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598 Unit 8 Solutions Manual Copyright © 2007 Thomson Nelson

CHAPTER 15 Equilibrium Systems

Starting Points

(Page 674)

(Students’ answers will reflect their understanding at this point. There are no ‘correct’ or‘incorrect’ answers. Students will revisit their answers at the end of the chapter.)

Exploration: Shakin’ the Blues

(Page 675)(a) The reaction in the flask is reversible; reactants change to products which then change back to

reactants.(b) The solution turns blue after being shaken and then returns colourless, after being allowed to

stand for a few moments. (This cycle can be repeated.)(c) The colour changes will continue forever.(d) The blue colourless cycle continues even after 10 cycles. But, after a considerable time, the solution

turns yellow and the colour changes stop. The prediction is falsified after a long time period.

15.1 EXPLAINING EQUILIBRIUM SYSTEMS

Investigation 15.1: The Extent of a Chemical Reaction

(Pages 676, 700)

PurposeThe purpose of this investigation is to test the validity of the assumption that chemical reactionsare quantitative.

ProblemWhat are the limiting and excess reagents in the chemical reaction of selected quantities ofaqueous sodium sulfate and aqueous calcium chloride?

PredictionAccording to the method of stoichiometry, which assumes that reactions are quantitative, ifdifferent stoichiometric reacting amounts of reagents are used, the reactant in lesser amount is thelimiting reagent and the reactant in the greater amount is the excess reagent. The particularreaction of this investigation is assumed to be quantitative. The mole ratio is 1:1, and the limitingand excess reagents are as shown, for the sample volumes chosen. All of the sodium sulfateshould react, and there should be no sulfate ions in the filtrate. Calcium chloride is in excess, sothere should be calcium ions in the filtrate.

CaCl 2(aq) + Na 2SO 4(aq) CaSO 4(s) + 2 NaCl(aq)10 mL 5 mL

1.0 mol/L 1.0 mol/L0.0010 mol 0.0005 molexcess limitingreagent reagent

Procedure1. Measure 10 mL of the excess reagent solution (CaCl 2(aq) in this example) in a graduated

cylinder.2. Pour the solution into a clean 50 mL or 100 mL beaker.

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Copyright © 2007 Thomson Nelson Unit 8 Solutions Manual 599

3. Measure 5 mL of the limiting reagent solution (Na 2SO 4(aq) in this example) in a cleangraduated cylinder.

4. Slowly add the excess reagent solution to the limiting reagent solution, while stirring.5. Allow the precipitate to settle, and filter the precipitate from the mixture.6. Collect about 5 mL of the filtrate into a small clean test tube.7. Test the filtrate by adding a few drops of Na 2CO 3(aq).8. Collect another 5 mL of the filtrate from step 5 into another small clean test tube and test with

a few drops of Ba(NO 3)2(aq).9. Repeat the experiment if time and supplies permit, changing the solutions used in excess and

limiting quantities.10. Dispose of solutions into the sink and solids into the solid waste container.

Evidence A white precipitate formed upon mixing the solutions of CaCl 2(aq) and Na 2SO 4(aq). White precipitates formed in both diagnostic tests of the filtrate. The same evidence was obtained when the experiment was repeated, this time using the

Na 2SO 4(aq) in excess.

Analysis

According to the evidence obtained, neither reactant appears to be limiting. Both reactants appearto be in excess. Positive tests are obtained for the presence of both calcium and sulfate ions in thefinal (filtrate) solution.

EvaluationThe experimental Design combining precipitation with diagnostic tests for excess ions isadequate, since clear evidence was obtained to answer the question. Because the experimentrequired little time, it was repeated. To make the testing complete, at least one diagnostic test forone ion from each reactant is performed. The Materials were adequate for performing theinvestigation—simple solutions of simple reagents are all that is necessary for both the reactionand the diagnostic tests. The Procedure appears to be adequate to answer the problem. Performingadditional trials would be an improvement; however, the results obtained agreed with those of therest of the class. The technological skills required to perform the test were adequate as no practicewas needed to obtain reliable results. Overall, I am very confident of the experimental results. Insome cases, the precipitate was a little slow in forming. However, it was never uncertain whethera precipitate formed. Assuming that the solutions were not contaminated, there are no othersignificant sources of uncertainty or error.

The prediction is falsified, since the evidence indicates the presence of both reactantsafter the reaction was completed. The assumption of a quantitative reaction is judged to beunacceptable for this reaction.

The purpose of this investigation was accomplished, but additional reactions need to bestudied to see if the reaction studied is an exception or if most reactions are not quantitative.

Practice

(Page 677)1. (a) Na 2SO 4(aq) + CaCl 2(aq) CaSO 4(s) + 2 NaCl(aq)(b) When solutions containing equal chemical amounts (25 mmol each) of sodium sulfate

and calcium chloride were mixed, a white precipitate of calcium sulfate was produced, asexpected. Diagnostic tests, however, indicated the presence of both unreacted sodiumsulfate and unreacted calcium chloride in the final filtrate.

(c) Ca 2+(aq) + SO 42– (aq) CaSO 4(s)

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(d) When equal amounts of aqueous calcium ions and aqueous sulfate ions were mixed, awhite precipitate of calcium sulfate was formed. Diagnostic tests, however, indicated the

presence of both aqueous calcium ions and aqueous sulfate ions in the filtrate.(e) According to collision-reaction theory, it is the collision of aqueous calcium ions with

aqueous sulfate ions that produces calcium sulfate precipitate. Aqueous sodium ions andaqueous chloride ions were not involved in the reaction. Since the statement from (d) isgiven in terms of the reacting ions, it is a more accurate description of the chemicalsystem.

2. Yes, other groups in the class collected similar evidence.

Web Activity: Simulation Equilibrium State

(Page 678)[No written response is required.]

Mini Investigation: Modelling Dynamic Equilibrium

(Page 678)(a) Both graphs have a similar shape. As the reactant amounts decrease, the product amounts

increase. Also, the rate of increase of the product decreases (slows down) over time. Thegraphs differ because more transfers are necessary to reach equilibrium with the smallerstraw.

(b) This would not affect the final volumes in the cylinders because volume is being plotted as afunction of the number of transfers, not as a function of time.

(c) Using an even larger straw would slightly decrease the final volume in the cylinder fromwhich the larger straw removed water, and increase the final volume in the cylinder to whichthe large straw delivered water.

(d) [Answers will vary depending on diameter of straws used.] When the liquid levels remainedconstant, the “reactant” cylinder held 5.2 mL (of the original 25.0 mL of water), and the“product” cylinder held 19.8 mL.

% yield =19.8 mL25.0 mL 100% 79.2%

(e) The volume of water in both cylinders will gradually become equal as the activity progresses .Therefore, the graph of volume for each cylinder will plateau (level off) at the same value.

Practice

(Page 682)3. (a) For a chemical system at equilibrium, some directly observable characteristics include

colour and volume of solid, liquid, or gas phases, size and shape of any solid particles,and odour of any vapours. All of these will appear to remain constant when the system isat equilibrium.

(b) At equilibrium, both the forward and reverse chemical processes are occurring, although

observation indicates that the reaction appears to have stopped.(c) The rates of the forward and reverse changes are assumed to be equal at equilibrium.4. (a) CH 4(g) + Cl 2(g) CH 3Cl(g) + HCl(g)

2.00 mol 10.00 molSince methane is the limiting reagent, the maximum possible yield of chloromethane

produced is:

3CH Cl

1 = 2.00 mol = 2.00 mol

1n

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(b)1.40 mol

percent reaction (yield) =2.00 mol

100% = 70.0%

Therefore, products are favoured.

5.

6. (Parts (a) and (b) need to be done together, or do part (b) first . )(a)

(b)[C 2H4(g)] [Br 2(g)] [C 2H4Br 2(g)]

Concentration (mol/L) (mol/L) (mol/L)Initial 4.0 2.50 0.0

Change –1.5 –1.5 +1.5Equilibrium 2.5 1.0 1.5

(c) From C 2H4(g) data, (since the volume of any gaseous substance in the container is alsonecessarily the container volume) the volume of the container is:

4.00 molV

/L4.00 mol

1.00 L

(d) (The maximum possible amount of dibromoethane is based on the stoichiometry and the fact that the bromine is the limiting reagent.)

2 4 2C H Br

1 = 2.50 mol = 2.50 mol

11.5 mol

percent yield =

n

2.50 mol 100% = 60%

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7. (a) + 2+Cu(s) + 2 Ag (aq) Cu (aq) + 2 Ag(s)

(b) Ca 2+ (aq) + SO 42 (aq)

50% CaSO 4 (s)

(c)50% + –

3 2 3 3CH COOH(aq) + H O(l) H O (aq) + CH COO (aq)

Lab Exercise 15.A: The Synthesis of an Equilibrium Law(Page 683)

PurposeThe purpose of this exercise is the synthesis of an equilibrium law.

ProblemWhat mathematical formula, using equilibrium concentrations of reactants and products, gives aconstant for the iron(III) thiocyanate reaction system?

AnalysisTest the following mathematical relationships for constancy:1. [Fe 3+(aq)][SCN – (aq)][FeSCN 2+(aq)]

2. [Fe3+

(aq)] + [SCN –

(aq)] + [FeSCN2-

(aq)]3. [FeSCN 2+(aq)]

[Fe 3+(aq)][SCN – (aq)]4. [Fe 3+(aq)]

[FeSCN 2- (aq)]5. [SCN – (aq)]

[FeSCN 2-(aq)]Let [Fe 3+(aq)] = a , [SCN – (aq)] = b, and [FeSCN 2+(aq)] = cPossible Relationships among [Fe 3+(aq)], [SCN – (aq)], and [FeSCN 2+(aq)]

a b c a + b + c c /(a b ) a/c b/c

Trial ( 10 -9 mol 3 /L3) ( 10 -2 mol/L) (mol/L) -1 ( 10 -2)

1 2.89 4.01 294 42.4 8.702 2.34 1.58 293 17.9 23.13 1.15 0.73 288 9.53 55.54 0.41 0.30 307 6.03 1525 0.34 0.27 296 5.51 190

According to the evidence and mathematical trial-and-error analysis, the thirdmathematical formula gives a relatively constant value within normal experimental uncertaintiesfor concentrations. The value, together with its uncertainty, can be expressed as 300 ± 7.

The relationship producing this constant value is shown below.2+

3+

[FeSCN (aq)] =

[Fe (aq)] [SCN (aq)] K

(Note that the reciprocal relationship would necessarily also give a constant value.)

Web Activity: Canadian Achievers—Paul Kebarle

(Page 683)1. Kebarle and his co-workers collected fundamental data on isolated molecules, such as proton

affinities, hydride ion affinities, and electron affinities, on isolated (gas phase) particles.These kinds of data, significantly expanded by other workers, now constitute a centraldatabase that is of fundamental importance in many diverse fields. More recently, Kebarlehas worked on electrospray mass spectrometry (ESMS), where electrolyte ions present in

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solution can be transferred to the gas phase and then subjected to mass spectrometric analysis.By means of this method, ESMS has become a prime analytical tool in biochemical and

biomedical laboratories of the chemical and pharmaceutical research laboratories.2. Kebarle’s work has aided research in protein molecular modelling, physical organic

chemistry, and many analytical methods, including biochemical mass spectrometry.

Lab Exercise 15.B: Determining an Equilibrium Constant(Page 686)

PurposeThe purpose of this investigation is to use the equilibrium law to determine the equilibriumconstant at 200 °C for the decomposition reaction of the molecular compound phosphorus

pentachloride.

ProblemWhat is the value of the equilibrium constant for the decomposition of phosphorus pentachloridegas to phosphorus trichloride gas and chlorine gas, at a temperature of 200 °C?

Analysis

PCl 5(g) PCl 3(g) + Cl 2(g)3 2

5

-4

[PCl (g)] [Cl (g)] =

[PCl (g)]

(0.014) (0.014)4.3 10

0.46

c K (Note that units can be ignored if all units are mol/L)

According to the evidence provided and the law of equilibrium, the (numerical) value ofthe equilibrium constant for the decomposition of phosphorus pentachloride gas to phosphorustrichloride gas and chlorine gas is 0.46 at 200 °C.

Web Activity: Simulation Writing Equilibrium Expressions

(Page 688)[No written response is required.]

Section 15.1 Questions

(Pages 688–689)

1. (a)2

2 2

[HCl(g)] =

[H (g)][Cl (g)]c K from H 2(g) + Cl 2(g) 2 HCl(g)

(b)

2

33

2 2

NH (g) =

H (g) N (g)c K from N 2(g) + 3 H 2(g) 2 NH 3(g)

(c)2

22

2 2

[H O(g)] =[H (g)] [O (g)]c

K from 2 H 2(g) + O 2(g) 2 H 2O(g)

(d)2

3 62+ 6

3

[Ni(NH ) (aq)] =

[Ni (aq)][NH (aq)]c K from Ni 2+(aq) + 6 NH 3(aq) Ni (NH 3)62+(aq)

(e) 2= CO (g)c K from CaCO 3(s) CaO(s) + CO 2(g)

(f) 2+ 24

1 =

[Ca (aq)][SO (aq)]c K from Ca 2+(aq) + SO 4

2-(aq) CaSO 4(s)

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(g) 22 3

[CO (g)] =

[H CO (aq)]c K from H 2CO 3(aq) H2O(l) + CO 2(g)

2. The addition reactions of halogens and alkenes are generally very fast, even at roomtemperature, and the colour of the bromine disappears completely. Therefore, for this reactionof bromine with ethylene, a >99% yield of products is predicted.

3. (a)2H 2.00 Ln 6.0 mol1 L

2I

12 mol

2.00 Ln4.0 mol

1 L8.0 mol

(b) H 2(g) + I 2(g) 2 HI(g)

[H2(g)] [I 2(g)] [HI(g)]Concentration (mol/L) (mol/L) (mol/L)

Initial 6.0 4.0 0.0Change –3.6 –3.6 +7.2Equilibrium 2.4 0.4 7.2

At equilibrium, HI 2.00 Ln 7.2 mol

1 L = 14 mol .

(c) Initially, hydrogen is reacting quickly as a result of numerous favourable collisions withiodine molecules. As the number of iodine and hydrogen molecules decreases, thefrequency of collisions producing product decreases. At the same time as the number of

product hydrogen iodide molecules increases, the rate at which they collide to break up back into hydrogen and iodine molecules will steadily increase. An observer will see therate of reaction of hydrogen and iodine gradually “slowing down.” Eventually, the rate atwhich hydrogen molecules react to form hydrogen iodide becomes equal to the rate atwhich hydrogen iodide molecules break apart to form hydrogen and iodine molecules. Anobserver will see the reaction as having “stopped,” because no macroscopic changes will

be observable.

4. (a) 3 H 2(g) + N 2(g)11%

2 NH 3(g)(b) C(s) + H 2O(g)

50% CO(g) + H 2(g)

(c) 2 Ag +(aq) + Cu(s) 2 Ag(s) + Cu 2+(aq)

(d) 2 SO 2(g) + O 2(g)65% 2 SO 3(g)

5.2

2 2

[HI(g)] =

[H (g)][I (g)]c K , from H 2(g) + I 2(g) 2 HI(g) t = 448 °C

System 1:

27.80

501.10 1.10c

K (Note that all units are the same, in mmol/L)

System 2: K c = 49; System 3: K c = 516. N 2(g) + 3 H 2(g) 2 NH 3(g)

23

32 2

[NH (g)] =

[N (g)] [H (g)]c K

[NH 3(g)]2 = K c [N 2(g)] [H 2(g)]

3

= (6.0 10 –2) (1.50) (0.50) 3

= 1.1 10 –2

[NH 3(g)]21.1 10 = 0.11 mol/L

(Units are ignored in K c calculation but required for final concentration answer.)

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7. (a) 2 HBr(g) H2(g) + Br 2(g)

(b) 2 22[H (g)] [Br (g)]

=[HBr(g)]

c K

(c) nHBr = 2.00 L 0.100 mol/L = 0.200 mol(d) nHBr reacted at equilibrium = 1.00 mol – 0.200 mol = 0.80 mol

(e) 2H1

produced at equilibrium = 0.80 mol = 0.40 mol2n

2Br

1 produced at equilibrium = 0.80 mol = 0.40 mol

2n

(f) [HBr(g)] = 0.100 mol/L

[H2(g)] =0.40 mol

= 0.20 mol/L2.00 L

[Br 2(g)] =0.40 mol

= 0.20 mol/L2.00 L

(g) 2(0.20 ) (0.20 )

= = 4.0(0.100 )c

K

8. H 2(g) + I 2(g) 2 HI(g)H2(g) I 2(g) HI(g)

Concentration (mmol/L) (mmol/L) (mmol/L)Initial 6.23 4.14 22.40Change -1.56 -1.56 +3.12Equilibrium 4.67 2.58 25.52

2 2

2 2

[HI(g)] (25.52 ) = = = 54.1

[H (g)] [I (g)] (4.67 ) (2.58 )c K

9. CO 2(g) + H 2(g) CO(g) + H 2O(g) (Assuming no CO 2 or H 2 added and a volume of 1.0 L)[CO 2(g)] [H 2(g)] [CO(g)] [H 2O(g)]

Concentration (mol/L) (mol/L) (mol/L) (mol/L)

Initial 0.00 0.00 0.20 0.25Change +0.10 +0.10 -0.10 -0.10Equilibrium 0.10 0.10 0.10 0.15

2

2 2

[CO(g)] [H O(g)] (0.10) (0.15) = = = 1.5

[CO (g)] [H (g)] (0.10) (0.10)c K

10. (a) 2 HBr(g) H2(g) + Br 2(g)

2 2

0.25 mol[H (g)] 0.50 mol/L [Br (aq)]

0.500 L

[HBr(g)] [H 2(g)] [Br 2(g)]Concentration (mol/L) (mol/L) (mol/L)

Initial 0.00 0.50 0.50Change +2x – x – xEquilibrium 2 x 0.50 – x 0.50 – x

2 22 2

[H (g)] [Br (g)] (0.50 - ) (0.50 - ) = = = 0.020

[HBr(g)] (2 )c x x

K x

2

2

(0.50 - ) = 0.020

(2 ) x

x

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0.500.020 0.14

2 x

x

(0.50 - x) = 0.28 x x = 0.39

[HBr(g)] [H 2(g)] [Br 2(g)]

Concentration (mol/L) (mol/L) (mol/L)Initial 0.00 0.50 0.50Change 0.78 -0.39 -0.39Equilibrium 0.78 0.11 0.11

At equilibrium, [H 2(g)] = [Br 2(g)] = 0.11 mol/L, and [HBr(g)] = 0.78 mol/L.

(b)2 2H Br

= = 0.500 Ln n0.11 mol

1 L 0.055 mol

HBr = 0.500 Ln0.78 mol

1 L = 0.39 mol

11. Atomic theory is necessary because equation reactions involve the collision of reactantsand products, which are made up of atoms, or ions, or combinations of atoms in

molecules. The kinetic molecular theory states that all matter is composed of tiny particles that are inconstant motion. Further, the extent of this motion is dependent upon the amount ofenergy possessed by the particles. Molecules with greater energy will move faster. Thismovement is essential for chemical reactions to occur.

According to collision reaction theory, collisions that take place between reactantmolecules must occur with enough energy and in the proper orientation for products toform. Only some percentage of reactant entity collisions called “effective”collisions will result in a change in the entities.

The position of equilibrium is dependent on reaction rates for the forward and reversereactions. Equilibrium is a state in which these rates have become equal.

Extension12. It is universally accepted that the concentration of carbon dioxide in the atmosphere is

increasing, and it is widely believed that this supports the process of global warming (byincreasing the “greenhouse effect”). Recent research suggests that ocean absorption of CO 2 ishaving significant acidifying effects. This could be a grave problem for marine life. Somescientists are considering ways to remove carbon dioxide from the atmosphere, to slow thegreenhouse effect. One suggestion is to “inject"” carbon dioxide deep into ocean waters— which would exacerbate the acidification of the oceans. Other significant effects on CO 2concentration include production by volcanoes and sequestration by marine organisms that

build calcium carbonate shells.

15.2 QUALITATIVE CHANGE IN EQUILIBRIUM SYSTEMS

Investigation 15.2: Equilibrium Shifts (Demonstration)

(Pages 691, 700–701)

PurposeThe purpose of this demonstration is to test Le Châtelier’s principle by studying two chemicalequilibrium systems: the equilibrium between two oxides of nitrogen, and the equilibrium ofcarbon dioxide gas and carbonic acid.

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