Power Point Lesson Notes- double click on the lesson number. Worksheets               Quiz

1.         Approaching Equilibrium                    WS  1                          Q1

2.         LeChatelier's Principle-1                    WS 2

3.         LeChatelier's Principle-2                    WS 3 & 4                    Q2

4.         LeChatelier's-3 & Start Lab                WS 5

5.         Lab Lechatelier's                                 Questions 1-10            Conclusion

6.         Haber/Graphing                                  WS 6 & 7                    Q3

7.         Equilibrium Constants                         WS 8                           Q4

8.         Keq Calculations                                 WS 9 & 10

9.         K-trial & Size Keq                              WS 11                         Q5

10.       Entropy & Enthalpy                             WS 12                         Q6

12.       Review                                                Practice Test 2

Worksheet #1 Approaching Equilibrium

1. What are the conditions necessary for equilibrium?

Must have a closed system.

Must have a constant temperature.

Ea must be low enough to allow a reaction.

2. What is a forward reaction versus a reverse reaction?

In a forward reaction, the reactants collide to produce products and it goes from left to right.

In a reverse reaction, the products collide to produce reactants and it goes form right to left.

3. Why does the forward reaction rate decrease as equilibrium is approached?

As the reaction goes to the right, the reaction concentration decreases and therefore, there are less reactant collisions causing the forward rate to decrease.

4. What are the characteristics of equilibrium?

Forward rate is equal to the reverse rate.

The concentration of reactants and products are constant.(not equal)

Macroscopic properties are constant (color, mass, density, pressure, concentrations).

5. Define equilibrium.

Equilibrium occurs when:

Forward rate is equal to the reverse rate.

The concentration of reactants and products are constant.(not equal)

Macroscopic properties are constant. (color, mass, density, pressure, concentrations)

6. Define the word dynamic and explain its relevance to the concept of equilibrium.

The word dynamic means that forward and reverse continue to occur.

7. Why does the reverse reaction rate increase as equilibrium is approached?

The reverse reaction rate increases as equilibrium is approached because as the reaction goes from left to right,

the concentrations of the products increases, therefore there are more product collisions causing the reverse reaction rate to increase.

As a reaction is approaching equilibrium describe how the following change. Explain what causes each change.

8. Reactant concentration.  As the reaction goes to the right, the reaction concentration decreases.

9. Products concentration. As the reaction goes from left to right, the concentration of the products increases.

10. Forward reaction rate. The reaction concentration decreases and therefore, there are less reactant collisions causing the forward rate to decreases.

11. Reverse reaction rate. The concentrations of the products increases, therefore there are more product collisions causing the reverse reaction rate to increase.

12. What is equal at equilibrium? The forward and reverse rates are equal.

13. What is constant at equilibrium? The reactant and product concentrations and the macroscopic properties are constant.

14. Sketch each graph to show how concentrations change as equilibrium is approached

[reactant]                                 [product]                                             Overall Rate   15. Label each graph with the correct description.

·       The forward and reverse rates as equilibrium is approached

·       The overall rate as equilibrium is approached

·       The reactant and product concentrations as equilibrium is approached (two graphs)    16. Draw a PE Diagram for the reaction if PE of the reactants is 100 KJ/mole N2O4  and

Ea = 110  KJ/mole N2O4.

N2O4 (g) <----->         2 N02 (g)            DH=  +58KJ

(colorless)                               (brown) If a catalyst was added to the reaction, what would happen to the PE Diagram, the forward rate, and the reverse rate?

PE Diagram     The activation energy would decrease

Forward rate     Increase

Reverse rate    Increase

One mole of very cold, colorless N2O4 (g) is placed into a 1.0L glass container of room temperature. The reaction:

N2O4 (g)               2 N02 (g)            DH=  +58KJ

(colorless)                   (brown)

proceeds to equilibrium. The concentration of each gas is measured as a function of time.

Time  (s)   0          5          10        15        20        25

[N2O4] (M)           1.0       0.83     0.81     0.80     0.80     0.80

[N02]  (M)   0.0     0.34     0.38     0.40     0.40     0.40

17. Plot concentration of N2O4 and N02 against time on the same graph below. 1.0 -

0.9 -

0.8 -

0.7 -

0.6 -

0.5 -

0.4 -

0.3 -

0.2 -

0.1 -

0.0 -

0            5              10            15            20            25            30             35

TIME (s)

18. After what time interval has equilibrium been established?          15 seconds

19. Describe the change in the appearance of the container over 25 seconds (describe the colour change and when it becomes constant).

The container will gradually increase the intensity of brown and then remain constant after 15 seconds.

20. Calculate the rate of N2O4 consumption in (M/s) over the first 5s period and then the second 5s period.

0-5 sec.                                                                                                     rate = 1.0 – 0.83 M    =   0.034 M / s

5.0 sec

5-10 sec.                                                                                                   rate = 0.83 – 0.81 M    =   0.004 M / s

5.0 sec

Why is the rate greater over the first five minutes compared to the second five minutes (think in terms of reactant and product concentrations?

The reactant concentration has decreased and the product concentration increased.

The forward rate has decreased and the reverse rate increased and because of this the overall net rate has decreased.

21. Calculate the rate of N02 production in (M/s) over the first 5s period and then the second 5s period.

0-5 sec.                                                                                                     rate = 0.34 – 0.00 M    =   0.068 M / s

5.0 sec

5-10 sec.                                                                                                   rate = 0.38 – 0.34 M    =   0.008 M / s

5.0 sec

How does the rate of formation of N02 compare to the rate of consumption of N2O4? Remember, if you measure the reactants or products, it is still the overall rate.

It is twice as great because of the stoichiometric relationship.       2moles NO2

1mole N2O4

22. What are the equilibrium concentrations of N2O4 and  N02?

[N2O4]= 0.80 M                                  Are they equal?  No!

[N02]  = 0.40 M

23. Is the reaction over, when equilibrium has been achieved? If not, explain.

No it is not. Although the concentrations are constant, the forward and reverse reactions continue forever.

24. What are the necessary conditions to establish equilibrium?

Must have a closed system.

Must have a constant temperature.

Ea must be low enough to allow a reaction.

25. What are the characteristics of an equilibrium?

Forward rate is equal to the reverse rate.

The concentration of reactants and products are constant.(not equal)

Macroscopic properties are constant. (color, mass, density, pressure, concentrations)

Worksheet #2 LeChatelier’s Principle

Describe the changes that occur after each stress is applied to the equilibrium.

N2 (g)    +     3H2 (g)         2NH3(g)    +   92 KJ

Shifts                        Shifts to the

Stress                                                      [N2]                  [H2]                [NH3]             Right or Left                Reactants or Product

1. [N2] is increased                             increases          decreases        increases           right                products

2. [H2] is increased                            decreases          increases         increases            right                products

3. [NH3] is increased                          increases          increases         increases             left                  reactants

4. Temp is increased                           increases          increases         decreases            left                  reactants

5. [N2] is decreased                            decreases         increases         decreases            left                  reactants

6. [H2] is decreased                            increases          decreases        decreases             left                  reactants

7. [NH3] is decreased                          decreases         decreases        decreases            right                products

8. Temp is decreased                          decreases         decreases        increases              right                products

9. A catalyst is added                          nochange          nochange         nochange             nochange        nochange

N2O4 (g)                     2NO2(g)                        DH =  +   92 KJ

Shifts                           Shifts to Favor the

Stress                                                    [N2O4]              [NO2]           Right or Left                Reactants or Products

1. [N2O4] is increased                         increases        increases        right                            products

2. [NO2] is increased                          increases        increases        left                              reactants

3. Temp is increased                           decreases       increases        right                            products

4. [N2O4] is decreased                        decreases       decreases       left                              reactants

5. [H2] is decreased                            nochange        nochange        nochange                    nochange

6. [NO2] is decreased                          decreases       decreases       right                            products

7. Temp is decreased                          increases        decreases       left                              reactants

4HCl (g)    +     O2 (g)         2H2O(g)    +     2Cl2 (g)   +   98 KJ

Shifts               Shifts to Favour the

Stress                                      [O2]                 [H2O]              [HCl]               Right or Left    Reactants or Products

1. [HCl] is increased               decreases       increases        increases        right                products

2. [H2O] is increased              increases        increases        increases        left                  reactants

3. [O2] is increased                 increases        increases        decreases       right                products

4. Temp is increased               increases        decreases       increases        left                  reactants

5. [H2O] is decreased              decreases       decreases       decreases       right                products

6. [HCl] is decreased              increases        decreases       decreases       left                  reactants

7. [O2] is decreased                decreases       decreases       increases        left                  reactants

8. Temp is decreased              decreases       increases        decreases       right                products

9. A catalyst is added              nochange        nochange        nochange        nochange            nochange

CaCO3 (s)   +   170 KJ        CaO (s)    +    CO2  (g)

Note :  Adding solids or liquids and removing solids or liquids does not shift the equilibrium. This is because you cannot change the concentration of a pure liquid or solid as they are 100% pure. It is only a concentration change that will change the # of collisions and hence shift the equilibrium.

Shifts                           Shifts to Favor the

Stress                                      [CO2]                           Right or Left                Reactants or Products

1. CaCO3 is added                   nochanges                   nochanges                   nochanges

2. CaO is added                      nochanges                   nochanges                   nochanges

3. CO2 is added                       increases                    left                              reactants

4. Temp is decreased              decreases                   left                              reactants

5. A catalyst is added              nochanges                   nochanges                   nochanges

6. [CO2] is decreased              decreases                   right                            products

7. Temp is increased               increases                    right                            products

8. CaO is removed                  nochanges                   nochanges                   nochanges

Worksheet #3            Applying Le Châtelier's Principle

The oxidation of ammonia is a reversible exothermic reaction that proceeds as follows:

4 NH3 (g) + 5 O2 (g)                      4 NO (g) + 6 H2O (g)

Le Châtelier’s Principle allows us to predict the changes that occur in an equilibrium reaction

to compensate for any stress that is placed upon the system. For each situation described in the table,

indicate an increase or decrease in overall concentration from before to after a new equilibrium has

been established.

Component Stress                        Equilibrium Concentrations

[NH3]              [O2]               [NO]               [H2O]

NH3         addition                       increases        decreases      increases         increases

removal                       decreases       increases       decreases        decreases

O2            addition                       decreases       increases       increases            increases

removal                       increases        decreases      decreases           decreases

NO           addition                       increases        increases         increases            decreases

removal                       decreases       decreases        decreases           increases

H2O          addition                       increases        increases         decreases            increases

removal                      decreases       decreases        increases            decreases

Increase in temperature:                 increases        increases         decreases            decreases

Decrease in temperature:               decreases       decreases         increases            increases

Increase Presssure:                        increases      increases            increases            increases

Decrease in pressure:                     decreases       decreases        decreases            decreases

We increased the volume- all concentrations go down

Addition of a catalyst:                     nochange        nochange          nochange            nochange

An inert gas is added:                    nochange        nochange            nochange            nochange

Worksheet #4            Le Chatelier’s Principle

State the direction in which each of the following equilibrium systems would be shifted upon the application

of the following stress listed beside the equation.

1.         2 SO2 (g) + O2 (g)                           2 SO3 (g) + energy         decrease temperature         right

2.         C (s) + CO2 (g) + energy                               2 CO (g)          increase temperature          right

3.         N2O4 (g)         2 NO2 (g)                                  increase total pressure                          left

4.         CO (g) + H2O (g)                           CO2 (g) + H2 (g)             decrease total pressure          nochange

5.         2 NOBr (g)                    2 NO (g) + Br2 (g)                     decrease total pressure            right

6.         3 Fe (s) + 4 H2O (g)                   Fe3O4 (s) + 4 H2 (g)         add Fe(s)                                       nochanges

7.         2SO2 (g) + O2 (g)                             2 SO3 (g)                      add catalyst                          nochanges

8.         CaCO3 (s)                        CaO (s) + CO2 (g)                      remove CO2 (g)                            right

9.         N2 (g) + 3 H2 (g)                              2 NH3 (g)                       increase [He (g)]                   no change

Consider the following equilibrium system:

3 H2 (g) + N2 (g)          2 NH3 (g) + Heat.

State what effect each of the following will have on this system:

10.       More N2 is added to the system                                                                           right

11.       Some NH3 is removed from the system                                                                right

12.       The temperature is increased                                                                                 left

13.       The volume of the vessel is increased                                                                     left

14.       A catalyst was added                                                                                            nochange

15.       An inert gas was added                                                                                         nochange

16.       If a catalyst was added to the above reaction and a new equilibrium was established.

Compare to the original system, the rates of the forward and reverse reactions of the new equilibrium.

Forward Rate has        increases                          Reverse Rate has               increases

17.       If the temperature was increased in the above reaction and a new equilibrium was established.

Compare to the original system, the rates of the forward and reverse reactions of the new equilibrium.

Forward Rate has        increases                         Reverse Rate has                increases

18.       If the volume of the container was increased in the above reaction and a new equilibrium

was established. Compare to the original system, the rates of the forward and reverse reactions

of the new equilibrium.

Forward Rate has        decreases                          Reverse Rate has              decreases                      .

Consider the following equilibrium system

H2 (g) + I2 (g)  2 HI (g)

State what effect each of the following will have on this system in terms of shifting.

19.       The volume of the vessel is increased             nochange

20.       The pressure is increased                               nochange

21.       A catalyst is added                                        nochange

Consider the following equilibrium system:

3 Fe (s) + 4 H2O (g)  <------> Fe3O4 (s) + 4 H2 (g)

State what effect each of the following will have on this system in terms of shifting.

21.       The volume of the vessel is decreased                          nochange

22.       The pressure is decreased                                           nochange

23.       More Fe is added to the system                                   nochange

24.       Some Fe3O4 is removed from the system                     nochange

25.       A catalyst is added to the system                                 nochange

Consider the following equilibrium:

2NO (g)   + Br2 (g)  + energy  <------>  2NOBr (g)

State what affect each of the following will have on this system in terms of shifting.

26.       The volume of the vessel is increased                         left

27.       The pressure is decreased                                          left

28.       More Br2 is added to the system                                right

29.       Some NO is removed from the system                        left

30.       A catalyst is added to the system                                nochange

Consider the following equilibrium:

Some CO was added to the system and a new equilibrium was established.

2CO (g)   + O2 (g)  <------>  2CO2 (g) + energy

31.       Compare to the original system, the rates of the forward and reverse reactions of the new equilibrium.

Forward Rate has        increases                         Reverse Rate has                increases

32.       Compared to the original concentrations, after the shift, have the new concentrations increased or decreased?

[CO]                increases                    [O2]                 decreases                   [CO2]               increases

33.       Did the equilibrium shift favour the formation of reactants or products?        products

A catalyst was added to the system at constant volume and a new equilibrium was established.

2CO (g)   + O2 (g)    2CO2 (g) + energy

34.       Compare to the original system, the rates of the forward and reverse reactions of the new equilibrium.

Forward Rate has        increases                          Reverse Rate has               increases

35.       Compared to the original concentrations, after the shift, have the new concentrations increased or decreased?

[CO]    no change                       [O2]             no change                   [CO2]               no change

36.       Did the equilibrium shift favour the formation of reactants or products?        Neither

The volume of the container was decreased and a new equilibrium was established.

2CO (g)   +   O2 (g)    2CO2 (g) + energy

37.       Compare to the original system, the rates of the forward and reverse reactions of the new equilibrium.

Forward Rate has increased                           Reverse Rate has         increased

38.       Compared to the original concentrations, after the shift, have the new concentrations increased or decreased?

[CO]                increased            [O2] increased                                [CO2]   increased

39.       Did the equilibrium shift favor the formation of reactants or products?          Products

Worksheet #5    Applying Le Châtelier's Principle

1.         The chromate and dichromate ions set up an equilibrium as follows:

energy + 2 CrO4 2-(aq) + 2 H+(aq)               Cr2O7 2-(aq) + H2O (l)

yellow                                           orange

Describe how the above equilibrium will shift after each stress below:

shift                  color change

Increase in [H+]                                               right                Orange

Increase in [CrO4 2-]                                        right

Increase in [Cr2O7 2-]                                       left

Decrease in [H+]                                             left                  Yellow

Decrease in [CrO42-]                                       left

Increase in temperature                                   right                Orange

Decrease intemperature                                   left                  Yellow

OH- reacts with H+ and lowers [H+] causing the reaction to shift left.

2.         The copper (II) ion andcopper (II) hydroxide complex exist in equilibrium as follows:

Cu(OH)2 (aq) +4 H2O (l)               Cu(H2O)4 2+(aq) +2 OH-(aq) + 215 kJ

violet                                                                 light blue

Describe how the above equilibrium will shift after each stressbelow:

shift                 colorchange

Increase in [Cu(H2O)4 2+]                                 left

Increase in [Cu(OH)2]                                     right

Decrease in [Cu(H2O)4 2+]                               right

Decrease in[Cu(OH)2]                                    left

Increase temperature                                       left                  Violet

Decrease temperature                                      right                Light  Blue

3.         Consider the equilibriumthat follows:

4 HCl (g)+ 2 O2 (g)         2 H2O (l) +      2 Cl2 (g)    +      98 kJ

(yellow)

Describe how the above equilibrium will shift after each stress below:

shift                        color change

Increase in temperature                                    left                  clear

Increase [HCl]                                                right                yellow

Decrease in [Cl2]                                            right

Decrease temperature                                      right                yellow

Add Ne at constant volume                             No Change

4.         Consider the equilibrium that follows:

Cu+ (aq)   +    Cl-(aq)         CuCl (s)           ΔH   =   + 98 kJ

(green)

Describe how the above equilibrium will shift after each stress below:

Cu+ is green

shift                 color change

Increase in temperature                                                           right               less green

Increase [HCl]                                                                        right                less green

Decrease temperature                                                              left                  green

(reacts with HCl)

(check your solubility table for a possible reaction)

Add CuCl(s)                                                     no change                   no change

(check your solubility table for a possible reaction)

Add CuNO3 (aq)                                                right because it contains the Cu+ ion.

Add Cu(NO3)2 (aq)                                            no change because the Cu2+ ion is a spectator.

Worksheet #6            Graphing and LeChatelier’s Principle

Consider the following equilibrium system.

I2(g)    +     Cl2(g)       2 ICl (g) + energy

Label the graph that best represents each of the following stresses and shift.

·       increasing the temperature

·       decreasing the pressure

·

 Increasing the temperature

removing Cl2(g) Worksheet #7    Maximizing Yield

1.         N2O4(g)     +    59  KJ         2 NO2(g)

Describe four ways of increasing the yield of for the reaction above.

Increase the temperature                 Increase the concentration of (N2O4)

Decrease the pressure                      Decrease the concentration of (NO2)

Describe three ways to increase the rate of the above reaction.

Increase the temperature                 Increase the concentration of (N2O4)

Add a catalyst                                    Increase the pressure

2.         2SO3(g)           2SO2(g)     +    O2(g)     +     215 KJ

Describe four ways of increasing the yield of for the reaction above.

Decrease the temperature                Increase the concentration of SO3

Decrease the pressure                      Decrease the concentration of SO2

Decrease the concentration of O2

Describe three ways to increase the rate of the above reaction.

Increase the temperature                 Increase the concentration of (SO3)

Increase the pressure

3.         H2O(g)  ®     H2O(l)         DH = -150 KJ

Describe three ways of increasing the yield of for the reaction above.

Decrease the temperature                Increase the concentration of H2O

Increase the pressure

Describe four ways to increase the rate of the above reaction.

Increase the temperature                 Increase the concentration of H2O

Add a catalyst                                    Increase the pressure

4.         In the Haber reaction:              3H2(g)   +    N2(g)           2NH3(g)        +          energy

Explain why each condition is used in the process to make ammonia.

A High pressure of 50 MP                              High yield shifts to right

The presence of Fe2O3                                     A catalyst increases the rate

Condensing NH3 to a liquid                             Removing the products shifts to the right increasing the yield

A relatively high temperature 500 oC              Even though the yield is lowered the rate is increased
Worksheet #8            Equilibrium Calculations

Solve each problem and show all of your work.

1.         SO3(g)         +     H2O(g)                    H2SO4(l)          Do not count the liquid!!

At equilibrium [SO3] = 0.400M             [H2O] = 0.480M                  [H2SO4] = 0.600M

Calculate the value of the equilibrium constant.

Keq =             1

[SO3] [H2O]

Keq =             1

[0.400] [0.480]

Keq = 5.21

2.         At equilibrium at 100oC, a 2.0L flask contains:

0.075 mol of PCl5             0.050 mol of H2O               0.750 mol 0f HCl           0.500 mol of POCl3

Calculate the Keq for the reaction:

PCl5 (s) + H2O (g) 2HCl (g) + POCl3 (g)

Keq = 1.4

3.         Keq= 798 at 25oC for the reaction:  2SO2 (g)   +  O2 (g)   2SO3 (g).

In a particular mixture at equilibrium, [SO2]= 4.20 M and [SO3]=11.0M. Calculate the equilibrium [O2] in this mixture at 25oC.

[O2] = 0.00860M

4.         Consider the following equilibrium:

2SO2 (g) + O2 (g) 2SO3 (g)

0.600 moles of SO2 and 0.600 moles of O2 are present in a 4.00 L flask at equilibrium at 100oC. If the Keq = 680, calculate the SO3 concentration at 100oC.

Keq  =   [SO3]2

[SO2]2[O2]

680  =    [SO3]2

[0.150]2[0.150]

[SO3]2 = (0.150)(0.150)2(680)

[SO3] = 1.51 M

5.         Consider the following equilibrium:

2 NO2(g)                   N2O4(g)

2.00 moles of NO2 and1.60 moles of N2O4 are present in a 4.00 L flask at equilibrium at 20oC. Calculate the Keq at 20oC .

Keq = 1.60

6.         2 SO3(g)                   2 SO2(g)           +          O2(g)

4.00 moles of SO2 and 5.00 moles O2 are present in a 2.00 L container at 100oC and are at equilibrium. Calculate the equilibrium concentration    of SO3 and the number of moles SO3 present if the Keq = 1.47  x  10-3.

[SO3] = 82.5 M                       165 moles SO3

7.         If  at equilibrium [H2] = 0.200M and [I2] = 0.200M and Keq=55.6 at 250oC, calculate the equilibrium concentration of HI.

H2 (g) + I2 (g) 2HI (g)

[HI] = 1.49 M

8.         1.60 moles CO, 1.60 moles H2O, 4.00 moles CO2, 4.00moles H2 are found in a 8.00L container at 690oC at equilibrium.

CO (g) + H2O (g) CO2 (g) + H2 (g)

Calculate the value of the equilibrium constant.

Keq = 6.25

Worksheet #9            Equilibrium Calculations

Solve each problem and show all of your work.

1.         At equilibrium, a 5.0L flask contains:

0.75 mol of PCl5       0.50 mol of H2O 7.50 mol of HCl          5.00 mol of POCl3

Calculate the Keq for the reaction:

PCl5 (s) + H2O (g) 2HCl (g) + POCl3 (g)

Keq = 23

2.         Keq= 798 for the reaction:

2SO2 (g) +  O2 (g)   2SO3 (g).

In a particular mixture at equilibrium, [SO2]= 4.20 M and [SO3]=11.0 M. Calculate the equilibrium [O2] in this mixture.

[O2] = 8.60 X 10-3 M

3.         Consider the following equilibrium:

2SO2 (g) + O2 (g) 2SO3 (g)

When a 0.600 moles of SO2 and 0.600 moles of O2 are placed into a 1.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.250M. Calculate the Keq value.

Keq =1.07

4.         Consider the following equilibrium:

2 NO2(g)                   N2O4(g)

If 2.00 moles of NO2 are placed in a 1.00 L flask and allowed to react. At equilibrium 1.80 moles NO2 are present. Calculate the Keq.

2 NO2(g)                   N2O4(g)

I           2.00                             0.00 C       -0.20                              0.10

E         1.80 M                        0.10 M

Note the loss of one sig fig

Keq     =          (0.10)

(1.80)2

Keq = 0.031

5.         2 SO2(g)           +          O2(g)                        2 SO3(g)

4.00 moles of SO2 and 5.00 moles O2 are placed in a 2.00 L container at 200oC and allowed to reach equilibrium. If the equilibrium         concentration of O2 is 2.00 M, calculate the Keq

Keq = 0.50

6.         If the initial [H2] = 0.200M, [I2] = 0.200M and Keq = 55.6 at 250oC calculate the equilibrium concentrations of all molecules.

H2 (g) + I2 (g) 2HI (g)

[HI] = 0.315 M                       [H2] = [I2] = 0.042 M

7.         1.60 moles CO and1.60 moles H2O are placed in a 2.00L container at 690 oC (Keq=10.0).

Calculate all equilibrium concentrations.

CO (g) + H2O (g) CO2 (g) + H2 (g)

[CO2] = [H2] = 0.608 M                     [CO] = [H2O] = 0.192 M

8.         SO3(g) +   NO(g)                     NO2(g)  +   SO2(g)

Keq = 0.800 at 100oC.     If 4.00 moles of each reactant are placed in a 2.00L container, calculate all equilibrium concentrations at 100oC.

[NO2] = [SO2] = 0.944 M                  [SO3] = [NO] = 1.06 M

9.         Consider the following equilibrium system:               2NO2(g)        N2O4

Two sets of equilibrium data are listed for the same temperature.

Container 1                  2.00 L              0.12 moles NO2           0.16 moles N2O4            0.060 M NO2              .080 M N2O4

Container 2                 5.00 L              0.26 moles NO2           ? moles N2O4               0.052 M NO2

Determine the number of moles N2O4 in the second container. Get a Keq from the first             container and use it for the second container.

Keq                 =          [N2O4]

[NO2]2

=          [0.080]                        =          22.22

[0.060]2

Keq                 =          [N2O4]

[NO2]2

22.22               =          [N2O4]

[0.052]2

[N2O4]            =          0.0600 M        5.00 L     x    0.0600 moles     =          0.30 moles

1 L

Worksheet #10          Equilibrium Calculations

Solve each problem and show all of your work in your portfolio.

1. At equilibrium, a 2.0 L flask contains:

0.200 mol of PCl5        0.30   mol of H2O        0.60   mol of HCl        0.300 mol of POCl3

Calculate the Keq for the reaction:

PCl5  (g) + H2O  (g) 2HCl  (g) + POCl3  (g)

Keq = 0.90

2. Keq= 798 for the reaction:

2SO2  (g) +  O2  (g)   2SO3  (g).

In a particular mixture at equilibrium, [SO2]= 4.20 M and [SO3]= 11.0M. Calculate the equilibrium [O2] in this mixture.

[O2] = 8.60 X 10-3 M

3. Consider the following equilibrium: 2SO2  (g) + O2  (g) 2SO3  (g)

When a 0.600 moles of SO2 and 0.600 moles of O2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO3] is to be 0.250M. Calculate the Keq value.

(3 marks)

2SO2 (g)           +          O2  (g)                                  2SO3  (g)

I           0.300                           0.300                                       0

C         0.250                           0.125                                       0.250 E         0.050                           0.175                                       0.250

Keq  =  (0.250)2

(0.050)2(0.175)

Note the loss of one sig fig!

Keq = 1.4 x 102

4. H2  (g) + S  (s) H2S  (g)     Keq= 14

0.60 moles of H2 and 1.4 moles of S are placed into a 2.0L flask and allowed to reach equilibrium. Calculate the [H2] at equilibrium. (4 marks)

Don’t count S. It is a solid!

[H2] = 0.02 M

5. Keq=0.0183 for the reaction:

2HI  (g) H2 (g) + I2  (g)

If 3.0 moles of HI are placed in a 5.00L vessel and allowed to reach equilibrium, what is the equilibrium concentration of H2?

[H2] = 0.064 M

6. Consider the equilibrium:

I2 (g) + Cl2  (g)   2ICl (g)    Keq= 10.0

The same number of moles of I2 and Cl2 are placed in a 1.0L flask and allowed to reach equilibrium. If the equilibrium concentration of ICl is 0.040 M, calculate the initial number of moles of I2 and Cl2.

I2 (g)      +          Cl2  (g)        2ICl  (g)            Keq = 10.0

I           x                      x                      0

C         0.020               0.020               0.040

E         x – 0.020         x - 0.020          0.040

(0.040)2          =             10.0

(x – 0.020)2

.04                  =             3.1622

(x – 0.020)

.04       =          -0.063244        +          3.1622x

0.103244                     =          3.1622x

x          =          0.033 M

1.0 L   x   0.033  mole   =   0.033 mole

L

7. Consider the equilibrium:   2ICl(g)    I2 (g) + Cl2  (g)   Keq= 10.0

If x moles of ICl were placed in a 5.0 L container at 10 oC and if an equilibrium concentration of I2 was found to be 0.60 M, calculate the number of moles ICl initially present.

2ICl(g)                          I2 (g)      +          Cl2  (g)              Keq= 10.0

I           x                                  0                      0

C         1.2                               0.60                 0.60

E         x – 1.2                         0.60                 0.60

(0.60)2      =     10.0

(x – 1.2)2

0.60          =     3.162

(x – 1.2)

0.60   =  3.162x  -  3.7944

4.3944  =  3.162x

x  = 1.3896 M

5.0 L  x  1.3896 moles  =  6.9 moles

L

8. A student places 2.00 moles SO3 in a 1.00 L flask. At equilibrium [O2] = 0.10 M at 130 oC. Calculate the Keq.

2SO2(g)                    +          O2(g)               2SO3(g)

I           0                      0                      2.0                   Note this reaction starts with a product and shifts left to go to equilibrium.

C     +.20                 +.10                  -  0.20                 So add on the left and subtract on the right.

E         .20                   .10                   1.8

Keq   =   (1.8)2    =  810

(0.1)(.2)2

Worksheet #11                      Review, Ktrial, & Size of Keq

1.         2 CrO4-2 (aq) + 2H+ (aq) Cr2O7-2 (aq) + H2O (l)

Calculate the Keq if the following amounts were found at equilibrium in a 2.0L volume.

CrO4-2 = .030 mol, H+ = .020 mol, Cr2O7-2 = 0.32 mol, H2O = 110 mol

Do not count water. It is a liquid!!

Keq  =  (0.16)

(0.015)2(0.010)2

Keq = 7.1 X 106

2.         PCl5 (s) + H2O (g) 2HCl (g) + POCl3 (g)  Keq= 11

At equilibrium the 4.0L flask contains the indicated amounts of the three chemicals.

PCl5        .012 mol          H2O     .016 mol          HCl     .120 mol

Calculate [POCl3].

Keq     =    [HCl]2[POCl3]

[H2O]

11        =    [0.030]2[POCl3]

[.0040]

[POCl3] =    (11)(0.0040)

[0.030]2

[POCl3] = 49

3.         6.0 moles H2S are placed in a 2.0L container. At equilibrium 5.0 moles H2 are present. Calculate the Keq

2H2S (g)                  2H2 (g)              +          S2 (g)

I           3.0                               0                                  0

C         2.5                               2.5                               1.25 E         0.5                               2.5                               1.25

Note the loss of 1 significant digit

Keq  =  (2.5)2(1.25)

(0.5)2

Keq = 3  x  101

4.         4.0 moles H2 and 2.0 moles Br2 are placed in a 1.0L container at 180oC. If the [HBr] = 3.0 M at equilibrium, calculate the Keq.

H2 (g)    +          Br2 (g)        2HBr (g)

I           4.0                   2.0                   0

C         -1.5                  -1.5                  +3.0

E         2.5                   0.5                   3.0 Keq =  (3.0)2                                      Note the loss of significant digits here

(2.5)(.5)

Keq = 7

5.         At 2000C Keq= 11.6 for 2NO(g) N2 (g) + O2 (g).    If some NO is placed in a 2.0 L vessel and the equilibrium [N2] = 0.120 M, calculate all other equilibrium concentrations

2NO(g)                    N2 (g)    +          O2 (g)

I           x                                  0                      0

C         0.240                           0.120               0.120

E         x – 0.240                     0.120               0.120

(0.120)2                       =          11.6

(x – 0.240)2

0.120               =          3.4058

x – 0.240

x  =  0.275 M

[N2] = [O2] = 0.120 M                        [NO] = 0.035 M

6.         At 800oC, Keq = 0.279  for CO2 (g) + H2 (g) CO (g) + H2O (g).

If 2.00 moles CO( g) and 2.00 moles H2O (g) are placed in a 500 ml container, calculate all equilibrium concentrations.

Note that when two products are placed in a container it shifts to the left to get to equilibrium.

CO2 (g)                         +          H2 (g)           CO (g) +          H2O (g).

I           0                                  0                      4.00                 4.00

C         x                                  x                      x                      x

E         x                                  x                      4.00 - x            4.00 - x

0.279   =  (4-x)2

(x)2

0.5282  =  4 - x

x

0.5282x  =  4 – x

1.5282x  =  4

[CO2]  =  [H2]  = x = 2.62 M

[CO] = [H2O] = 4.00  -  x  = 1.38M

7.         CO (g) + H2O (g) CO2 (g) + H2 (g)       Keq= 10.0 at 690oC.      If at a certain time [CO] = 0.80M, [H2O] = 0.050M, [CO2] = 0.50M and [H2] = 0.40M, is the reaction at equilibrium? If not, how will it shift in order to get to equilibrium

Ktrial  = 5       Keq = 10         -therefore the reaction is not at equilibrium and shifts right

8.         For the reaction: CO (g) + H2O (g) CO2 (g) + H2 (g)    Keq= 10.0 at 690oC.  The following concentrations were observed: [CO] =2.0M, [H2] = 1.0M, [CO2]=2.0M, [H2O] = 0.10M. Is the reaction at equilibrium? If not, how will it shift in order to get to equilibrium?

Ktrial = 10      Keq = 10         - therefore the reaction is at equilibrium

9.         For the same equation above the following concentrations were observed: [CO] = 1.5M, [H2] = 1.2, [CO2] = 1.0M, [H2O] = .10M. Is the reaction at equilibrium? If not, how will it shift in order to get to equilibrium?

Ktrial  = 8       Keq = 10         -therefore the reaction is not at equilibrium and shifts right

10.       At a certain temperature the Keq for a reaction is 75.  2O3(g)     3O2(g)

Predict the direction in which the equilibrium will proceed, if any, when the following amounts are introduced to a 10 L vessel.

a)  0.60 mole of O3 and 3.0 mol of O2

Ktrial = (0.30)3           =          7.5       <          Keq     Therefore the reaction will shift to the right to reach equilibrium.

(0.060)2

b) 0.050 mole of O3 and 7.0 mol of O2

Ktrial =   (0.70)3         =          13720  >   Keq            Therefore the reaction will shift to the left to reach equilibrium.

(0.0050)2

) 1.5 mole of O3 and no O2

Ktrial =           (0)3     =          0          <          Keq     Therefore the reaction will shift to the right to reach equilibrium.

(0.15)2

11. Consider the following equilibrium:

a) 2NO2 (g)      N2O4 (g)                                                           Keq = 2.2

b) Cu2+(aq)   +   2Ag(s)  Cu(s)    +   2Ag+ (aq)                             Keq = 1 x 10-15                    Favors reactants to the greatest extent

c) Pb2+ (aq)    +   2  Cl- (aq)        PbCl2(s)                                 Keq = 6.3 x  104                     Favors products to the greatest extent

d) SO2(g)    +   O2 (g)      SO3 (g)                                             Keq = 110

i)         Which equilibrium favors products to the greatest extent?

ii)        Which equilibrium favors reactants to the greatest extent?

12.       What is the only way to change the value of the Keq?

The only way to change the value of the Keq is by changing the temperature.

13.       In the reaction: A + B C + D + 100kJ, what happens to the value of Keq if we increase the temperature?

The Keq will decrease.

14.       If the value of Keq decreases when we decrease the temperature, is the reaction exothermic or endothermic?

The reaction is endothermic.

15.       In the reaction; W + X + 100kJ Y + Z, what happens to the value of Keq if we increase the [X]? Explain your answer.

The Keq will remain the same because the only way to change Keq is by changing the temperature.

16.       If the value of Keq increases when we decrease the temperature, is the reaction exothermic or endothermic?

The reaction is exothermic.

17.       Predict whether reactants of products are favored in the following equilibrium systems

(a)       CH3COOH(aq) H+(aq) + CH3COO-(aq)                                                            Keq = 1.8  x 10-5                    Reactants

(b)       H2O2(aq)       H+(aq) + HO2(aq)                                                Keq = 2.6  x 10-12                    Reactants

(c)       CuSO4(aq) (+ Zn(s)   Cu(s) + ZnSO4(aq)                                    Keq = 1037                               Products

18.)      What effect will each of the following have on the Keq of the reaction shown below:

2NO2(g)                   +          heat            N2O4(g)                                                            Keq = 2.2

(a)       adding a catalyst                                              Remains constant

(b)       increasing the concentration of a reactant        Remains constant

(c)       increasing the concentration of a product        Remains constant

(d)       decreasing the volume                                     Remains constant

(e)       decreasing the pressure                                               Remains constant

(f)        increasing the temperature                               Increases

(g)       decreasing the temperature                              Decreases

Worksheet #12          Enthalpy & Entropy

For each of these processes, predict if Entropy increases or decreases.

1. 2H2(g)   +    O2(g)         2H2O(g)                              decreases

2. 2SO3(g)           2SO2(g)     +    O2(g) increases

3.  Ag+(aq)      +      Cl-(aq)        AgCl(s)           decreases

4.   Cl2(g)        2Cl(g)  increases

5.   H2O(l)        H2O(g)            increases

6.   CaCO3(s)     +    180 KJ      CaO(s)     +    CO2(g) increases

7.    I2(s)     +    608 KJ        I2(aq)                   increases

8.  4Fe(s)    +    3O2(g)       2Fe2O3(s)    +    1570 KJ   decreases

Consider both Enthalpy and Entropy and determine if each reaction will

a) go to completion

b) not occur or

c) go to equilibrium

9.         H2O(l)        H2O(g)         DH = 150 KJ

min enthalpy max entropy

Equilibrium

10.   CaCO3(s)     +    180 KJ      CaO(s)     +    CO2(g)

min enthalpy max entropy

Equilibrium

11.      I2(s)          I2(aq)      +    608 KJ

max entropy      min enthalpy

Completion

12.   4Fe(s)    +    3O2(g)       2Fe2O3(s)    ΔH  =    1570 KJ

max entropy     min enthalpy

Does not Occur

13.     Cl2(g)        2Cl(g)              DH = +26.8 KJ

min enthalpy   ⇄  max entropy

Equilibrium

14.   Ag+(aq) +   Cl-(aq)     AgCl(s)   +    86.2 KJ

min entropy min enthalpy

Equilibrium

Considerboth Enthalpy and Entropy and determine if each reaction will

a) have a large Keq

b) have a small Keq

c) have a Keq about equal to 1

15.H2SO4(aq)    +   Zn(s)       ZnSO4(aq)    + H2(g)    DH +207 KJ

16.NH4NO3(s)      NH4+(aq)     +   NO3-(aq)      DH  =  -30 KJ Large Keq

17.N2(g)   +   3H2(g)   +  92 KJ      2NH3(g) Small Keq

18.  H2O(l)  + 150 KJ        H2O(g) 19.Ca(s)     +    H2O(l)      Ca(OH)2(aq)     +    H2(g)      DH  =    +210 KJ 