Chemistry 123 Problem Set #1

CHEMISTRY 123 Problem Set #1 Phase Equilibria 120. Below is the phase diagram for phenol, which has two solid phases, SI and SII, both more dense than the liquid. The transition SI SII is exothermic.

For questions 16 and 1120 use: (1) vapour or gas (2) liquid (3) solid SI (4) solid SII 1. Which phase is stable in region A? 2. Which phase is stable in region D? 3. Which phases are stable at 61 atm and 64 C? 4. Which phases are in equilibrium at 2000 atm and 64 C? 5. Which phases are in equilibrium at 3 103 atm and 40 C? 6. Which phase is the most dense? 7. Which phase is more dense: B or C? (1) B (2) C 8. The normal boiling point is: (1) 40 C (2) 64 C (3) 420 C (4) none of these 9. The critical temperature is: (1) 40 C (2) 64 C (3) 420 C (4) none of these 10. The transition SI liquid is exothermic. (1) True (2) False 11. A tube is half filled with solid phenol, then the air is removed and the tube sealed. It is kept at a

constant temperature of 35 C, and the phenol comes to equilibrium. What phases are present in the tube at equilibrium?

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Identify the phase(s) present in each of the numbered regions on the following graphs. 1216. Isotherm of phenol at 50 C

1720. Isobaric heating curve for phenol at 1 atm

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21. A substance has the following properties: boiling point = 75 C and melting point = 15 C. Sketch a heating curve for the substance from 50 C to 100 C in the diagram shown below. Clearly label the phases present at each individual step (five steps).

Time [min]

T [K]

373

223

273

2230. For H2O, the normal boiling point is 100 C; the normal melting point is 0 C; the triple point is at 0.0098 C and 0.006 atm; the critical point is at 374 C and 220 atm. Only three phases are present at the temperatures and pressures shown on the graph.

T(C)

P(atm)

220

1.0

0.006

0.0

0.0098

100 374

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22. Draw the phase diagram for H2O using the graph above. 23. Label the regions S (solid), L (liquid), and V (vapor). 24. Circle and label the triple point (T), the critical point (C), the normal boiling point (bp), and the

normal melting point (mp). 25. At 70 C, a rigid container is first evacuated, then half filled with water, sealed, and allowed to reach

equilibrium. Which phases are present in the end? 26. One of the unique properties of water is the relative density of the solid and liquid phases. Which

phase of H2O is more dense? 27. On the phase diagram, identify the curve that indicates the relative densities of the liquid and solid

phases, and label it with a D. 28. Is the slope of line D positive or negative? Explain in terms of the effect of pressure at constant

temperature. 29. Benzene freezes at +5.5 C. Although you can easily skate on a frozen lake of water at 5 C, you

would not be able to skate very well on frozen benzene at 0 C (i.e., also about 5 C below its melting point). Explain why not by discussing the fact that the phase diagram for benzene is normal. (Ignore toxicity.)

30. If youre skiing at the top of Whistler and youre cooking spaghetti in boiling water, why does it take a bit longer than usual (i.e., compared to at home, at sea level) to cook? (Consider only the time after which the spaghetti is added to the boiling water, not the time needed to heat the water to a boil.)

3134. A phase diagram for an unknown substance is shown below. This substance has a vapour and liquid phase, and three solid phases (SI, SII, and SIII). At point A on the diagram, SIII, SII, and the liquid phase are in equilibrium. SII is denser than SIII.

B

T [K]

A

P [atm]

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31. Label the states present in each of the five regions of the phase diagram. 32. Label all triple points with TP and the critical point with CP. 33. Which phase present at point B has the higher density? 34. Sketch (not to scale) on the axis below the effect of decreasing pressure on the volume of the

substance along the dashed line on the phase diagram on the previous page.

V

P [atm] 35. Which of the following compounds is expected to have the highest vapor pressure?

(1) CH3CH2CH3 (2) CH3OCH3 (3) CH3CH2OH (4) CH3CH2CH2CH3 (5) CH3CH2CH2Cl

36. The vapor pressure of a liquid increases with an increase in temperature. Which of the following

statements best explains this increase? (1) The average kinetic energy of molecules is greater; thus more molecules can enter the gaseous

state. (2) The average kinetic energy of molecules is less. (3) The faster-moving molecules in the liquid exert a greater pressure. (4) All the molecules have greater kinetic energies. (5) The intermolecular forces between the molecules decrease at higher temperatures.

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37. Which compound should have the highest vapor pressure at room temperature? (1) C4H10 (2) C5H12 (3) C6H14 (4) C7H16 (5) C8H18

38. Which of the following indicates the existence of strong intermolecular forces of attraction in a

liquid? (1) a very low boiling point (2) a very low vapor pressure (3) a very low critical temperature (4) a very low viscosity (5) a very low heat of vaporization

39. Which of the following compounds would be expected to have the highest vapor pressure?

(1) CH3CH2CH2F (2) CH3CH2CH2OH (3) CH3CHCH3OH (4) CH3CH2COOH (5) CH3CH2CH2NH2

40. Which of the following has the lowest vapor pressure?

(1) H2O (2) H2S (3) H2Se

41. Which of the following accounts for the relatively high vapor pressure of diethyl ether liquid, (CH3CH2)2O, at 25 C? (1) density of the liquid (2) strength of the intermolecular forces (3) strength of the intramolecular forces

42. Of the following substances, which is expected to have the lowest normal boiling point?

(1) CH3OH (2) CH3Cl (3) CH4 (4) C2H6 (5) C3H8

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CHEMISTRY 123 Problem Set #1 ANSWERS Phase Equilibria 1. 4 2. 1 3. 2 4. 2,3,4 5. 1,2,3 6. 4 7. 1 8. 4 9. 3 10. 2

11. 1,3 12. 1 13. 1,2 14. 2 15. 3 16. 4 17. 3 18. 2,3 19. 2 20. 1,2

35. 1 36. 1 37. 1 38. 2 39. 1 40. 1 41. 2 42. 3

Detailed Solutions

17.

A

C

B

D

P(atm.)

2000

61

1

0.003

40 64 420 T(C)

Increase P keeping T constant at 50 C If a phase transition occurs when the pressure is increased on a substance, the transition is from a lower density phase to a higher density phase. Starting in region D at a temperature of 50 C, consider increasing the pressure on the system. As the pressure increases, there are the following phase transitions in the following order: (1) D C (2) C B (3) B A Hence, the order of densities of the phases is D < C < B < A.

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The two solid phases, SI and SII, are denser than the liquid phase. Also, the liquid phase is denser than the vapor phase. So the lowest density phase, D, must be vapor; C must be liquid; and B and A must be SI and SII (but which is which is not yet determined).

The transition SI SII is exothermic:

SI SII + heat When both solid forms are in equilibrium (along the A:B equilibrium line) and the heat in the system is reduced by decreasing the temperature, the equilibrium will shift towards the SII side to generate heat and relieve the stress on the equilibrium. Thus, SII is A, the lower temperature solid phase, and SI is B, the higher temperature solid phase.

A

C

B

D

P(atm.)

2000

61

1

0.003

40 64 420 T(C)

Solid SII

Solid SI

liquid

vapour

Note that 2000 atm and 64 C is a triple point where liquid, SI, and SII are all in equilibrium. Note that 3 103 atm and 40 C is a triple point where vapour, liquid and SI are all in equilibrium.

8. A liquid boils when its vapour pressure equals the external pressure. The normal boiling point is the

temperature where the liquid phase boils when the external pressure is 1 atm. Therefore, the vapor pressure of the liquid is also 1 atm at the normal boiling point. The horizontal line at 1 atm on the phase diagram intersects the liquid-vapor equilibrium line (the vapor-pressure curve for the liquid) at T = 200300 C. Thus, the normal boiling point is in the range of 200300 C (note: the b.p. is only estimated because the temperature scale on the phase diagram is non-linear, which makes it difficult to accurately read).

9. The critical point is the point (Pc, Tc) on the phase diagram where the liquid-vapor equilibrium curve

ends. This is at P = 61 atm and T = 420 C for this substance.

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10. SI is region B, and liquid is region C. Starting on the SI:liquid equilibrium line, increasing the temperature (i.e. adding heat) causes a shift to the liquid. Therefore, SI liquid is endothermic:

SI + heat liquid 11. Vapor and SI would be in equilibrium. Starting at (P = 1 atm, T = 35 C), the phenol will be in the SI

form. If the tube could be pumped out very quickly and sealed, the pressure above the solid would initially be very low and the system would not be at equilibrium. At a temperature of 35 C, equilibrium would be attained by the solid subliming until the pressure above the solid was the vapor-pressure of solid (SI) phenol at 35 C. On the phase diagram, the system would end up on the SI : vapor equilibrium (SI vapor-pressure) line at a temperature of 35 C.

1216. An isotherm displays the path of an isothermal process (a vertical line on a PT phase diagram).

In this case, V vs. P is plotted as the pressure on the system is increased at 50 C. The path is upward along the line T = 50 C on the diagram.

12. The initial conditions are P < 0.04 atm and T = 50 C. Under these conditions, the phenol is all in

the vapor phase. Considering the vapor to be an ideal gas with T, n, and R constant, the volume is proportional to 1/P; hence the isotherm is hyperbolic in the vapour region.

13. As the pressure on the vapor is increased, eventually the path will reach the vapor:liquid

equilibrium line. The pressure holds constant for a while until all the vapour is converted to liquid, which occupies much less volume than the vapour. Hence, the isotherm for this region is a straight vertical line.

14. In this region, we are compressing the liquid. Since liquids are not very compressible, a large

change in pressure results in only a small change in volume; hence, the isotherm can be approximated as a horizontal line in this region.

15. In this region of the path, we are simply compressing the solid, SI. Since solids are not very

compressible, a large change in pressure results in only a small change in volume; hence, the isotherm can be approximated as a horizontal line in this region.

16. In this region of the path, we are simply compressing the solid, SII. Since solids are not very

compressible, a large change in pressure results in only a small change in volume; hence, the isotherm can be approximated as a horizontal line in this region.

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1720.

10020

1918

1743

T(C)

(3)solid SI

(2) liquid (3) solid SI

(2) liquid

(1) vapour or gas (2) liquid

(1) vapour or gas

The path of an isobaric (constant pressure) process is a horizontal straight line on the PT diagram; in this case at P = 1 atm. Since this is a heating curve, the path is towards higher temperature; i.e. from left to right.

17. With initial conditions of P = 1 atm and T ~ 20 C, the phenol will all be in the SI solid state.

Over this first region of the path we are simply heating the SI phenol, so the temperature rises linearly (in all cases in this question, we are assuming that the heat capacities of phenol in its various forms are not functions of temperature).

18. When the temperature of the SI phenol reaches 43 C, the SI : liquid equilibrium line is reached.

At this point both SI and liquid will be in equilibrium. If the temperature is increased by a very small amount, all of the SI solid will smoothly melt. The heat entering the system is used to bring about the endothermic SI liquid transition, so the temperature of the system remains constant during the phase change. It remains constant until all the SI is converted to liquid.

19. In this region, we are simply heating the liquid phenol and the temperature increases linearly. 20. When the temperature reaches 100 C, the system will be at the liquid:vapor equilibrium line. At

this point liquid and vapor will be in equilibrium. If the temperature is increased a very small amount, all of the liquid will be converted to vapor. The heat entering the system is used in the endothermic liquid vapor transition, so the temperature remains constant during the phase change. It remains constant until all the liquid is converted to vapour. After all of the liquid has been converted to vapor, the temperature will rise again.

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21.

T [K]

273

V L V

L

S L

S

373

223

Time 2224,27.

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25. Liquid & Vapor 26. Liquid 28. Negative. The melting point decreases with increased P. At constant T, compressing the solid

(higher P) causes it to melt. Or: The melting point increases with decreased P. At constant T, decreasing the pressure over the liquid causes it to freeze.

29. For a normal substance, its fusion curve (or line D) has positive slope. Hence, its melting point

increases with increased P. At constant T, compressing the solid (higher P) does not cause it to melt. Without the lubrication of the liquid (due to melting), one cannot skate well.

30. At a higher altitude (e.g., at the top of Whistler above the sea level), the air pressure is lower than

that at sea level. According to the vapor-pressure curve, the boiling point of water decreases with decreased P. Thus, water boils at a lower T at the the top of Whistler. At the vapor-liquid equlibirum, T does not change from the boiling point. Hence, with this lower cooking T, more time is needed to cook the spaghetti.

3132.

solid III

solid II

solid I

vapour

TP B

CP

TP

TP

liquid

P [atm]

A

T [K] 33. Liquid

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34.

35. CH3CH2CH3 has the smallest intermolecular forces. Recall that the intermolecular forces of the

liquid must be overcome for molecules to go from the liquid to the gas phase. The smaller these intermolecular forces, the more molecules will be in the gas phase and thus the higher the vapor pressure. CH3CH2OH can have hydrogen bonding and thus it has very strong intermolecular forces (and hence low vapor pressure). CH3CH2CH2Cl and CH3OCH3 have dipole-dipole interactions (they are polar molecules) and thus have stronger intermolecular forces than CH3CH2CH3 and CH3CH2CH2CH3 (nonpolar molecules). While CH3CH2CH3 and CH3CH2CH2CH3 are both nonpolar molecules and have only weak London (dispersion) forces as intermolecular forces, CH3CH2CH2CH3 has stronger London interactions than CH3CH2CH3 because it has a higher molecular weight and contains more electrons.

36. Vapor pressure depends on the number of gaseous molecules present. At higher temperatures, a

greater proportion of the molecules present will have sufficient kinetic energy to be in the gas phase. 37. C4H10 has the weakest intermolecular forces of the molecules in this question because it has the

smallest molecular weight. Thus, it has the highest vapor pressure. 38. Something with strong intermolecular forces would be expected to have a high boiling point, low

vapour pressure, high critical temperature, high viscosity, and high heat of vaporization. 39. CH3CH2CH2F has the highest vapor pressure because it has the weakest intermolecular forces of the

molecules in this question. CH3CH2CH2F has dipole-dipole intermolecular forces, but the other molecules can have intermolecular hydrogen bonding, a stronger interaction.

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40. H2O undergoes hydrogen bonding (H2S and H2Se do not because S and Se are not electronegative enough) and therefore has the strongest intermolecular forces and lowest vapor pressure.

41. The lower the strength of the intermolecular forces, the higher the vapor pressure. (CH3CH2)2O

doesnt have particularly high intermolecular forces (dipole-dipole interactions occur, but no hydrogen bonding), so its vapor pressure is high at 25 C.

42. The smaller the intermolecular forces in the liquid, the more readily it will become vapor and thus

the lower its boiling point. CH4 has the weakest intermolecular forces of the molecules in this question. C2H6 and C3H8 have higher molecular weights than CH4 and hence greater intermolecular forces. CH3Cl has strong dipole-dipole intermolecular interactions in the liquid state while CH3OH has very strong hydrogen bonding intermolecular interactions. CH4 is not capable of dipole-dipole interactions or hydrogen bonding.

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CHEMISTRY 123 Problem Set #2 Thermodynamics 1. Given the enthalpy changes at a certain temperature for reactions (a) and (b), calculate the enthalpy

change for reaction (c). Is reaction (c) endothermic or exothermic at this temperature?

(a) 3H2(g) + N2(g) 2NH3(g) H = 92.4 kJ mol1 (b) 2H2(g) + O2(g) 2H2O(g) H = 483 kJ mol1 (c) 4NH3(g) + 3O2(g) 2N2(g) + 6H2O

2. Given the following data, evaluate the standard enthalpy of formation of ICl(g) (in kJ mol1).

Cl2(g) 2Cl(g) H = 242.0 kJ mol1 I2(g) 2I(g) H = 151.0 kJ mol1 ICl(g) I(g) + Cl(g) H = 211.2 kJ mol1 Hsublimation of I2(s) Hsub = 63.7 kJ mol1

(1) 244.5 (2) 244.5 (3) 211.2 (4) 17.1 (5) 17.1 3. The standard enthalpies of formation of CO2(g) and H2O(l) are 393.5 kJ mol1 and 285.8 kJ mol1

at 298 K, repectively. The heat of combustion of liquid pentane, C5H12(l), according to the following equation is

C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(l) H = 3536 kJ

What is the standard enthalpy of formation of C5H12(l) (in kJ mol1)?

(1) 254.0 (2) 146.3 (3) +146.3 (4) +254.0

4. Carbon disulphide is made industrially by passing sulphur vapour slowly through red-hot carbon. In a calorimeter, this is not an easy reaction to carry out, and the determination of the standard enthalpy of formation of CS2 is best made indirectly. Given that the standard enthalpies of formation of gaseous CO2 and SO2 at 25 C are 393.13 kJ mol1 and 296.53 kJ mol1, respectively, and that the standard enthalpy of combustion (at 25 C) of gaseous CS2 to CO2 and SO2 is 1103.4 kJ mol1, calculate the standard enthalpy of formation of gaseous CS2 at 25 C.

5. Which of the following statements is correct?

(1) The standard enthalpy of formation of an element in its normal state is negative at 298 K. (2) When a system does work on the surroundings, the internal energy of the system is increased. (3) q and w are in general independent of path. (4) S is not a state function (property of the system), since it is always increasing. (5) A given system in a given state has a definite, fixed value for S.

69. When 44.0 g of solid carbon dioxide at 1.00 atm pressure are reversibly converted to vapour at

1 atm pressure and 78 C, 25.33 kJ of heat are absorbed by the system. The density of solid CO2 is 1.56 g cm3.

6. H for this process is (in kJ): (1) 23.71 (2) +23.71 (3) 25.33 (4) +25.33 (5) +26.95

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7. S for the process is (in J K1): (1) 130.0 (2) 130.0 (3) 121.5 (4) 121.5 (5) 138.1 8. G for the process is (in kJ): (1) 50.66 (2) 50.66 (3) 46.71 (4) 46.71 (5) 0 9. If S of CO2(g) is 197.7 J K1 mol1 at 78 C, the standard entropy of solid CO2 is (in J K1): (1) 67.70 (2) 67.70 (3) 327.7 (4) 0 (5) 130.0 10. At 279 K and 1 atm pressure, pure benzene crystallizes reversibly with an entropy change of

35.66 J mol1 K1. What is the molar heat of fusion of benzene (in kJ mol1) at its melting point? 1113. Consider the following processes at constant T and P:

Process H S (1) positive positive (2) positive negative (3) negative positive (4) negative negative

Answer the following questions, using the process numbers as the answer choices:

11. Which processes are ALWAYS spontaneous? 12. Which processes can be both spontaneous or non-spontaneous (i.e. the spontaneity depends

on the temperature)? 13. Which processes are NEVER spontaneous?

1418. You are given the following data at 298 K:

Ag(s) H2S(g) O2(g) Ag2S(s) H2O(g) Hf (kJ mol1) 20.5 242 S (J K1 mol1) 42.6 206 205 189

For the reaction

4Ag(s) + 2H2S(g) + O2(g) 2Ag2S(s) + 2H2O(g) at a temperature of 25 C, H = 507 kJ and S = 119 J K1.

14. Hf of Ag2S (s) is (in kJ mol1):

(1) +475 (2) 32 (3) 64 (4) 285.5 (5) 728.5 15. S of Ag2S (s) is (in J K1 mol1):

(1) 290.4 (2) 264.2 (3) 145.6 (4) 145.2 (5) 383.6 16. G for the reaction is (in kJ):

(1) + 396.3 (2) 471.5 (3) 43302 (4) 75300 (5) 35.6

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17. K for the reaction is: (1) 2.2 1083 (2) 1.9 102 (3) 3.4 1070 (4) 5.25 (5) 4.5 1082

18. If the reaction were to occur at a higher temperature, more Ag2S(s) would be formed.

(1) True (2) False 19. For the reaction Cl2(g) 2Cl(g), S is likely to be: (1) positive (2) zero (3) negative 20. For the reaction,

N2(g) + O2(g) NO(g), K = 1.11 102 at 1800 K, and 2.02 102 at 2000 K. The enthalpy of reaction at 2000 K is nearest

to (in kJ): (1) 89.6 (2) 90.3 (3) 90.3 (4) 89.6 (5) 0.0896 21. Consider the following data (values are tabulated at 25 C and 1 atm pressure):

NO(g) O3(g) NO2(g) SO2(g) O2(g) SO3(g) Hf(kJ mol1) +90.39 +142.1 +33.9 296.8 0.0 394.6 Gf(kJ mol1) +86.5 +163.0 +51.8 S (J mol1K1) +248.3 +204.8 +255.8

(a) Calculate G, H, S, and K at 25 C for the reaction, NO (g) + O3 (g) NO2 (g) + O2 (g) (b) For the reaction, SO2(g) + O2(g) SO3(g) calculate H, S, G, and K at 25 C. Assuming that H and S do not change with

temperature, also calculate K at 327 C. 22. Suppose you have to investigate the thermodynamic feasibility of converting coal and hydrogen at

25 C either to a synthetic natural gas such as propane, C3H8, or to the industrially useful acetylene, C2H2. For simplicity, the properties of coal are assumed to be the same as those of graphite. The relevant thermodynamic data are given in the following table:

C(s,graphite) H2(g) C3H8(g) C2H2(g) Hf,298 (kJ mol1) 0 0 103.75 226.51 S298 (J mol1 K1) 5.68 130.46 269.65 200.64

(a) Calculate the enthalpies, entropies, and free energies of formation of (i) propane and (ii)

acetylene from graphite and hydrogen. (b) Calculate the equilibrium constants at 25 C for the reactions of graphite and hydrogen to form

propane and to form acetylene. (c) From the results of (a) and (b) decide whether either of these processes is thermodynamically

feasible, and, if so, discuss what other factors would be involved in making them practical.

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23. Dinitrogen tetroxide, N2O4, dissociates reversibly into nitrogen dioxide, NO2, according to the

following reaction:

N2O4(g) 2NO2(g) Some thermodynamic data for these compounds at 25 C are as follows:

Compound Hf (kJ mol1) S (J mol1 K1) N2O4 9.66 303.9 NO2 33.15 240.4

(a) Calculate the standard free energy change, G, at 25 C for the above reaction. (b) Calculate the equilibrium constant for the reaction at 25 C. (c) At what temperature will the equilibrium constant for the reaction be equal to one? (Assume H

and S do not vary with temperature.) (d) Calculate G for one mole of reaction if each gas were at 100 atmospheres partial pressure

(assume ideal behaviour). 24. Assume that all gases are ideal. For ideal gases, PV = nRT, and R = 8.314 J K1 mol1. One possible way to make dinitrogen trioxide (N2O3) is through the following reaction:

NO(g) + NO2(g) N2O3(g) At 298.15 K, the basic thermodynamic quantities are given below:

Substance Hf (kJ mol1) S (J mol1 K1) Gf (kJ mol1) NO(g) 91.3 210.8 87.6 NO2(g) 33.2 240.1 51.3 N2O3(g) 82.8 314.6

(a) What is the heat of reaction under constant pressure at 298.15 K under standard conditions? (b) Calculate S, E, and G for this reaction at 298.15 K? (c) Evaluate the standard Gibbs free energy of formation, Gf, of N2O3(g) at 298.15 K. (d) Evaluate the equilibrium constant, K, at 298.15 K. (e) Write an expression for the equilibrium constant in terms of activities and in terms of partial

pressures. (f) Could this reaction be used to mass produce N2O3 at 298.15 K? Give reasoning for your answer. (g) Will increasing the temperature favour the production of N2O3? Give reasoning for your answer. (h) If a chemical factory operating under standard conditions is built in a desert near the equator in

Australia, where the temperature stays above 35 C during the entire year, will N2O3 be produced spontaneously by using this reaction? Assume that S and H do not change with the temperature. Give reasoning for your answer.

25. The standard free energy change (G) for the hydrogenation of ethylene, shown below, is 101 kJ

at 25 C. Assume all gases are ideal; PV = nRT. CH2=CH2(g) + H2(g) CH3CH3(g)

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(a) By inspection of the above reaction, predict the sign of the standard entropy change (S) for

this reaction at 25 C: (1) negative (2) positive (3) zero (4) cannot be determined from the given data (b) The above reaction is: (1) endothermic (2) exothermic (3) neither (4) cannot be determined from the given data (c) If the temperature is increased, the equilibrium will shift: (1) to the right (2) to the left (3) neither way (4) cannot be determined from the given data (d) Calculate the Q for the reaction when the three gases are present at a partial pressure of 1 atm

each. (e) Calculate the free energy change (G) for the reaction when the three gases are present at a

partial pressure of 1 atm each. (f) Under the conditions of 25(e): (1) The reaction is at equilibrium. (2) The reaction will shift to the right. (3) The reaction will shift to the left. (4) Shifts cannot be determined from the given data. 26. The white pigment TiO2(s) can be prepared by the hydrolysis of titanium tetrachloride in the gas

phase according to the reaction: TiCl4(g) + 2H2O(g) TiO2(s) + 4HCl(g)

At 25 C, its equilibrium constant K = 1.55 1015, and the Hf data are given below:

TiCl4(g) H2O(g) TiO2(s) HCl(g) Hf (kJ mol1) 763.2 241.8 944.7 92.3

This reaction is carried out at 25 C. Assume all gases are ideal (PV = nRT).

(a) Calculate H (b) Calculate G (c) Calculate E (d) Calculate S (e) Calculate K for this reaction at 100 C. What assumptions have you made in your calculation? (f) Write the expression for K in terms of activities. (g) At constant temperature, if 0.3 atm HCl(g) are added to the system after equilibrium is reached,

K will: (1) increase (2) decrease (3) stay the same (4) cannot be determined from the given data

Give reasoning for your answer.

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CHEMISTRY 123 Problem Set #2 ANSWERS Thermodynamics 1. Reaction (c) is exothermic.

[2 (a)] + [3 (b)] = (c)

4NH3 6H2 + 2N2 6H2 + 3O2 6H2O 4NH3 + 3O2 2N2 + 6H2O (c) Hc = 2Ha + 3Hb = 2(92.4) + 3(483) = 1264.2 kJ

2. The answer is (5).

The formation reaction of ICl(g) is:

I2(s) + Cl2(g) ICl(g)

Using the reactions given, the formation reaction can be arrived at via: Cl2(g) Cl(g) H = (242.0 kJ mol1) I2(g) I(g) H = (151.0 kJ mol1) I(g) + Cl(g) ICl(g) H = 211.2 kJ mol1 I2 (s) I2 (g) Hsub = (63.7 kJ mol1)

I2(s) + Cl2(g) ICl(g) Htot = Hf(ICl(g))

= 1molkJ)7.63(212.211)0.151(

21)0.242(

21

++

= +17.15 kJ mol1

Note: (1) Hesss Law has been used: the total H is the sum of the individual H. (2) When a reaction is multiplied by a constant (e.g. ) the H is multiplied by that

constant. (3) Reversing the reaction (e.g. the third reaction listed in the question has been reversed

above to get the answer) changes the sign of H. This is equivalent to multiplying the reaction by the constant 1.


Hrxn = molesHf(products) molesHf(reactants)

3536 kJ = 5 mol (393.5 kJ mol1) + 6 mol (285.8 kJ mol1) 1 mol Hf(C5H12(l)) 8 mol (0 kJ mol1)

Page 1 of 11

Hf(C5H12(l)) = 146.3 kJ mol1

4. CS2 + 3O2 CO2 + 2SO2 H = 1103.4 kJ H = Hf(CO2) + 2Hf(SO2) Hf(CS2) 3Hf(O2) 1103.4 = 393.13 + 2(296.53) Hf(CS2) 3(0) Hf(CS2) = +117.2 kJ mol1 5. The answer is (5).

(1) The standard enthalpy of formation of an element in its standard state is defined to be zero. (2) When the system does work on the surroundings, the internal energy of the system decreases.

(3) The values of q and w depend upon the path. (4) S is a state function. (5) Each unique state of a system must have a unique value of S.


H is the heat transferred into or out of the system in a constant pressure process.

H = qp = +25.33 kJ Note that H is positive because heat is absorbed by the system.


For a process carried out reversibly:

S = T

q rev

= TH rev (for a constant pressure process, q = H)

S = K195

J25330+ = +130.0 J K1

(Entropy has increased, since the final state is a gas, which is more disordered than the initial state solid form.)


G = 0 if the process is carried out reversibly; i.e., incremental steps are made between the initial and final states such that the system is always in equilibrium.

Page 2 of 11


S = S(g) S(s) S(s) = S(g) S = (197.7 130.0) J K1 = +67.7 J K1

10. C6H6 (l) C6H6 (s) at 1atm and 279 K, S = 35.66 J K1 mol1

TH

TqS rev ==

H = TS (note: G = H TS = 0) = 279 K (35.66 103 kJ mol1 K1) = 9.95 kJ mol1 Therefore, for C6H6 (s) C6H6 (l), H = +9.95 kJ mol1

11-13. 11: The answer is (3) 12. The answer is (1, 4) 13. The answer is (2)

G = H T S

(1) +, + + + non-spontaneous at low T, spontaneous at high T (2) + + + never spontaneous

(3) + + always spontaneous (4) ,+ + spontaneous at low T, non-spontaneous at high T

G < 0 for a spontaneous process. G > 0 for a non-spontaneous process. T is in Kelvin and is always positive.


H = molesHf(products) molesHf(reactants) 507 kJ = 2 mol Hf(Ag2S) + 2 mol (242 kJ mol1)

4 mol (0 kJ mol1) 2 mol (20.5 kJ mol1) 1 mol (0 kJ mol1)

Hf(Ag2S) = 32 kJ mol1 15. The answer is (4).

S = molesS(products) molesS(reactants)

Page 3 of 11

119 J K1 = 2 mol S(Ag2S) + 2 mol (+189 J mol1 K1)

4 mol (42.6 J mol1 K1) 2 mol (206 J mol1 K1) 1 mol (205 J mol1 K1)

S(Ag2S) = +145.2 J mol1 K1 16. The answer is (2).

G = H TS = 507,000 J (298K)(119 J K1) = 471,538 J = 471.5 kJ


G = RT lnK

471,538 J = (8.314 J mol1 K1)(298K) ln K ln K = 190.32 K = 4.5 1082


H < 0. The reaction is exothermic:

4Ag(s) + 2H2S(g) + O2(g) 2Ag2S(s) + 2H2O(g) + heat 19. The answer is (1).

For each mole of Cl2(g) reacted, two moles of Cl(g) atoms are produced. A system with more particles will be more disordered than one with fewer particles; hence, S is positive.

20. The answer is: (4).

G = H TS = RT lnK

Assume H and S are independent of T; i.e. S(T1) = S(T2).

ln K1 = RS

RTH

1

+ ln K2 = RS

RTH

2

+

ln K1 ln K2 =

12 T

1T1

RH

Page 4 of 11

ln

2

1

KK =

12 T1

T1

RH

H = 12

2

1

T1

T1

KK

lnR

=

K18001

K20001

1002.21011.1ln

J1000kJ1KmolJ314.8 2

211

= 89.6 kJ mol1

21.(a) NO(g) + O3(g) NO2(g) + O2(g)

G = +51.8 + 0 86.5 163.0 = 197.7 kJ H = +33.9 + 0 90.39 142.1 = 198.5 kJ G = H TS

1

1

KJ0.3KkJ00299.0

298)7.197()5.198(

TGHS

==

=

=

G = RT lnK

8.79)298)(314.8(

)107.197(RT

GKln

3

+=

=

=

K = 4.5 1034

(b) SO2(g) + O2(g) SO3(g)

H = 394.6 (296.8) (0) = 97.8 kJ S = +255.8 (+248.3) (204.8) = 94.9 J K1

Page 5 of 11

At 25 C (298 K): G = H TS = 97.8 298(94.9 103) = 69.5 kJ

05.28)298)(314.8()105.69(

RTGKln

3

+=

=

=

K = 1.52 1012 At 327 C (600 K): G = H TS = 97.8 600(94.9 103) = 40.86 kJ

19.8)600)(314.8(

)1086.40(RT

GKln

3

+=

=

=

K = 3.6 103

22.(a) (i) 3C(s) + 4H2(g) C3H8(g) H = 103.75 kJ mol1 S = +269.65 3(5.68) 4(130.46) = 269.23 J K1 mol1 G = H TS = 103.75 298(269.23 103) = 23.52 kJ mol1 (ii) 2C(s) + H2(g) C2H2(g)

H = +226.51 kJ mol1 S = +200.64 2(5.68) (130.46) = +58.82 J K1 mol1 G = H TS = +226.51 298(58.82 103) = +208.98 kJ mol1

Page 6 of 11

(b) (i) 3C(s) + 4H2(g) C3H8(g)

493.9)298)(314.8(

)1052.23(RT

GKln

3

+=

=

=

K = 1.33 104

(ii) 2C(s) + H2(g) C2H2(g)

35.84)298)(314.8(

)1098.208(RT

GKln

3

=

+=

=

K = 2.33 1037

(c) Formation of propane is thermodynamically feasible (G < 0). Formation also depends on

kinetics (relative to formation of other possible products). Formation of acetylene is not thermodynamically feasible (G > 0).

23.(a) N2O4(g) 2NO2(g)

H = 2(33.15) (9.66) = +56.64 kJ S = 2(240.4) (303.9) = +176.9 J K1 G = H TS = 55.64 298(176.9 103) = +3.92 kJ

(b) G = RT lnK

58.1)298)(314.8()1092.3(

RTGKln

3

=

+=

=

K = 0.205

(c) K = 1, lnK = 0, G = RT lnK = 0

Also, G = H TS = 0

Page 7 of 11

K320KJ9.176

J1064.56SHT

1

3

=

==

(d)

kJ3.1541.1192.3

100)100(ln)298)(10314.8(99.3

PP

lnRTG

QlnRTGG

23

ON

2NO

42

2

=++=

++=

+=+=

24.(a) qp = H H = molesHf(products) molesHf(reactants) = 82.8 (91.3 + 33.2) = 41.7 kJ (b) S = molesS(products) molesS(reactants) = 314.6 (210.8 + 240.1) = 136.3 J K1 G = H TS = 41.7 298.15 (136.3 103) = 1.1 kJ E = H (PV) = H (nRT) = H (n)RT = 41.7 (1) 8.314 298.15 103 = 39.2 kJ (c) G = molesGf(products) molesGf(reactants) 1.1 = Gf(N2O3) (87.6 + 51.3) Gf(N2O3) = 137.8 kJ mol1 (d) G = RT lnK lnK = G/(RT) lnK = (1.1)/(8.314 298.15 103) 0.44 K e0.44 1.6 (e) aN2O3(g) PN2O3(g) K = aNO(g)aNO2(g) PNO(g)PNO2(g)

Page 8 of 11

(f) Yes. Because G = 1.1 kJ < 0. (g) No. H = 41.7 kJ < 0 Decreasing T favours exothermic reaction. (h) G = H TS < 0 for spontaneous process. H < TS, S = 136.3 J K1 < 0 T < H/S = T0 for spontaneous process. T0 = H/S = 41.7/(136.3 103) 305.9 K = 32.8 C When T = 35 C > T0 = 32.8 C, G > 0 Thus, non-spontaneous. Alternatively, use the vant Hoff equation and equilibrium constant. G = RT lnK < 0 for spontaneous process Hence, lnK = G/(RT) > 0 for spontaneous process For lnK > 0, K > 1 for spontaneous process Lets calculate K2 at T2 = 35 C = 308.15 K, with K1 at T1 = 25 C = 298.15 K and H = 41.7 kJ, using the vant Hoff equation: ln(K1/K2) = (H/R) (T21 T11) ln(K1/K2) = 41.7/(8.314 103) (308.151 298.151) 0.55 K1/K2 e0.55 1.73 K2 K1/1.73 1.6/1.73 0.9 < 1 Thus, non-spontaneous. 25.(a) The answer is (1). Two units of gas turn into one unit of gas. Thus, the entropy decreases. (b) The answer is (2). G = 101 kJ < 0 S < 0 (result from (a)) Thus, H = G + TS < 0, exothermic (c) The answer is (2). Increasing T favours endothermic reaction. Because the forward reaction is exothermic, the

reverse reaction is endothermic. Then, the equlibirum will shift to the left (in the reverse direction) as T increases.

Page 9 of 11

(d) aC2H6(g) Q = aC2H4(g)aH2(g) PC2H6(g) 1 = = 1 PC2H4(g)PH2(g) 11 (e) G = G + RT lnQ = G + RT ln1 = G = 101 kJ (f) The answer is (2). G = 101 kJ < 0, the forward reaction is spontaneous. 26.(a) H = molesHf(products) molesHf(reactants) = (92.34 944.7) (241.82 763.2) = 67.1 kJ (b) G = RT lnK = 8.314 103 298.15 ln(1.551015) = 86.7 kJ (c) H = E + (PV) E = H (PV) (ignoring volume change of solid) = H (nRT) (assuming ideal gas) = H (n)RT (at constant T) = 67.1 (43) 8.314 298.15 103 = 69.6 kJ (d) G = H TS H G S = T 67.1 (86.7) = = 0.0657 kJ K1 = 65.7 J K1 298.15 (e) Assume H and S to be independent on the change of T. At 25 C, T1 = 298.15 K, K1 = 1.551015 At 100 C, T2 = 373.15 K, K2 = ??? ln(K1/K2) = (H/R) (T21 T11) = (67.1 103 8.314) (373.151 298.151) = 5.441 K1/K2 = e5.441 K2 = K1 e5.441 = (1.551015) e5.441 = 6.7 1012

Page 10 of 11

Or: at 100 C, T = 373.15 K, G = RT lnK = H TS = 67.1 373.15 65.7 103 = 91.6 kJ G 91.6 lnK = = = 29.5 RT 8.314 373.15 103 K = e29.5 = 6.7 1012 (f) aTiO2(s)a4HCl(g) K = aTiCl4(g)a2H2O(g) (g) The answer is (3) K is a function of T only. If T stays the same, K remains the same.

Page 11 of 11

CHEMISTRY 123 Problem Set #3 Chemical Equilibrium (Review) 1. At 1000 K, it is found that an equilibrium mixture consists of H2S(g), H2(g), and S2(g) at partial

pressures of 41 atm, 8.2 atm, and 33 atm, respectively. The equation is:

2H2S(g) 2H2(g) + S2(g) K for the equilibrium is: (1) 0.13 (2) 0.76 (3) 1.3 (4) 2.6 (5) 6.6 27. A sample initially containing 0.84 mole PCl5 and 0.18 mole PCl3 is mixed in a 10 L flask at

340 K and allowed to come to equilibrium according to

PCl5(g) PCl3(g) + Cl2(g) At equilibrium, there are 0.72 moles of PCl5. 2. The number of moles of PCl3 at equilibrium is:

(1) 0.18 (2) 0.30 (3) 0.12 (4) 0.72 (5) none of these 3. The partial pressure of Cl2 at equilibrium is:

(1) 2.0 atm (2) 0.84 atm (3) 1.26 atm (4) 0.33 atm (5) none of these 4. The equilibrium constant K is:

(1) 0.050 (2) 0.083 (3) 0.14 (4) 1.4 (5) 7.2 5. The total pressure at equilibrium is:

(1) 2.0 atm (2) 0.84 atm (3) 0.33 atm (4) 1.17 atm (5) 3.18 atm

6. If the number of moles of PCl3 increases when the temperature increases the reaction is: (1) exothermic as written (2) endothermic as written 7. If the flask was expanded to 20 L at constant temperature, the number of moles of PCl5 at

new equilibrium would: (1) increase (2) decrease (3) stay the same

89. A 20.0 g sample of BaO2 is heated at 794 C in a closed evacuated flask of volume 3.00 L.

It decomposes according to:

BaO2(s) BaO(s) + O2(g) K for the reaction is 0.71. The volume of the solids can be neglected. 8. The total pressure at equilibrium is: (1) 0.71 atm (2) 0.50 atm (3) 0.75 atm (4) 1.06 atm (5) 1.25 atm 9. The weight of BaO2(s) remaining in the flask at equilibrium is: (1) 2.8 g (2) 5.7 g (3) 17.2 g (4) 14.2 g (5) none of these

Page 1 of 2

10. For the reaction,

H2(g) 2H(g) K is 7.0 1018 at 1000 C and 3.1 106 at 2000 C. Calculate the pressure of H atoms at equilibrium at a total pressure of 0.1 atm and temperature of

(a) 1000 C, and (b) 2000 C. 11. At 298 K, the value of the equilibrium constant is 2.0 1012 for the reaction, 2NO(g) + O2(g) 2NO2(g) In a particular equilibrium mixture of NO, NO2, and O2 at 298 K, the number of molecules of

NO2 equals the number of molecules of NO. What is the pressure of O2 in the mixture? 12. At 1000 K, the pressure of CO2(g) in equilibrium with CaCO3(s) and CaO(s),

CaCO3(s) CaO(s) + CO2(g) is equal to 3.9 102 atm. The equilibrium constant for the reaction, C(s) + CO2(g) 2CO(g) is 1.9 at 1000 K when pressures are in atmospheres. Solid carbon, CaO, and CaCO3 are mixed

and allowed to come to equilibrium at 1000 K in a closed vessel. What is the pressure of CO(g) at equilibrium?

13. Five grams of ammonium hydrosulphide, NH4HS, are placed in a one litre vessel and heated at

24 C. The final pressure attained is 0.618 atm.

(a) Calculate the equilibrium constant at 24 C for the reaction,

NH4HS(s) NH3(g) + H2S(g) (b) If 50 g of NH4HS are placed in a one litre flask, what is the final pressure at 24 C?

(c) If 0.078 g of ammonia and 5.0 g of NH4HS are initially present in a one litre flask, what

are the final pressures of ammonia and hydrogen sulphide at equilibrium at 24 C?

Page 2 of 2

CHEMISTRY 123 Problem Set #3 ANSWERS Chemical Equilibrium (Review) 1. The answer is (3).

3.1)41(

)33()2.8(2

2

2

2

2

22 ===SH

SH

P

PPK


PCl5(g) PCl3(g) + Cl2(g) initial (moles) 0.84 0.18 0.00 change (moles) 0.12 +0.12 +0.12 equilibrium (moles) 0.72 0.30 0.12 3. The answer is (4).

atm33.0L10

)K340)(KmolatmL08206.0)(mol12.0(VRTn

P

11

ClCl

2

2

==

=


atm01.2L10

)K340)(KmolatmL08206.0)(mol72.0(V

RTnP

11

PClPCl

5

5

==

=

atm84.0L10

)K340)(KmolatmL08206.0)(mol30.0(V

RTnP

11

PClPCl

3

3

==

=

14.001.2

)33.0)(84.0(5

23

==

=PCl

ClPCl

PPP

K

Page 1 of 4

5. The answer is (5). Assuming the components of the equilibrium mixture to be ideal gases, the total pressure at

equilibrium is equal to the sum of the equilibrium partial pressures: PT = PPCl3 + PCl2 + PPCl5 = (0.84 + 0.33 + 2.01) atm = 3.18 atm 6. The answer is (2). In response to the increased temperature (heat), the equilibrium shifts in a direction that reduces heat

in the system. The equilibrium is endothermic and shifts towards the right: PCl5 + heat PCl3 + Cl2 7. The answer is (2). When the volume is increased, there is a shift of the equilibrium towards the side with the most

number of moles. In this case, that is the product side (equilibrium shifts towards the right). Therefore, the number of moles of PCl5 decreases.

8. The answer is (2). K = PO2

1/2 = 0.71 PO2 = (0.71 atm)

2 = 0.50 atm 9. The answer is (4). The number of moles of O2 at equilibrium is:

mol017.0)K1067)(KmolatmL08206.0(

)L00.3)(atm50.0(RT

VPn

11

OO

2

2

==

=

initial moles BaO2 = 20.0 g / 169.3 g mol1 = 0.118 mol

BaO2(s) BaO(s) + O2(g) initial (moles) 0.118 0.000 0.000 change (moles) 0.034 +0.034 +0.017 equilibrium (moles) 0.084 0.034 0.017 final mass BaO2 = 0.084 mol 169.3 g mol1 = 14.2 g

Page 2 of 4

10. H2(g) 2H(g)

Ptot = PH2 + PH PH2 = Ptot PH

Htot

2H

H

2H

PPP

PPK

2==

If PH

13. NH4HS(s) NH3(g) + H2S(g) 5g 0 0 (a) Ptot = PNH3 + PH2S = 0.618 atm

Also, PNH3 = PH2S PNH3 = 0.309 atm, PH2S = 0.309 atm There must be some solid remaining to allow equilibrium. Is this the case? initial moles NH4HS = 5.0 g / 51.12 g mol1 = 0.098 mol NH4HS(s) NH3(g) + H2S(g) initial moles 0.098 0 0

x +x +x final moles 0.098 x x x x = moles NH3 = moles H2S = PV/RT (by ideal gas law)

x = RTPV =

K298KmolatmL0821.0L0.1atm309.0

11

= 0.013 mol

moles NH4HS at equilibrium = 0.098 0.013 = 0.085 mol solid remaining = 0.085 mol 51 g mol1 = 4.34 g equilibrium is possible K = PNH3PH2S = (0.309)

2 = 0.0955 (b) Final pressure is the same (0.618 atm), whether we start with 5g or 50g, so long as solid remains at

equilibrium. (c) initial moles NH4HS = 5.0 g / 51.12 g mol1 = 0.098 mol initial moles NH3 = 0.078 g / 17.034 g mol1 = 0.00458 mol

initial PNH3 = VRTn

3NH = L0.1

K298KmolatmL0821.0mol00458.0 11 = 0.112 atm Although this problem could also be solved by calculating the change in moles needed to establish

equilibrium, the fastest route to the answer involves letting x be the change in pressure needed to establish equilibrium (x = P = nRT/V):

NH4HS(s) NH3(g) + H2S(g)

Initial 0.098 mol 0.112 atm 0 xV/RT mol +x atm +x atm Final (0.098 xV/RT) mol (0.112 + x) atm x atm

K = PNH3PH2S = (0.112 + x)(x) = 0.0955 x2 + 0.112x 0.0955 = 0 Solving the quadratic equation gives x = 0.258 atm PNH3 = (0.112 + 0.258) atm = 0.370 atm PH2S = 0.258 atm

Page 4 of 4

CHEMISTRY 123 Problem Set #4 Electrochemistry

Note: Refer to the table at the end for value when necessary.

1. Explain how electrochemical measurements could be used to determine the change in Gibbss Free Energy G for the redox process:

Zn + Cu2+ Cu + Zn2+

2. Calculate the equilibrium constant at standard conditions for the reaction: Cu(s) + Cl2(g) CuCl2(aq)

3. Calculate the G(298 K) for the following reactions based on the electrode potentials in the

table: (a) Pb(s) + ZnCO3(aq) PbCO3(aq) + Zn(s) (b) K(s) + H2O(l) KOH(aq) + H2(g)

(c) K2S2O8(aq) + 2KI(aq) I2(s) + 2K2SO4(aq) 4. A disproportionation reaction involves a substance that is both oxidized and reduced. Is the

disproporionation reaction HClO2(aq) ClO3(aq) + HClO(aq) (note: this is not balanced)

spontaneous under standard conditions? Use: ClO3 + 3H+ + 2e HClO2 + H2O = +1.21 V HClO2 + 2H+ + 2e HClO + H2O = +1.65 V

5. Consider the following Galvanic cell at T = 25 C Pt | Cr2+(0.3M), Cr3+(2.0M) || Co2+(0.2M) | Co The overall reaction is: 2Cr2+(aq) + Co2+(aq) 2Cr3+(aq) + Co(s) which has an equilibrium constant K = 2.79 107.

Calculate the cell potential and G for the cell reaction at these conditions.

6. A cell has the following reaction Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) Increasing the concentration of Ag+ by a factor of 10 will change the cell potential by how many

volts? The temperature is 25 C. (1) 0.0295 V (2) 0.059 V (3) 10.00 V (4) +0.059 V (5) the cell potential is unchanged

7. For a certain redox reaction the standard cell potential ( ) is positive. Which of the following

is correct? (1) G > 0, K > 1 (2) G > 0, K < 1 (3) G < 0, K > 1 (4) G < 0, K < 1 (5) none of the above

Page 1 of 3

8. Given the following two half-cell reactions at 25 C: Fe2+(aq) + 2e Fe(s) = 0.440 V Fe3+(aq) + e Fe2+(aq) = +0.771 V Calculate the half-cell potential for the following half-cell reaction at 25 C:

Fe3+(aq) + 3e Fe(s)

9. Consider the galvanic cell based on the following half-cell reactions at 25 C: Cr3+(aq) + e Cr2+(aq) = 0.407 V Cr2+(aq) + 2e Cr(s) = 0.913 V

(1) Write the overall cell reaction in the direction of spontaneous change. (2) Calculate the standard cell potential and G. (3) Calculate the cell potential when the concentration of Cr2+(aq) and Cr3+(aq) are both

0.001 M. (4) Qualitatively (that is increase, decrease, or no change), what happens to the cell potential of a

cell operating under standard conditions, when some NaOH solution is added to the compartment containing Cr(s) and Cr2+(aq)? The compound Cr(OH)2 precipitates: it is a sparingly soluble substance. Provide a brief explanation of your choice.

(5) Given that the standard cell potential goes down on raising the temperature, what is the sign of the enthalpy change for the cell reaction at 25 C? Provide a brief explanation of your choice.

(6) Calculate the half-cell potential for the following half-cell reaction at 25 C: Cr3+(aq) + 3e Cr(s)

10. Consider the galvanic cell based on the following half-cell reactions

Br2(l) + 2e 2Br(aq) = +1.08 V Fe3+(aq) + e Fe2+(aq) = +0.77 V

(1) Write the overall cell reaction in the direction of spontaneous change. (2) Calculate the standard cell potential and G. (3) If the cell is operated with the [Fe3+] concentration set at 1.00 M and the [Fe2+] concentration

set at 1.00 104 M in the Fe2+/Fe3+ electrode, what concentration of Br is required in the Br2/Br electrode so that Fe3+ will be spontaneously reduced?

Page 2 of 3

CHEMISTRY 123 Problem Set #4 ANSWERS Electrochemistry 1. Because of the relation between G and the cell potential , one can use an electrochemical

measurement of to determine G via G = nF . For the given reaction, we must first use the two half-cell reactions:

Cu2+(aq) + 2e Cu(s) Zn(s) Zn2+(aq) + 2e to determine that n = 2 mol, that is 2 mol of e are transferred for each mole of the reaction. One can

then set up a standard galvanic cell, with the species under standard conditions, i.e. 1 Molar solutions and pure metal electrodes. Thus, we obtain G without ever measuring H or S.

2. Cu(s) + Cl2(g) CuCl2(aq)

Half-cell reactions based on the overall redox reaction:

Cu2+(aq) + 2e Cu(s) = 0.340 V Cl2(g) + 2e 2Cl(aq) = 1.358 V The standard cell potential = 1.358 0.340 = 1.018 V G = nF = 2 mol 96500 J mol1 V1 1.018 V = 196.5 kJ (for the reaction as written) Because: G = RT lnK lnK = G / (RT) = (196474 J) / (8.134 J mol1 K1 298 K) = 79.30 K = e79.30 = 2.75 1034 K is huge, much larger than 1, so the equilibrium lies far to the right (as written). The reaction

essentially goes to completion and we can neglect the reverse reaction. 3. a) Pb(s) + ZnCO3(aq) PbCO3(aq) + Zn(s)

Half-cell reactions based on the overall redox reaction:

Zn2+(aq) + 2e Zn(s) Pb(s) Pb2+(aq) + 2e Refer to the table at the end, we get Zn2+(aq) + 2e Zn(s) = 0.763 V Pb2+(aq) + 2e Pb(s) = 0.125 V = 0.763 (0.125) = 0.638V G = nF = 2 mol 96500 J mol1 V1 (0.638 V) = +123 kJ G is positive, so this is a non-spontaneous reaction under standard conditions. b) K(s) + H2O(l) KOH(aq) + H2(g) Half-cell reactions based on the overall redox reaction: H2O(l) + e OH(aq) + H2(g) K(s) K+(aq) + e

Page 1 of 6

Referring to the table at the end, we get H2O(l) + e OH(aq) + H2(g) = 0.828 V K+(aq) + e K(s) = 2.924 V = 0.828 (2.924) = +2.096 V G = nF = 1 mol 96500 J mol1 V1 2.096 V = 202 kJ c) K2S2O8(aq) + 2KI(aq) I2(s) + 2K2SO4 (aq) Half-cell reactions based on the redox reaction: S2O82(aq) + 2e SO42(aq) 2I(aq) I2 + 2e

Referring to the table at the end, we get S2O82(aq) + 2e SO42(aq) = 2.01 V I2 + 2e 2I(aq) = 0.535 V

= 2.01 0.535 = +1.475V G = nF = 2 mol 96500 J mol1 V1 1.475 V = 284.7 kJ 4. The two half-cell reactions: HClO2 + 4H2O ClO3(aq) + 3H3O+ + 2e oxidization HClO2 + 2H3O+ + 2e HClO + 3H2O reduction Overall reaction: HClO2 + H2O ClO3(aq) + H3O+ + HClO Under standard state, = 1.65 1.21 = +0.44 V > 0, So the disproportionation of the HClO2 occurs spontaneously under standard conditions. 5. Net redox reaction: 2Cr2+(aq) + Co2+(aq) 2Cr3+(aq) + Co(s) Half-cell reactions: 2Cr2+(aq) 2Cr3+(aq) + 2e Co2+(aq) + 2e Co(s) Also, G = nF = RT lnK Therefore, = RT/(nF) lnK = 1 mol 8.314 J mol1 K1 298 K / (2 mol 96500 J mol1 V1) ln(2.79 107) = 0.22 V Apply Nernsts equation to calculate the cell potential for these non-standard conditions: = RT/(nF) lnQ Q = [Cr3+]2 / ([Cr2+]2 [Co2+]) = 22 / (0.32 0.2) = 222.22

Page 2 of 6

Thus, = RT/(nF) lnQ = 0.22 V 1 mol 8.314 J mol1 K1 298 K / (2 mol 96500 J mol1 V1) ln222.22 = 0.15 V 6. The answer is (4).

= RT/(nF) lnQ =

nV0.0258 lnQ

= n

V0.0258 lnQ = 2

0.0258 V ln

2+

+2

][Ag][Cu

f i = 2

0.0258 V ln

2f

+f

+2

][Ag][Cu +

20.0258 V ln

2i

+i

+2

][Ag][Cu

= 2

0.0258 V ln

2f

+f

+2

][Ag][Cu +

20.0258 V ln

2i

+i

+2

][Ag][Cu

= 2

0.0258 V ln

2f

+f

+2

][Ag][Cu

20.0258 V ln

i+2

2i

+

][Cu][Ag

= 2

0.0258 V

i+2

2i

+

2f

+f

+2

][Cu][Ag

ln+][Ag][Cu

ln

= 2

0.0258 V ln

2f

+i

+2

2i

+f

+2

][Ag][Cu][Ag][Cu

= 2

0.0258 V ln

2f

+

2i

+

][Ag][Ag since [Cu2+]i = [Cu2+]f

= 2

0.0258 V ln

2

2

101

= 2

0.0258 V ln ( )210 = 0.059 V 7. The answer is (3). G = nF = RT lnK If > 0, then G < 0 and K > 1.

Page 3 of 6

8. This half-cell reaction is a sum of the two half-cell reactions provided above. Fe2+(aq) + 2e Fe(s) 1 = 0.440 V (1) +) Fe3+(aq) + e Fe2+(aq) 2 = +0.771 V (2)

Fe3+(aq) + 3e Fe(s) 3 1 + 2 ??? (3) The available data can be summarized in the following table:

n1 = 2 1 = 0.440 V G1 = n1F1 = 2F1 n2 = 1 2 = +0.771 V G2 = n2F2 = F2 n3 = 3 3 = ???

3 = 1 + 2 3 = 0.036 V

G3 = n3F3 = 3F3 G3 = G1 + G2 3F3 = 2F1 F2

For the third half-cell reaction, G3 = G1 + G2 3F3 = 2F1 F2 Thus, 3 = 1 + 2 = (0.440) + (+0.771) = 0.036 V. 9. (1) has to be positive for spontaneous change. This requires that the second equation be

subtracted from the first that has to be multiplied by 2 to balance electrons. 2 { Cr3+(aq) + e Cr2+(aq) } = 0.407 V +) Cr(s) Cr2+(aq) + 2e = +0.913 V

2Cr3+(aq) + Cr(s) 3Cr2+(aq) = 0.913 0.407 > 0 = 0.506 V (2) = 0.913 0.407 = 0.506 V G = nF = 2 96500 0.506 = 97658 J 97.7 kJ (3) = RT lnQ/(nF) = 0.506 [8.314 298.15/(2 96500)] ln{[Cr2+]3/[Cr3+]2} = 0.506 [8.314 298.15/(2 96500)] ln(0.001) 0.506 (0.0887) 0.595 V (4) = RT lnQ/(nF) Q = [Cr2+]3/[Cr3+]2 The removal of Cr2+(aq) as Cr(OH)2(s) will decrease [Cr2+], causing Q to decrease. Then, lnQ will

decrease, RT lnQ/(nF) will increase accordingly, and will increase as a result. (5) Lets use the vant Hoff equation for this question.

We are given T1 < T2 and 1 > 2, with

T1 K1 1 T2 K2 2

Page 4 of 6

vant Hoff equation: ln(K1/K2) = (H/R) (T21 T11)

Also, 1 = RT1 lnK1/(nF), 2 = RT2 lnK2/(nF),

ln(K1/K2) = lnK1 lnK2 = (nF/R) [1/T1 2/T2]

Hence, nF (1/T1 2/T2) = H (T21 T11) ()

T1 < T2, then T11 > T21, thus (T21 T11) < 0

T11 > T21 and 1 > 2, thus (1/T1 2/T2) > 0

Now, look at Eq. (), all the terms on the left-hand side are positive, the term in the brackets on the right-hand side is negative. Therefore, H must be negative: H < 0.

(6) This half-cell reaction is a sum of the two half-cell reactions provided above.

Cr3+(aq) + e Cr2+(aq) 1 = 0.407 V (1) +) Cr2+(aq) + 2e Cr(s) 2 = 0.913 V (2)

Cr3+(aq) + 3e Cr(s) 3 1 + 2 ??? (3) The available data can be summarized in the following table:

n1 = 1 1 = 0.407 V G1 = n1F1 = F1 n2 = 2 2 = 0.913 V G2 = n2F2 = 2F2 n3 = 3 3 = ???

3 = 1 + 2 3 = 0.744 V

G3 = n3F3 = 3F3 G3 = G1 + G2 3F3 = F1 2F2

For the third half-cell reaction, G3 = G1 + G2 3F3 = F1 2F2 Thus, 3 = 1 + 2 = (0.407) + (0.913) = 0.744 V. 10. (1) has to be positive for spontaneous change. This requires that the second equation be

multiplied by 2 and be subtracted from the first to balance electrons. Br2(l) + 2e 2Br(aq) = +1.08 V +) 2 { Fe2+(aq) Fe3+(aq) + e } = 0.77 V

Br2(l) + 2Fe2+(aq) 2Br(aq) + 2Fe3+(aq) = 1.08 0.77 > 0 = 0.31 V (2) = 1.08 0.77 = 0.31 V G = nF = 2 96500 0.31 = 59.83 kJ mol1 (or: 59830 J mol1) (3) = {RT/(nF)} lnQ = {8.314 298.15/ (2 96500)} lnQ = 0.31 0.01284 lnQ Q = [Fe3+]2 [Br]2 / [Fe2+]2 = 1.00 [Br]2 / (1.00104)2 = 108 [Br]2

Page 5 of 6

For Fe3+ to be spontaneously reduced, < 0: the reverse reaction in part (1). = 0.31 0.01284 lnQ < 0 lnQ > 0.31/0.01284 = 24.14 Q > 3.05 1010 Q = 108 [Br]2 > 3.05 1010 [Br]2 > 3.05 102 [Br] > 17.46 M The Br concentration must be bigger than 17.46 M.

Page 6 of 6

CHEMISTRY 123 Problem Set #5 Acids and Bases

pH = log10[H3O+] pOH = log10[OH] pKa = log10Ka pKb = log10Kb pKw = log10Kw[H3O+] = 10pH [OH] = 10pOH Ka = 10pKa Kb = 10pKb Kw = 10pKw

1. Which of the following has the largest dissociation constant in water?

(1) NH3 (2) HClO4 (3) H2O (4) HSO4 (5) HCN 2. If the pH of an aqueous solution is 3.75, the [H3O+] is

(1) 6.6 103 M (2) 1.8 103 M (3) 5.6 104 M (4) 1.8 104 M (5) 5.6 103 M 3. Kw for water at normal body temperature, 37 C, is 2.42 1014. What is the pH for a neutral

solution at this temperature? (1) 0 (2) 6.8 (3) 7.0 (4) 7.2 (5) 14

4. The pH of a 0.20 M solution of a weak base (Kb = 1.8 105) is

(1) 2.4 (2) 2.6 (3) 4.2 (4) 11.3 (5) 12.1 5. The acid ionization constant for HSO3 is 6.24 108; therefore the base ionization constant for

SO32 is (1) 1.7 102 (2) 2.5 104 (3) 3.1 108 (4) 1.6 107 (5) 1.6 107

6. The pH of a 0.40 M NH4Br solution is (Kb = 1.8 105 for NH3):

(1) 2.7 (2) 4.8 (3) 7.0 (4) 9.2 (5) 12.1 7. What is the pH of a solution of 0.300 mole of acetic acid (Ka = 1.80 105) and 1.50 moles of

sodium acetate in 1.00 liter of water? (1) 0.52 (2) 4.05 (3) 5.44 (4) 7.00 (5) 8.56

8. When 10 mL of 5 M HCl are added to the solution from question 7, the change in pH is

(1) +0.08 (2) 0.08 (3) +0.18 (4) 0.18 (5) +0.28 9. An equimolar solution of chloroacetic acid (Ka = 1.4 103) and its potassium salt in water has an

approximate pH of (1) 1.4 (2) 2.9 (3) 7.0 (4) 8.4 (5) 9.9

10. Consider the following bases and their pKb values:

(a) NH3 (4.74) (b) acetate ion (9.26) (c) CN (4.60) (d) N2H4 (5.52) The order of increasing strength as a base is

(1) a, d, c, b (2) c, a, d, b (3) b, d, a, c (4) d, a, c, b (5) c, b, d, a 11. Which of the following salts dissolve in water to give acid solutions?

(1) KClO4 (2) NaHSO4 (3) Ca(CN)2 (4) NH4NO3 (5) Na2CO3 12. The pH of a 0.158 M solution of NH3 is 11.22. What is the percent dissociation of NH3 in this

solution? (1) 0.0105% (2) 1.05% (3) 10.5% (4) 0.166% (5) 9.45%

Page 1 of 3

13. Given Ka(HCN) = 4.0 1010; Ka(CH3COOH) = 1.8 105, then a 1 M solution of sodium acetate

is more basic than a 1 M solution of sodium cyanide. (1) true (2) false

14. The pH of a 5.0 108 M HCl solution is

(1) 6.72 (2) 6.89 (3) 7.00 (4) 7.30 (5) 7.51 15. The pH of a 5.0 108 M Ca(OH)2 solution is

(1) 6.70 (2) 7.30 (3) 7.00 (4) 7.21 (5) 7.47 16. The pH of a 1.0 104 M CH3CO2H solution (Ka = 1.82 105) is

(1) 4.28 (2) 4.37 (3) 4.00 (4) 4.82 (5) 4.46 17. The addition of sodium acetate to a solution of acetic acid in water will

(1) reduce the pH of the solution. (2) increase the degree of ionization of the acetic acid. (3) increase the H+ concentration in the solution. (4) cause Ka for acetic acid to decrease. (5) represse the ionization of acetic acid.

18. A solution contains 0.20 mol per liter of NH3 and 0.10 mol per liter of NH4Cl. What is the OH ion

concentration in this solution? Kb for ammonia is 1.8 105? (1) 1.8 103 M (2) 3.6 105 M (3) 9.0 103 M (4) 3.6 106 M (5) 6.0 103 M

19. Hypochlorous acid (HClO) is a weak acid. It ionizes slightly in water to produce hydrogen ions and

hypochlorite ions (ClO). Which one of the following substances could be used with hypochlorous acid to make a buffer solution? (1) NaClO4 (2) KaClO (3) KNO3 (4) NaCl (5) HClO4

20. What is the pH of a buffer solution that contains 0.20 M formic acid and 0.20 M sodium formate?

The Ka of monoprotic formic acid is 6.0 104. (1) 4.96 (2) 2.34 (3) 4.56 (4) 3.22 (5) 1.96

21. What mass of sodium hydroxide should be added to 100 mL of a 0.0100 M acetic acid solution in

order that the solution should have a pH of 4.74? Ka for acetic acid is 1.8 105. Assume that the volume does not change when this sodium hydroxide is added. (1) 5.0 104 g (2) 1.0 103 g (3) 8.00 mg (4) 20.0 mg (5) 80 mg

22. A buffer solution is prepared from the weak base ammonia and its conjugate acid such that the

concentration of the base is exactly ten times the concentration of its conjugate acid. What is the pH of this buffer solution if Kb for ammonia is 1.8 105? (1) 3.74 (2) 10.26 (3) 8.26 (4) 5.74 (5) 9.26

23. Given 100.0 mL of a buffer that is 0.5 M in HOCl and 0.40 M in NaOCl, what is the pH after 10.0

mL of 1.0 M NaOH has been added? (Ka for HOCl is 3.5 108) (1) 6.45 (2) 6.64 (3) 7.36 (4) 7.45 (5) 7.55

Page 2 of 3

24. How many millimoles of HCl must be added to 100 mL of a 0.100 M solution of methylamine

(pKb = 3.36) to give a buffer having a pH of 10.0? Assume no volume changes. (1) 8.1 (2) 18.7 (3) 20.0 (4) 6.1 (5) 12.7

25. Calculate the pH of pure water at 45 C (Kw = 4.20 1014 at 45 C). Is this solution acidic, neutral,

or basic? Explain your answer. 26. To 200 mL of 0.100 M benzoic acid (C6H5COOH) solution is added 110 mL of 0.100 M NaOH

solution. Calculate the pH for the new solution (Ka for benzoic acid = 6.3 105). If 2.00 103 mol of HCl are added to the above solution and equilibrium is reached. Calculate the pH of the resulting solution.

Page 3 of 3

CHEMISTRY 123 Problem Set #5 ANSWERS Acids and Bases 1. 2 2. 4 3. 2 4. 4 5. 5 6. 2 7. 3 8. 2

9. 2 10. 3 11. 2,4 12. 2 13. 2 14. 2 15. 4 16. 5

17. 5 18. 2 19. 2 20. 4 21. 4 22. 2 23. 5 24. 1

Detailed Solutions 1. HClO4 is the strongest acid. Hence, it has the highest dissociation constant.

2. Note: throughout this problem set, H+ and H3O+ are used interchangeably.

pH = log[H+] 3.75 = log[H+] [H+] = 1.8 104 M.

3. H2O(l) H+(aq) + OH(aq) x x

Kw = [H+][OH] = 2.42 1014 (x)(x) = 2.42 1014 x2 = 2.42 1014 x = [H+] = 1.56 107 M pH = log(1.56 107) = 6.81

4. Let B be the base:

B + H2O BH+(aq) + OH(aq) 0.20 x x x

Kb = ]B[]OH][BH[ + = 1.8 105

x20.0)x)(x(

= 1.8 105

Assume x

pH + pOH = 14

pH = 14 2.72 = 11.28. Check the assumption:

M20.0

x 100% = M20.0

M109.1 3 100% = 0.95% x < 1% of 0.20 M, therefore the assumption is OK. 5. Kw = KaKb 1 1014 = (6.24 108) Kb Kb = 1.6 107 6. Kw = KaKb 1 1014 = Ka (1.8 105) Ka = 5.6 1010 NH4+ + H2O NH3(aq) + H3O+(aq) 0.40 x x x

Ka = ]NH[

]OH][NH[

4

33+

+ = 5.6 1010

x40.0)x)(x(

= 5.6 1010

Assume that x

7. [H3O+] = [base][acid]Ka

So, pH = pKa + log

]acid[]base[

pH = log(1.80 105) + log

300.050.1 = 5.44

8. HCl is a strong acid, so the moles of H+ added is: moles H+ added = 5 M 0.010 L = 0.05 mol The H+ added reduce the amount of Ac in solution by protonating it to HAc. Therefore, we have:

[Ac] = L010.1

mol05.050.1 = 1.44 M

[H+] = L010.1

mol05.0300.0 + = 0.35 M (Note: the total volume is now 1.010 L since 10 mL of acid has been added.) To calculate the new pH:

Because: [H3O+] = [base][acid]Ka

pH = pKa + log

]acid[]base[ = log(1.80 105) + log

35.044.1 = 5.36

pH = 5.36 5.44 = 0.08

9. [H3O+] = [base][acid]Ka

So, pH = pKa + log

]acid[]base[

= log(1.4 103) + log 1 (equimolar: [salt] = [acid]) = 2.9

Page 3 of 10

10. The strongest base has the highest Kb. Since pKb = log Kb, the highest Kb (greatest base strength)

results in the smallest pKb: Kb(acetate ion) < Kb(N2H4) < Kb(NH3) < Kb(CN). 11. (1) KClO4 K+(aq) + ClO4(aq) ClO4 is the weak conjugate base of the strong acid HClO4. Because it is weak, it does not react

with H2O appreciably. All the H+ and OH in solution are therefore from water dissociation only. The resulting solution is neutral.

(2) NaHSO4 Na+(aq) + HSO4 (aq) HSO4 + H2O H3O+ (aq) + SO42(aq) HSO4 is a strong acid. It reacts with H2O to give H3O+ and the resulting solution is acidic. (3) Ca(CN)2 Ca2+(aq) + 2CN(aq) CN(aq) + H2O HCN + OH(aq) CN is the strong conjugate base of the weak acid HCN. It reacts with H2O to give OH and the

resulting solution is basic. (4) NH4NO3 NH4+(aq) + NO3(aq) NH4+(aq) + H2O H3O+(aq) + NH3 NH4+ is the strong conjugate acid of the weak base NH3. It reacts with H2O to give H3O+ and the

resulting solution is acidic. (5) Na2CO3 2Na+(aq) + CO32(aq) CO32(aq) + H2O HCO3(aq) + OH(aq) CO32 is the strong conjugate base of the weak acid HCO3. It reacts with H2O to give OH and

the resulting solution is basic.

12. NH3(aq) + H2O NH4+(aq) + OH(aq) pOH = 14 11.22 = 2.78 [OH] = 102.87 = 1.66 103 M One mole of NH3 produces one mole of OH; hence, in 1 L of this solution, 1.66 103 moles of

NH3 have dissociated.

% dissociation = 3

3

NHmolesinitialddissociateNHmoles

100% = mol158.0

mol1066.1 3 100% = 1.05% 13. Ka (HCN) < Ka (CH3COOH), so HCN is a weaker acid than CH3COOH. The smaller Ka is for an acid, the larger Kb is for its conjugate base: Kb (CN) > Kb (CH3COO), so CN is a stronger base than CH3COO. Therefore, a 1 M solution of NaCN (1 M CN) be more basic than a 1 M solution of NaCH3COO

(1 M CH3COO).

Page 4 of 10

14. HCl will dissociate completely: HCl H+ (aq) + Cl (aq) 5.0 108 M 5.0 108 M Thus, the [H+] from HCl dissociation is 5.0 108 M. However, this [H+] is low enough that the H+

from water dissociation is also important and must be taken into account: H2O(l) H+(aq) + OH(aq) x x where x = amount of dissociation of H2O The water dissociation constant Kw is: Kw = [H+][OH] = 1.0 1014 (x + 5.0 108)(x) = 1.0 1014 Note that there are two sources of H+: from HCl and from H2O dissociation (this is essentially an

example of the common ion effect, where the common ion is H+). x2 + 5.0 108 x 1.0 1014 = 0 Solution of the quadratic equation gives: x = 7.81 108 M = [H+] from H2O dissociation. Thus, the total [H+] is the sum of the H+ from water dissociation and from HCl dissociation: [H+]tot = 7.81 108 M + 5.0 108 M = 1.281 107 M pH = log(1.281 107 M) = 6.89 15. Assume complete dissociation of Ca(OH)2 (strong base): Ca(OH)2(s) Ca2+(aq) + 2OH(aq) 5.0 108 M 1.0 107 M The [OH] from Ca(OH)2 is very low, so [OH] from H2O dissociation must also be taken into

account: H2O(l) H+(aq) + OH(aq) x x where x = amount of dissociation of H2O x2 + 1.0 107 x 1.0 1014 = 0 Solving quadratic equation gives: x = 6.18 108 M = [OH] from H2O dissociation Thus, the total [OH] is the sum of the contributions from H2O dissociation and from Ca(OH)2

dissociation: [OH]tot = 6.18 108 M + 1.0 107 M = 1.618 107 M pOH = log (1.618 107) = 6.79 pH = 14 pOH = 14 6.79 = 7.21

Page 5 of 10

16. CH3CO2H CH3CO2(aq) + H+(aq) 1.0 104 M x x x

Ka = ]HCOCH[]H][COCH[

23

23+

1.82 105 = x100.1

)x)(x(4

x2 + 1.82 105x 1.82 109 = 0 Solving quadratic equation gives: x = 3.45 105 M = [H+] pH = log(3.45 105 M) = 4.46 Note: it is not valid to assume 1.0 104 x is 1.0 104 for this situation. This leads to the

incorrect answer of 4.37. 17. If initial concentration of acetic acid, CH3COOH, is A then we have the equilibrium: CH3COOH CH3COO(aq) + H+(aq) A x x x where x is the degree of dissociation of the acid. If the acetate ion, CH3COO, and H+ were present only because of the acetic acid dissociating, then

we would have:

Ka = xA)x)(x(

]COOHCH[]H][COOCH[

3

3

=+

= constant

However, when a salt containing the acetate ion is added, the [CH3COO] is larger than x. For

example, sodium acetate completely dissociated when added to the solution. If the initial concentration of NaCH3COO is B, then the concentration of the CH3COO produced from the salt dissociating is also B:

NaCH3COO(s) CH3COO(aq) + Na+(aq) B B This is a case of the common ion effect. The CH3COO comes from two sources (acid dissociation

and salt dissociation) and the total [CH3COO] is B + x. Now the form for Ka is:

Ka = xA)x)(xB(

]COOHCH[]H][COOCH[

3

3

+=

+ = constant

Page 6 of 10

Since [CH3COO] increases when it comes from two sources, then [H+] must decrease so that

Ka remains constant: if the concentration of one ion involved in the acetic acid equilibrium goes up, the concentration of the other ion must go down. Therefore, [H+] = x is less, i.e. the degree of ionization of the acetic acid is repressed. In addition, the pH wil be higher since [H+] is lowered.

18. NH3 + H2O NH4+(aq) + OH(aq) 0.20 x x x NH4Cl(s) NH4+(aq) + Cl(aq) 0.10 M 0.10 M 0.10 M

Kb = ]NH[]OH][NH[

3

4+

1.8 105 = x20.0

)x)(x10.0(

+ Assume 0.20 x 0.20 and 0.10 + x 0.10:

1.8 105 = 20.0

)x)(10.0(

x = 3.6 105 M = [OH]. Was assumption valid?

10.0x 100% =

10.0106.3 5 100% = 0.036%

20. For buffer, [H3O+] = [base][acid]Ka

So, pH = pKa + log

]acid[]base[ = log(6.0 104) + log

20.020.0 = 3.22

21. First react base with acid. Assume all base added reacts completely and is used up. Let y = [OH] added (since we assume no volume change, we can use concentration here instead of

#moles) CH3COOH + OH CH3COO + H2O initial conc. 0.0100 y 0 0 change y y +y +y final conc. 0.0100y 0 y y

[H3O+] = [base][acid]Ka

pH = pKa + log

]acid[]base[

4.74 = log (1.8 105) + log

y0100.0

y

log

y0100.0

y = 0

y0100.0

y = 1

y = 0.00500 M = [OH] added mass NaOH = 0.00500 M 0.100 L 39.9971 g/mol = 0.0200 g = 20.0 mg

22. [OH] = [acid][base]Kb

So, pOH = pKb + log

[base][acid] = log(1.8 105) + log

101 = 3.74

pH = 14 pOH = 10.26

Page 8 of 10

23. initial moles HOCl = 0.5 M 0.100 L = 0.05 mol

initial moles OH = 1.0 M 0.0100 L = 0.01 mol initial moles OCl = 0.4 M 0.100 L = 0.04 mol Assume all added base reacts completely with acid:

HOCl + OH OCl + H2O initial moles 0.05 0.01 0.04 0.00 change 0.01 0.01 +0.01 +0.01 final moles 0.04 0.00 0.05 0.01 Therefore, the final concentrations are:

[HOCl] = L110.0

mol04.0 = 0.364 M

[OCl] = L110.0

mol05.0 = 0.455 M

Note: one cannot assume the volume change is negligible. Change in volume is 10% when NaOH is added.

[H3O+] = [base][acid]Ka

So, pH = pKa + log

]acid[]base[ = log (3.5 108) + log

M364.0M455.0 = 7.55

24. CH3NH2 = methylamine. (Note: we do not need to know the formula to do the question! One could

simply assume that A is methylamine and HA+ is its conjugate acid.)

pOH = 14 pH = 14 10.0 = 4.0

[OH] = [acid][base]Kb

So, pOH = log Kb + log

[base][acid]

4.0 = 3.36 + log

]base[]acid[

]base[]acid[ = 4.365

initial moles CH3NH2 = 0.100 M 0.100 L = 0.01 mol

Page 9 of 10

Let x = moles of HCl added. Assume all the added HCl reacts with CH3NH2 (the weak conjugate

base) to produce CH3NH3+ (the weak acid): CH3NH2 + HCl CH3NH2+ + Cl initial moles 0.01 x 0 0 change x x +x +x final moles 0.01x 0 x x

x01.0x = 4.365

x = 0.0081 mol = 8.1 mmol

25. [OH][H3O+] = Kw [OH] = [H3O+] for pure water [H3O+]2 = 4.20 1014 [H3O+] = 2.049 107 pH = log10[H3O+] = 6.69 This solution is neutral, as [OH] = [H3O+] (despite the pH being below 7). 26. In the beginning, moles of HA = 0.200L 0.100M = 0.020, moles of OH = 0.110L 0.100M = 0.011 At equilibrium, moles of HA = 0.020 0.011 = 0.009, moles of A = 0.011 [H3O+]= Ka[HA]/[A] = 6.3 105 0.009/0.011 = 5.15 105 pH = log10[H3O+] = log10(5.15 105) = 4.29 Once more HCl is added, at new equilibrium, moles of HA = 0.009 + 0.002 = 0.011, moles of A = 0.011 0.002 = 0.009 [H3O+] = Ka[HA]/[A] = 6.3 105 0.011/0.009 = 7.70 105 pH = log10[H3O+] = log10(7.70 105) = 4.11

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Chemistry 123 Problem Set #1