Equilibrium Worksheet Questions
Power Point Lesson Notes- double click on the lesson number.
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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
7. Equilibrium Constants WS 8 Q4
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
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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)
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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
removal increases decreases decreases 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
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
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 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
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 AgNO3 (aq) left green
(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.
· adding I2(g)
· increasing the temperature
· decreasing the pressure
·
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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.
[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 =
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
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
Keq about 1
16.NH4NO3(s) ⇋ NH4+(aq) + NO3-(aq) DH = -30 KJ
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Large Keq
17.N2(g) + 3H2(g) + 92 KJ ⇋ 2NH3(g)
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Small Keq
18. H2O(l) + 150 KJ ⇋ H2O(g)
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Keq about equal to 1
19.Ca(s) + H2O(l) ⇋ Ca(OH)2(aq) + H2(g) DH = +210 KJ
Keq about 1