Equilibrium Worksheet Questions

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

11. Review Web Review Practice Test 1

12. Review Practice Test 2 Quizmebc

Read Hebdon Unit 2

Worksheet #1 Approaching Equilibrium

Read unit II your textbook. Answer all of the questions. Do not start the questions until you have completed the reading. Be prepared to discuss your answers next period.

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 N₂O_4.

N₂O_{4 (g) <----->} 2 N0_{2 (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 N₂O_{4 (g)} is placed into a 1.0L glass container of room temperature. The reaction:

N₂O_{4 (g)} ⇋ 2 N0_{2 (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.

N_{2 (g)} + 3H_{2 (g)} ⇋ 2NH_3(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

N₂O_{4 (g)} ⇋ 2NO_2(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) + O_{2 (g)} ⇋ 2H₂O_(g) + 2Cl_{2 (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

CaCO_{3 (s)} + 170 KJ ⇋ CaO _(s) + CO_{2 (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 NH_{3 (g)} + 5 O_{2 (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 SO_{2 (g)} + O_{2 (g)} ⇋ 2 SO_{3 (g)} + energy decrease temperature right

2. C _(s) + CO_{2 (g)} + energy ⇋ 2 CO _(g) increase temperature right

3. N₂O_{4 (g)} ⇋ 2 NO_{2 (g)} increase total pressure        left

4. CO _(g) + H₂O _(g) ⇋ CO_{2 (g)} + H_{2 (g)} decrease total pressure nochange

5. 2 NOBr _(g) ⇋ 2 NO _(g) + Br_{2 (g)} decrease total pressure right

6. 3 Fe _(s) + 4 H₂O _(g) ⇋ Fe₃O_{4 (s)} + 4 H_{2 (g)} add Fe_(s) nochanges

7. 2SO_{2 (g)} + O_{2 (g)} ⇋ 2 SO_{3 (g)} add catalyst nochanges

8. CaCO_{3 (s)} ⇋ CaO _(s) + CO_{2 (g)} remove CO_{2 (g)} right

9. N_{2 (g)} + 3 H_{2 (g)} ⇋ 2 NH_{3 (g)} increase [H_{e (g)}] no change

Consider the following equilibrium system:

3 H_{2 (g)} + N_{2 (g)} ⇋ 2 NH_{3 (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

H_{2 (g)} + I_{2 (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) <------> Fe₃O_{4 (s)} + 4 H_{2 (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) + Br_{2 (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) + O_{2 (g)} <------> 2CO_{2 (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) + O_{2 (g)} ⇋ 2CO_{2 (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) + O_{2 (g)} ⇋ 2CO_{2 (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 [CrO₄^2-] left

Increase in temperature right Orange

Decrease intemperature left Yellow

Add HCl _(aq) right Orange

Add NaOH 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

Add NaOH left Violet

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

AddKCl _(aq) no change nochange

AddHCl _(aq) right Light Blue

3. Consider the equilibriumthat follows:

4 HCl _(g)+ 2 O_{2 (g)} ⇋ 2 H₂O _(l) + 2 Cl_{2 (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

Add NaCl right less green

Decrease temperature left green

Add NaOH _(aq) left clear

(reacts with HCl)

(check your solubility table for a possible reaction)

Add CuCl_(s) no change no change

Add AgNO_{3 (aq)} left green

(check your solubility table for a possible reaction)

Add CuNO_{3 (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.

I_2(g) + Cl_2(g) ⇋ 2 ICl _(g) + energy

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

· adding I_2(g)

· increasing the temperature

· decreasing the pressure

·

Increasing the temperature

removing Cl_2(g)

Worksheet #7 Maximizing Yield

1. N₂O_4(g) + 59 KJ ⇋ 2 NO_2(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. 2SO_3(g) ⇋ 2SO_2(g) + O_2(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: 3H_2(g) + N_2(g) ⇌ 2NH_3(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. SO_3(g) + H₂O_(g) ⇄ H₂SO_4(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 100^oC, 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 25^oC 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 25^oC.

[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 100^oC. If the Keq = 680, calculate the SO₃ concentration at 100^oC.

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 NO_2(g) ⇄ N₂O_4(g)

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

Keq = 1.60

6. 2 SO_3(g) ⇄ 2 SO_2(g) + O_2(g)

4.00 moles of SO₂ and 5.00 moles O₂ are present in a 2.00 L container at 100^oC 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 250^oC, 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 690^oC 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:

PCl_{5 (s)} + H₂O _(g) ⇄ 2HCl _(g) + POCl_{3 (g)}

Keq = 23

2. Keq= 798 for the reaction:

2SO_{2 (g)} + O_{2 (g)} ⇄ 2SO_{3 (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 NO_2(g) ⇄ N₂O_4(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 K_eq.

2 NO_2(g) ⇄ N₂O_4(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 SO_2(g) + O_2(g) ⇄ 2 SO_3(g)

4.00 moles of SO₂ and 5.00 moles O₂ are placed in a 2.00 L container at 200^oC 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 250^oC 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. SO_3(g) + NO_(g) ⇄ NO_2(g) + SO_2(g)

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

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

9. Consider the following equilibrium system: 2NO_2(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:

PCl_{5 (g)} + H₂O _(g) ⇄ 2HCl _(g) + POCl_{3 (g)}

Keq = 0.90

2. Keq= 798 for the reaction:

2SO_{2 (g)} + O_{2 (g)} ⇄ 2SO_{3 (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: 2SO_{2 (g)} + O_{2 (g)} ⇄ 2SO_{3 (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)

2SO_{2 (g)} + O_{2 (g)} ⇄ 2SO_{3 (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. H_{2 (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) ⇄ H_{2 (g)} + I_{2 (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:

I_{2 (g)} + Cl_{2 (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.

I_{2 (g)} + Cl_{2 (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) ⇄ I_{2 (g)} + Cl_{2 (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) ⇄ I_{2 (g)} + Cl_{2 (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 K_eq.

2SO_2(g) + O_2(g) ⇄ 2SO_3(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. PCl_{5 (s)} + H₂O _(g) ⇄ 2HCl _(g) + POCl_{3 (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) ⇄ 2H_{2 (g)} + S_{2 (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 180^oC. If the [HBr] = 3.0 M at equilibrium, calculate the Keq.

H_{2 (g)} + Br_{2 (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) ⇄ N_{2 (g)} + O_{2 (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) ⇄ N_{2 (g)} + O_{2 (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 800^oC, Keq = 0.279 for CO_{2 (g)} + H_{2 (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.

CO_{2 (g)} + H_{2 (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) ⇄ CO_{2 (g)} + H_{2 (g)} Keq= 10.0 at 690^oC. 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) ⇄ CO_{2 (g)} + H_{2 (g)} Keq= 10.0 at 690^oC. 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. 2O_3(g) ⇄ 3O_2(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) 2NO_{2 (g)} ⇄ N₂O_{4 (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) ⇄ PbCl_2(s) Keq = 6.3 x 10⁴ Favors products to the greatest extent

d) SO_2(g) + O_{2 (g)} ⇄ SO_{3 (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) H₂O_2(aq) ⇋ H⁺_(aq) + HO_2(aq) Keq = 2.6 x 10^-12 Reactants

(c) CuSO_4(aq) (+ Zn_(s) ⇋ Cu_(s) + ZnSO_4(aq) Keq = 10³⁷ Products

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

2NO_2(g) + heat ⇋ N₂O_4(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. 2H_2(g) + O_2(g) ⇋ 2H₂O_(g) decreases

2. 2SO_3(g) ⇋ 2SO_2(g) + O_2(g) increases

3. Ag⁺_(aq) + Cl^-_(aq) ⇋ AgCl_(s) decreases

4. Cl_2(g) ⇋ 2Cl_(g) increases

5. H₂O_(l) ⇋ H₂O_(g) increases

6. CaCO_3(s) + 180 KJ ⇋ CaO_(s) + CO_2(g) increases

7. I_2(s) + 608 KJ ⇋ I_2(aq) increases

8. 4Fe_(s) + 3O_2(g) ⇋ 2Fe₂O_3(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. CaCO_3(s) + 180 KJ ⇄ CaO_(s) + CO_2(g)

                        min enthalpy ⇄ max entropy

Equilibrium

11. I_2(s) ⇄ I_2(aq) + 608 KJ

⇄ max entropy      min enthalpy

Completion

12. 4Fe_(s) + 3O_2(g) ⇄ 2Fe₂O_3(s) ΔH = 1570 KJ

max entropy     min enthalpy ⇄

Does not Occur

13. Cl_2(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.H₂SO_4(aq) + Zn_(s) ⇋ ZnSO_4(aq) + H_2(g) DH +207 KJ

Keq about 1

16.NH₄NO_3(s) ⇋ NH₄⁺_(aq) + NO₃^-_(aq) DH = -30 KJ

Large Keq

17.N_2(g) + 3H_2(g) + 92 KJ ⇋ 2NH_3(g)

Small Keq

18. H₂O_(l) + 150 KJ ⇋ H₂O_(g)

Keq about equal to 1

19.Ca_(s) + H₂O_(l) ⇋ Ca(OH)_2(aq) + H_2(g) DH = +210 KJ

Keq about 1