

x M solubilitymolar COAg of solubility then,

solubilitymolar ]CO[ 10x 1.29 x

10x 8.5 4x ; 10x 8.5 (x)(2x) ; 10x 5.8 ]CO[][Ag K

10x 5.8 x x 2 ICE

]CO[][Ag K; CO Ag 2 COAg

32

32

4

COAg32

2-

3

4

123122122-

3

2

sp

12

-2

3

2

sp(aq)-2

3(aq)32

32