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Chemical Equilibrium
Section 17.1 Equilibrium: A State of Dynamic Balance
In your textbook, read about chemical equilibrium.
Complete each statement.
1. When a reaction results in almost complete conversion of reactants to products, chemists say the reaction goes to _____________________.
2. A reaction that can occur in both the forward and the reverse directions is called a(n) _____________________.
3. _____________________ is a state in which the forward and reverse reactions balance each other because they take place at equal rates.
4. At equilibrium, the concentrations of reactants and products are ___________________ but that does not mean that the amounts or concentrations are ___________________.
5. Equilibrium is a state of _____________________, not one of _____________________.
In your textbook, read about equilibrium expressions and constants.
For each statement below, write true or false.
__________________ 6. The law of chemical equilibrium states that at a given pressure, a chemical system may reach a state in which a particular ratio of reactant to product concentrations has a constant value.
__________________ 7. The equation H2(g) ( I2(g) (( 2HI(g) is an example of a homogeneous equilibrium.
__________________ 8. If an equilibrium constant has a value less than one, the reactants are favored at equilibrium.
__________________ 9. The value for Keq is constant only at a specific volume.
__________________ 10. If the equilibrium constant for a reaction at 300 K is 49.7, the concentration of the reactants will be greater than the concentration of the products.
__________________ 11. A heterogeneous equilibrium means that reactants and products are present in more than one state.
__________________ 12. The product of the forward chemical reaction is HI, for the equilibrium expression:
[pic]
Section 17.1 continued
In your textbook, read about determining equilibrium constants.
A chemist did two experiments to determine the equilibrium constant for the reaction of sulfur dioxide with oxygen to form sulfur trioxide. Use the table showing the results of the experiments to answer the following questions.
|2SO2(g) ( O2(g) ( 2SO3(g) at 873 K |
|Experiment 1 |Experiment 2 |
|Initial concentrations |Equilibrium concentrations |Initial concentrations |Equilibrium concentration |
|[SO2] ( 2.00M |[SO2] ( 1.50M |[SO2] ( 0.500M |[SO2] ( 0.590M |
|[O2] ( 1.50M |[O2] ( 1.26M |[O2] ( 0M |[O2] ( 0.0450M |
|[SO3] ( 3.00M |[SO3] ( 3.50M |[SO3] ( 0.350M |[SO3] ( 0.260M |
13. Write the equation to calculate the equilibrium constant for the reaction.
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14. Is this reaction an example of a homogeneous or heterogeneous equilibrium?
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15. Calculate the equilibrium constant from the data obtained in experiment 1.
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16. What is the equilibrium constant for the reaction in experiment 2?
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17. Was it necessary to calculate both equilibrium constants? Why or why not?
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18. What does this experiment show about the initial concentrations of products and
reactants in a reversible reaction?
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Section 17.2 Factors Affecting Chemical Equilibrium
In your textbook, read about Le Châtelier’s Principle.
Answer the following questions.
1. What does Le Châtelier’s Principle say?
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2. What are three kinds of stresses that can be placed on a system?
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For each reaction below, state the direction, left or right, in which the equilibrium will shift when the indicated substance is added. Identify one other way in which the reaction could be shifted in the same direction you indicated. (Hint: There may be more than one way to do this.)
3. Reaction: N2(g) ( 3H2(g) ( 2NH3(g); NH3 added
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4. Reaction: H2(g) ( I2(g) ( 2HI(g); H2 added
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5. Reaction: CO(g) ( H2O ( CO2(g) ( H2(g); H2O added
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6. Reaction: 2SO2(g) ( O2(g) ( 2SO3(g); SO3 added
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7. Reaction: 2SO2(g) ( O2(g) ( 2SO3(g); SO2 added
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8. Reaction: 2NCl3(g) ( N2(g) ( 3Cl2(g); NCl3 added
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Section 17.2 continued
In your textbook, read about factors affecting chemical equilibrium.
Use each of the terms below just once to complete the passage.
|right |exothermic |increase |stress |catalyst |energy |
|smallest |change |reverse |constant |forward | |
When you decrease the volume of a reaction vessel, you (9) ________________________ the pressure. This causes a reaction at equilibrium to shift to the side with the
(10) ________________________ number of moles. If the reaction has an equal number of moles of reactants and products, changing the volume of the reaction vessel causes no
(11) ________________________ in the equilibrium.
Changing the temperature of a reaction at equilibrium alters both the equilibrium
(12) ________________________ and the equilibrium position. When a reaction is
(13) ________________________, which means it releases energy, lowering the temperature shifts the equilibrium to the (14) ________________________ because the forward reaction liberates heat and removes the (15) ________________________.
A (16) ________________________ speeds up a reaction by lowering the
(17) ____________________ requirements for the reaction, but it does so equally in both the
(18) _____________________ and the (19) _____________________ directions. The reaction will reach equilibrium more quickly, but with no change in the amount of product formed.
For each reaction below, indicate in which direction the equilibrium shifts when the stated stress is applied to the system. Write R if the reaction shifts to the right, L if it shifts to the left, or NC if there is no change.
Reaction Stress
_________ 20. PCl5(g) ( PCl3(g) ( Cl2(g) ( heat
_________ 21. CO(g) ( Fe3O4(s) ( CO2(g) ( 3FeO(s)
_________ 22. C2H2(g) ( H2O(g) ( CH3CHO(g) ( heat
_________ 23. 2NO(g) ( H2(g) ( N2O(g) ( H2O(g) ( heat
_________ 24. Heat ( H2(g) ( I2(g) ( 2HI(g)
_________ 25. H2(g) ( Cl2(g) ( 2HCl(g) ( heat
temperature increase
volume increase
temperature decrease
volume decrease
temperature decrease
volume decrease
Section 17.3 Using Equilibrium Constants
In your textbook, read about calculating equilibrium concentrations.
Answer the following questions.
1. What can you use the equilibrium constant to do?
______________________________________________________________________________________________
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2. Given the reaction: N2 ( O2 ( 2NO for which the Keq at 2273 K is 1.2(10(4
a. Write the equilibrium constant expression for the reaction.
___________________________________________________________________________________________
b. Write the equation that would allow you solve for the concentration of NO.
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c. What is the concentration of NO if [N2] ( 0.166M and [O2] ( 0.145M?
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3. What is the solubility product constant?
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4. What is the solubility product constant expression for the reaction:
Mg3(PO4)2(s) ( 3Mg2((aq) ( 2PO43((aq)
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5. Given the equilibrium BaSO4(s) ( Ba2((aq) ( SO42((aq), what is the solubility product constant expression?
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6. The solubility product constant for BaSO4 at 298 K is 1.1(10(10. Calculate the
solubility of BaSO4 in mol/L at 298 K.
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Section 17.3 continued
In your textbook, read about predicting precipitates.
The solubility product constant can be used to determine if a precipitate will form when two aqueous solutions are mixed together. First, calculate the concentrations of the ions in the final solution. Use the solubility product constant expression to calculate the ion product (Qsp ) for the substance that might precipitate. Compare the result with the Ksp of the substance.
7. What can you say about a solution when
a. Qsp is greater than Ksp?
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b. Qsp is equal to Ksp?
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c. Qsp is less than Ksp?
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8. Predict whether a precipitate of AgBr will form if 100 mL of 0.0025M AgNO3 and
100 mL of 0.0020M NaBr are mixed.
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9. Explain briefly why Ag3PO4 might be more soluble in water than in the same volume of a solution containing Na3PO4.
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Study Guide - Chapter 17 – Chemical Equilibrium
Section 17.1 Equilibrium: A State of Dynamic Balance
1. completion
2. reversible reaction
3. Chemical equilibrium
4. constant, equal
5. action, inaction
6. false
7. true
8. true
9. false
10. false
11. true
12. true
13. Keq ’ [SO3]2/[SO2]2[O2]
14. homogeneous
15. Keq ’ 4.32
Solution: [3.50]2/[1.50]2[1.26] ’ 4.32098 ’ 4.32
16. Keq ’ 4.32
Solution: [0.260]2/[0.590]2[0.0450]
’ 4.31549 ’ 4.32
17. No; because both experiments were done at the
same temperature, the equilibrium constant would
be the same.
18. This experiment shows that it does not matter what the initial concentrations of products and reactants are in a reversible reaction. When the reaction is at equilibrium, they will have the same ratio to one another.
Section 17.2 Factors Affecting Chemical Equilibrium
1. If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.
2. Answers may vary, but should include changes in concentration, volume, and temperature. Some students may list pressure in place of volume.
3. left; remove N2 or H2
4. right; add I2 or remove HI
5. right; add CO or remove CO2 or H2
6. left; remove SO2 or O2
7. right; remove SO3
8. right; remove N2 or Cl2
9. increase
10. smallest
11. change
12. constant
13. exothermic
14. right
15. stress
16. catalyst
17. energy
18. forward
19. reverse
20. L
21. NC
22. R
23. R
24. L
25. NC
Section 17.3 Using Equilibrium Constants
1. The equilibrium constant is used to calculate the equilibrium concentration of a substance if the concentrations of all other reactants and products are known.
2. a. Keq ’ [NO]2/[N2][O2]
b. [NO]2 ’ Keq ( [N2][O2]
c. 1.7 10–3M
Solution: [NO2 ’ (1.2×10–4) ( (0.166M) (
(0.145M) ’ 2.8884(10–6M
Taking the square root gives 1.6995(10–3, which rounds to ’ 1.7(10–3M
3. The solubility product constant is an equilibrium constant for the dissolving of a sparingly soluble ionic compound in water.
4. Ksp ’ [Mg2+]3[PO43–]2
5. Ksp ’ [Ba2+][SO42–]
6. Ksp ’ [Ba2+][SO42–] ’ 1.1(10–10
[Ba2+][SO42] ’ (s)(s) ’ s2 ’ 1.1(10–10
s ’[pic] s2 ’ 1.1(10-5 mol/L
The solubility of BaSO4 is 1.0(10–5 mol/L
at 298 K.
7. a. This indicates that a precipitate will form. The precipitate will reduce the ion concentrations
in the solution until the Ksp expression is
satisfied.
b. This indicates that the solution is saturated so no further change will occur.
c. This indicates that the solution is unsaturated and no precipitate will form.
8. AgBr(s) ( Ag+(aq) + Br–(aq)
Qsp ’ [Ag+]/2 [Br–]/2
Qsp ’ (0.00125)(0.0010) ’ 1.3(10–6
Qsp (1.3(10–6) > Ksp (5.4(10–13
A precipitate will form.
9. Not as much Ag3PO4 would dissolve in a Na3PO4 solution because of the common ion effect. The substances have PO43– ions in common. Because of the high concentration of PO43– ions, only a small concentration of Ag+ can be present in the equilibrium mixture. The solubility of Ag3PO4 is equal to the Ag+ ion concentration.
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