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Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
1
Thermodynamics: Spontaneity,Entropy, and Free Energy
Chapter SeventeenChapter Seventeen
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
2
• Thermodynamics examines the relationship between heat and work.
• Spontaneity is the notion of whether or not a process can take place unassisted.
• Entropy is a mathematical concept describing the distribution of energy within a system.
• Free energy is a thermodynamic function that relates enthalpy and entropy to spontaneity, and can also be related to equilibrium constants.
Introduction
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
3
• With a knowledge of thermodynamics and by making a few calculations before embarking on a new venture, scientists and engineers can save themselves a great deal of time, money, and frustration.– “To the manufacturing chemist thermodynamics gives information
concerning the stability of his substances, the yield which he may hope to attain, the methods of avoiding undesirable substances, the optimum range of temperature and pressure, the proper choice of solvent.…” - from the introduction to Thermodynamics and the Free Energy of Chemical Substances by G. N. Lewis and M. Randall
• Thermodynamics tells us what processes are possible.– (Kinetics tells us whether the process is practical.)
Why Study Thermodynamics?
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
4
• A spontaneous process is one that can occur in a system left to itself; no action from outside the system is necessary to bring it about.
• A nonspontaneous process is one that cannot take place in a system left to itself.
• If a process is spontaneous, the reverse process is nonspontaneous, and vice versa.
• Example: gasoline combines spontaneously with oxygen.
• However, “spontaneous” signifies nothing about how fast a process occurs.
• A mixture of gasoline and oxygen may remain unreacted for years, or may ignite instantly with a spark.
Spontaneous Change
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
5
• Thermodynamics determines the equilibrium state of a system.
• Thermodynamics is used to predict the proportions of products and reactants at equilibrium.
• Kinetics determines the pathway by which equilibrium is reached.
• A high activation energy can effectively block a reaction that is thermodynamically favored.
• Example: combustion reactions are thermodynamically favored, but (fortunately for life on Earth!) most such reactions also have a high activation energy.
Spontaneous Change (cont’d)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
6
Example 17.1Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made.
(a) The action of toilet bowl cleaner, HCl(aq), on “lime” deposits, CaCO3(s).
(b) The boiling of water at normal atmospheric pressure and 65 °C.
(c) The reaction of N2(g) and O2(g) to form NO(g) at room temperature.
(d) The melting of an ice cube.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
7Example 17.1
Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made.
(a) The action of toilet bowl cleaner, HCl(aq), on “lime” deposits, CaCO3(s).(b) The boiling of water at normal atmospheric pressure and 65 °C.(c) The reaction of N2(g) and O2(g) to form NO(g) at room temperature.(d) The melting of an ice cube.
Solution(a) When we add the acid, the fizzing that occurs—the escape of a gas—indicates
that the reaction occurs without any further action on our part. The net ionic equation is
CaCO3(s) + 2 H3O+(aq) Ca2+(aq) + 3 H2O(l) + CO2(g)
The reaction is spontaneous.
(b) We know that the normal boiling point of a liquid is the temperature at which the vapor pressure is equal to 1 atm. For water, this temperature is 100 °C. Thus, the boiling of water at 65 °C and 1 atm pressure is nonspontaneous and not possible.
(c) Nitrogen and oxygen gases occur mixed in air, and there is no evidence of their reaction to form toxic NO at room temperature. This makes their reaction to form NO(g) appear to be nonspontaneous. However, this could be an example of a spontaneous reaction that occurs extremely slowly, like that of H2 and O2 to form H2O. We need additional criteria before we can answer this question.
(d) We know that at 1 atm pressure ice melts spontaneously at temperatures above its normal melting point of 0 °C. Below this temperature, it does not. To answer this question, we would have to know the temperature.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
8Example 17.1 continued
Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made.
(a) The decay of a piece of wood buried in soil.(b) The formation of sodium, Na(s), and chlorine, Cl2(g), in a vigorously stirred
aqueous solution of sodium chloride, NaCl(aq).(c) The ionization of hydrogen chloride when HCl(g) dissolves in liquid water.
Exercise 17.1A
Indicate whether each of the following processes is spontaneous or nonspontaneous. Comment on cases where a clear determination cannot be made.
(a) The dissolution of 1.00 mL of liquid ethanol in 100 mL of liquid water.
(b) The formation of lime, CaO(s), and carbon dioxide gas at 1 atm pressure from
limestone, CaCO3(s), at 600 °C.
(c) The condensation of CO2(g) at 0.50 atm to produce CO2(s).
(d) The displacement of H2(g) at 1 atm from 1 M HCl(aq) by Cu(s).
Exercise 17.1B
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
9
• Early chemists proposed that spontaneous chemical reactions should occur in the direction of decreasing energy.
• It is true that many exothermic processes are spontaneous and that many endothermic reactions are nonspontaneous.
• However, enthalpy change is not a sufficient criterion for predicting spontaneous change …
Spontaneous Change (cont’d)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
10
Spontaneous Change (cont’d)
Water falling (higher to lower potential energy) is
a spontaneous process.
H2 and O2 combine spontaneously to form water
(exothermic) BUT …… liquid water vaporizes
spontaneously at room temperature; an
endothermic process.
Conclusion: enthalpy alone is not a sufficient criterion for prediction of spontaneity.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
11
The Concept of Entropy
When the valve is opened … … the gases mix spontaneously.
• There is no significant enthalpy change.
• Intermolecular forces are negligible.
• So … why do the gases mix?
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
12
• The other factor that drives reactions is a thermodynamic quantity called entropy.
• Entropy is a mathematical concept that is difficult to portray visually.
• The total energy of the system remains unchanged in the mixing of the gases …
The Concept of Entropy (cont’d)
• … but the number of possibilities for the distribution of that energy increases.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
13
Benzene and toluene have similar intermolecular forces, so there is no
enthalpy change when they are mixed.
They mix completely because entropy of the mixture is
higher than the entropies of the two substances separated.
Formation of an Ideal Solution
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
14
Increase in Entropy in theVaporization of Water
Evaporation is spontaneous because of the increase in entropy.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
15
• The spreading of the energy among states, and increase of entropy, often correspond to a greater physical disorder at the microscopic level (however, entropy is not “disorder”).
• There are two driving forces behind spontaneous processes: the tendency to achieve a lower energy state (enthalpy change) and the tendency for energy to be distributed among states (entropy).
• In many cases, however, the two factors work in opposition. One may increase and the other decrease or vice versa. In these cases, we must determine which factor predominates.
The Concept of Entropy
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
16
• The difference in entropy (S) between two states is the entropy change (S).
• The greater the number of configurations of the microscopic particles (atoms, ions, molecules) among the energy levels in a particular state of a system, the greater is the entropy of the system.
• Entropy generally increases when:– Solids melt to form liquids.– Solids or liquids vaporize to form gases.– Solids or liquids dissolve in a solvent to form
nonelectrolyte solutions.– A chemical reaction produces an increase in the
number of molecules of gases.– A substance is heated.
Assessing Entropy Change
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
17
Example 17.2Predict whether each of the following leads to an increase or decrease in the entropy of a system. If in doubt, explain why.
(a) The synthesis of ammonia:N2(g) + 3 H2(g) 2 NH3(g)
(b) Preparation of a sucrose solution:
C12H22O11(s) C12H22O11(aq)
(c) Evaporation to dryness of a solution of urea, CO(NH2)2, in water:
CO(NH2)2(aq) CO(NH2)2(s)
H2O(l)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
18
Solution(a) Four moles of gaseous reactants produce two moles of gaseous product.
Because the reaction leads to a decrease in the number of gas molecules, we predict a decrease in entropy.
(b) The sucrose molecules are confined in highly ordered crystals in the solid state, whereas they become widely distributed in the aqueous solution. The number of available energy levels for dispersal of the system’s energy increases, and we therefore predict an increase in entropy.
(c) Crystallization of the urea from an aqueous solution is the reverse of the process described in part (b), thus indicating a decrease in entropy. At the same time, however, in the evaporation of water a liquid is converted to a gas, suggesting an increase in entropy (recall Figure 17.4). Without further information, we cannot say whether the entropy of the system increases or decreases.
Example 17.2
Predict whether each of the following leads to an increase or decrease in the entropy of a system. If in doubt, explain why.
(a) The synthesis of ammonia:
(b) Preparation of a sucrose solution:
(c) Evaporation to dryness of a solution of urea, CO(NH2)2, in water:
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
19
• Sometimes it is necessary to obtain quantitative values of entropy changes.
S = qrxn/T• where qrxn is reversible heat, a state function.
Entropy Change
A reversible process can be reversed by a very small
change, as in the expansion of this gas. A reversible
process is never more than a tiny step from equilibrium.
The expansion can be reversed by allowing the sand to return, one grain at a
time.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
20
Entropy as a Function of Temperature
Entropy always increases with
temperature …
… and it increases dramatically during
a phase change.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
21
• According to the Third Law of Thermodynamics, the entropy of a pure, perfect crystal can be taken to be zero at 0 K.
• The standard molar entropy, S°, is the entropy of one mole of a substance in its standard state.
• Since entropy increases with temperature, standard molar entropies are positive—even for elements.
S = vp S°(products) – vr S°(reactants)
Standard Molar Entropies
Does the form of this equation look familiar? (remember
calculating enthalpy change from ∆Hf°?)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
22
Example 17.3Use data from Appendix C to calculate the standard entropy change at 25 °C for the Deacon process, a high-temperature, catalyzed reaction used to convert hydrogen chloride (a by-product from organic chlorination reactions) into chlorine:
4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
23Example 17.3
Use data from Appendix C to calculate the standard entropy change at 25 °C for the Deacon process, a high-temperature, catalyzed reaction used to convert hydrogen chloride (a by-product from organic chlorination reactions) into chlorine:
4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)
StrategyThis problem requires a straightforward application of Equation (17.2), keeping in mind that we should expect a decrease in entropy (∆S° < 0) because five moles of gaseous reactants yield four moles of gaseous products.
SolutionWe relate ∆S° for the reaction to the standard molar entropies, S°, of reactants and products as follows:
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
24Example 17.3 continued
AssessmentJust as we expected, ∆S° is indeed negative; entropy decreases.
Use data from Appendix C to calculate the standard molar entropy change at 25 °C for the reaction
CO(g) + H2O(g) CO2(g) + H2(g)
Note that this is reaction (c) in Exercise 17.2A, for which your answer should have beenthat you could not predict whether entropy increases or decreases.
Exercise 17.3A
The following reaction may occur in the high-temperature, catalyzed oxidation of NH3(g). Use data from Appendix C to calculate the standard molar entropy change for this reaction at 25 °C.
NH3(g) + O2(g) N2(g) + H2O(g) (not balanced)
Exercise 17.3B
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
25
• Entropy can be used as a sole criterion for spontaneous change …
• … but the entropy change of both the system and its surroundings must be considered.
• The Second Law of Thermodynamics establishes that all spontaneous processes increase the entropy of the universe (system and surroundings).
• If entropy increases in both the system and the surroundings, the process is spontaneous.– Is it possible for a spontaneous process to exhibit a decrease in
entropy? Yes, if the surroundings ____________________.
The Second Law of Thermodynamics
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
26
Therefore:
Since heat is exchanged
with the surroundings:
For a spontaneous process:
Multiply by –T:
The Second Law of Thermodynamics
∆Stotal = ∆ Suniverse = ∆ Ssystem + ∆ Ssurroundings
∆ Suniverse > 0
qsurr = –qp = ∆ Hsys
qsurr –∆Hsys∆ Ssurr = –––– = –––––– T T
and:
∆Hsys∆ Suniv = ∆ Ssys – –––––– T
–T ∆ Suniv = ∆ Hsys – T ∆ Ssys
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
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• What is the significance of: –T ∆ Suniv = ∆ Hsys – T ∆ Ssys ?
• The entropy change of the universe—our criterion for spontaneity—has now been defined entirely in terms of the system.
• The quantity –T ∆ Suniv is called the free energy change (G).
• For a process at constant temperature and pressure:
Gsys = Hsys – TSsys
Free Energy and Free Energy Change
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
28
• If G < 0 (negative), a process is spontaneous.
• If G > 0 (positive), a process is nonspontaneous.
• If G = 0, neither the forward nor the reverse process is favored; there is no net change, and the process is at equilibrium.
Free Energy and Free Energy Change
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
29
Case 3 illustrated
∆ H is (+) and is more-or-less constant with T.
Since ∆ S is (+), the slope T ∆ S is also (+).
At low T, the size of T ∆ S is small, and ∆ H (+)
predominates.
At high T, the size of T ∆ S is large, and –T ∆ S
predominates.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
30
Example 17.4Predict which of the four cases in Table 17.1 you expect to apply to the following reactions.(a) C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)
H = -2540kJ(b) Cl2(g) 2 Cl(g)
Example 17.5 A Conceptual ExampleMolecules exist from 0 K to a few thousand kelvins. At elevated temperatures, they dissociate into atoms. Use the relationship between enthalpy and entropy to explain why this is to be expected.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
31Example 17.4
Predict which of the four cases in Table 17.1 you expect to apply to the following reactions.
(a) C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) ∆H = –2540 kJ
(b) Cl2(g) 2 Cl(g)
Solution(a) The reaction is exothermic (∆H < 0). Because 12 moles of gaseous products
replace 6 moles of gaseous reactant, we expect ∆S > 0. Because ∆H < 0 and ∆S > 0, we expect this reaction to be spontaneous at all temperatures (case 1 of Table 17.1).
(b) One mole of gaseous reactant produces two moles of gaseous product, and therefore we expect ∆S > 0. No value is given for ∆H. However, we can determine the sign of ∆H by examining what occurs during the reaction. Bonds are broken, but no new bonds are formed. Because molecules must absorb energy for bonds to break, the reaction must be endothermic. Thus, ∆H must be positive. Because ∆H and ∆S are both positive, this reaction is an example of case 3 in Table 17.1. We expect the reaction to be nonspontaneous at low temperatures and spontaneous at high temperatures.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
32Example 17.4 continued
Predict which of the four cases in Table 17.1 is likely to apply to each of the followingreactions.
(a) N2(g) + 2 F2(g) N2F4(g) ∆H = –7.1 kJ (b) COCl2(g) CO(g) + Cl2(g) ∆H = +110.4 kJ
Exercise 17.4A
Predict which of the four cases in Table 17.1 is likely to apply to the following reaction.Use the value –93.85 kJ/mol for as necessary.
Br2(g) + BrF3(g) 3 BrF(g)
Exercise 17.4B
for BrF(g) and additional data from Appendix C
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
33
• The standard free energy change, G°, of a reaction is the free energy change when reactants and products are in their standard states.
• The standard free energy of formation, Gf°, is the free energy change for the formation of 1 mol of a substance in its standard state from the elements in their standard states.
G° = vp Gf°(products) – vr Gf°(reactants)
Standard Free Energy Change, ∆G°
The form of this equation should appear very familiar by now!
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
34
Example 17.6Calculate G at 298 K for the reaction
4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)
∆H° = -114.4 kJ
(a) using the Gibbs equation (17.8) and (b) from standard free energies of formation.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
35Example 17.6
Calculate ∆G° at 298 K for the reaction
4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g) ∆H° = –114.4 kJ
(a) using the Gibbs equation (17.8) and (b) from standard free energies of formation.
Strategy(a) To use the standard-state form of the Gibbs equation, we need values of both
∆H° and ∆S°. The value of ∆H° is given, and we can obtain ∆S° from standard molar entropies, which we did in Example 17.3.
(b) We must look up standard free energies of formation in Appendix C and use them in Equation (17.9).
Solution(a) The value of ∆S° we determined in Example 17.3 was –128.8 J K–1. To express
∆H° and ∆S° in the same energy unit, we can convert –128.8 J K–1 to –0.1288 kJ K–1. Then
∆G° = ∆H° – T∆S° = –114.4 kJ – [298 K x (–0.1288 kJ K–1)]
= –114.4 kJ + 38.4 kJ = –76.0 kJ
(b) Equation (17.9) takes the form
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
36Example 17.6 continued
AssessmentThe principal check on part (a) is that the T∆S° product should be about 40 kJ (that is, 300 x 0.13), which, when added to the value of ∆H°, should yield a value of ∆G° ≈ –75 kJ. The principal check on part (b) is that it should give the same answer as part (a), and it does.
Use the standard-state form of the Gibbs equation to determine the standard free energychange at 25 °C for these reactions:
(a) 2 NO(g) + O2(g) 2 NO2(g) ∆H° = –114.1 kJ ∆S° = –146.2 J K–1
(b) 2 CO(g) + 2 NH3(g) C2H6(g) + 2 NO(g) ∆H° = +409.0 kJ∆S° = –129.1 J K–1
Exercise 17.6A
Use standard free energies of formation from Appendix C to determine the standard freeenergy change at 25 °C for these reactions:
(a) CS2(l) + 2 S2Cl2(g) CCl4(l) + 6 S(s)
(b) NH3(g) + O2(g) N2(g) + H2O(g) (not balanced)
Exercise 17.6B
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
37
• At equilibrium, G = 0 (reaction is neither spontaneous nor nonspontaneous).
• Therefore, at the equilibrium temperature, the free energy change expression becomes:
H = TSand S = H/T
• Trouton’s rule states that the entropy change is about the same when one mole of a substance is converted from liquid to vapor (at the normal boiling point).
• S°vapn for many substances is about 87 J mol–1 K–1.
• This rule works best with nonpolar substances.
• It generally fails for liquids with a more ordered structure, such as those with extensive hydrogen bonding.
Free Energy Change and Equilibrium
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
38
Illustrating Trouton’s Rule
The three substances have different entropies and different boiling
points, but S of vaporization is about the same for all three.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
39
Example 17.7At its normal boiling point, the enthalpy of vaporization of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What should its approximate normal boiling point temperature be?
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
40Example 17.7
At its normal boiling point, the enthalpy of vaporization of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What should its approximate normal boiling point temperature be?
StrategyTrouton’s rule provides an approximate value (87 J mol–1 K–1) for the molar entropy of vaporization of a nonpolar liquid at its normal boiling point. Pentadecane, an alkane, is a nonpolar liquid. When we apply Equation (17.10), we will solve for the boiling point, Tbp.
Solution
AssessmentTo test the validity of our estimate, we could consult a chemistry handbook. We would find the experimentally determined boiling point of pentadecane to be 543.8 K. Our estimate is within about 5% of the true value.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
41
Raoult’s Law Revisited
Entropy of vaporization of the solvent is about the same in each case,
which means …
A pure solvent has a lower entropy than a solution containing
the solvent.
… higher entropy for the vapor from the
solution than from the pure solvent.
Entropy of a vapor increases if the vapor expands into a
larger volume—lower vapor pressure.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
42
• G = 0 is a criterion for equilibrium at any temperature.
• G° = 0 is a criterion for equilibrium at a single temperature, that temperature at which the equilibrium state has all reactants and products in their standard states.
G and Go are related through the reaction quotient, Q:
G = G° + RT ln Q
• When G = 0, then Q = Keq, and the equation becomes:
G° = RT ln Keq
Relationship of G° to the Equilibrium Constant, Keq
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
43
• The concentrations and partial pressures we have used in Keq are approximations.
• Activities (a) are the correct variables for Keq. But activities are very difficult to determine. That is why we use approximations.
• For pure solid and liquid phases: a = 1.
• For gases: Assume ideal gas behavior, and replace the activity by the numerical value of the gas partial pressure (in atm).
• For solutes in aqueous solution: Replace solute activity by the numerical value of the solute molarity (M).
The Equilibrium Constant, Keq
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
44
Example 17.8Write the equilibrium constant expression, Keq, for the oxidation of chloride ion by manganese dioxide in an acidic solution:
MnO2(s) + 4 H+(aq) + 2 Cl–(aq)
Mn2+(aq) + Cl2(g) + 2 H2O(l)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
45
StrategyFirst, we can write the Keq expression in terms of activities (a), and then we can substitute the numerical values of the appropriate quantities for these activities in accordance with the conventions just introduced.
Example 17.8
Write the equilibrium constant expression, Keq, for the oxidation of chloride ion by manganese dioxide in an acidic solution:
SolutionWe start with the equation
In substituting for these activities according to our conventions,
1. No term will appear for MnO2(s) because it is a pure solid phase with a = 1. 2. For Mn2+, H+, and Cl–, ionic species in dilute aqueous solution, activities are
replaced by the numerical values of the molarities.3. For Cl2(g), the activity is replaced by the numerical value of the partial pressure.4. Because H2O is the preponderant species in the dilute aqueous solution, the
activity of H2O is essentially the same as in pure liquid water, which means that a = 1.
Making these substitutions, we get
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
46
• Rearranging Equation (17.12):
Calculating Equilibrium Constants
G°ln Keq = ––––– RT
G° = RT ln Keq
• The units of G° and R must be consistent; both must use kJ or both must use J.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
47
Example 17.9Determine the value of Keq at 25 °C for the reaction
2 NO2(g) N2O4(g)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
48
StrategyOur first task is to obtain a value of ∆G°, which we can do from tabulated standard free energies of formation in Appendix C. Then we can calculate Keq using Equation (17.12).
Example 17.9
Determine the value of Keq at 25 °C for the reaction
SolutionThe standard free energies of formation of N2O4(g) and NO2(g) from Appendix C are substituted into Equation (17.9).
Next we list the data required for Equation (17.12), first changing the unit of ∆G° from kJ to J/mol. Then we can rearrange Equation (17.12) to solve for Keq.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
49Example 17.9 continued
AssessmentIn this problem, the value of ∆G° is negative, and so ln Keq must be greater than 1. If we had made a rather common error and forgotten about the minus sign on the right in Equation (17.12), the value of Keq would have been less than 1 (that is, e–1.93 = 0.145)—a tip-off that an error had been made.
Use data from Appendix C to determine Keq at 25 °C for the reaction
Exercise 17.9A
Use data from Appendix C to determine Keq at 25 °C for the reaction If NO(g) at 1 atm is allowed to react with an
excess of Br2(l), what will be the partial pressures of NO(g) and NOBr(g) at equilibrium?
Exercise 17.9B
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
50
The Sign and Magnitude of G°
Large, negative G°; equilibrium lies far to right.
Large, positive G°; equilibrium
lies far to left.
Intermediate G°;
equilibrium lies in intermediate position.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
51
• To obtain equilibrium constants at different temperatures, it will be assumed that H° does not change much with temperature.
• To obtain Keq at the desired temperature, the van’t Hoff equation is used:
The Dependence of G° and Keq on Temperature
H°ln Keq = ––––– + constant or RT
K2 H° 1 1ln ––– = –––– ––– – ––– K1 R T1 T2
[ ]• The form used depends on whether we have a single value
of Keq available, or multiple values.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
52
Example 17.10Consider this reaction at 298 K:
CO(g) + H2O(g) CO2(g) + H2(g) ∆H°298 = – 41.2 kJ
Determine Keq for the reaction at 725 K.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
53
StrategyThis is the reaction for which Keq is plotted as a function of 1/T in Figure 17.13. We cannot simply use the graph to obtain ln Keq when 1/T = 1/725 K because this point falls outside the range of data plotted. However, we can select a data point listed in the caption of Figure 17.13 as our K1 and T1 for use in Equation (17.14). The other required data are given in the statement of the problem.
Example 17.10
Consider this reaction at 298 K:
Determine Keq for the reaction at 725 K.
SolutionSuppose these are the data we choose for the van’t Hoff equation:
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
54
Solution continued
Example 17.10 continued
The solution of Equation (17.14) forK2 then follows:
AssessmentNotice that in the data table accompanying Figure 17.13, the value of Keq becomes progressively smaller as the temperature increases. Because T2 is greater than the highest temperature in the table, K2 must be somewhat less than 16. The calculated value of 6.6 seems reasonable. Another point to notice in this problem is that in order to avoid an inadvertent loss of significant figures, it is best to store the value of (1/T1 – 1/T2) in your calculator rather than to write intermediate results, such as the 1.56 x 10–3 and 1.38 x 10–3 shown in the solution.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
55
Cumulative ExampleWaste silver from photographic solutions or laboratory operations can be recovered using an appropriate redox reaction. A 100.0-mL sample of silver waste is 0.200 M in Ag+(aq) and 0.0200 M in Fe3+(aq). Enough iron(II) sulfate is added to make the solution 0.200 M in Fe2+(aq). When equilibrium is established at 25 °C, how many moles of solid silver will be present?
Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
56
StrategyThe Ag(s) must come at the expense of some of the Ag+(aq) originally present. We can determine [Ag+] at equilibrium, and the amount of Ag(s) that forms will be related to the difference [Ag+]init – [Ag+]equil. The core of the equilibrium calculation will be the ICE tabular format introduced in Chapter 14, but first we need an equilibrium constant for the reaction. Because this information is not given, we must turn to Equation (17.12) relating ln Keq to ∆G° for the reaction. We can establish ∆G° from data in Appendix C. In writing out the solution to this problem, we will reverse the order of the steps described here.
SolutionTo determine ∆G°, we write the appropriate data from Appendix C under the chemical equation and then apply Equation (17.9).
Cumulative Example
Waste silver from photographic solutions or laboratory operations can be recovered using an appropriate redox reaction. A 100.0-mL sample of silver waste is 0.200 M in Ag+(aq) and 0.0200 M in Fe3+(aq). Enough iron(II) sulfate is added to make the solution 0.200 M in Fe2+(aq). When equilibrium is established at 25 °C, how many moles of solid silver will be present?
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
57Cumulative Example continued
Solution continuedThen we can use this value of ∆G° in Equation (17.12) to solve for ln Keq.
We solve for Keq by taking the exponential of both sides of the previous expression.
We could establish the direction of net change by calculating Qc, but that is not necessary. Because there is no silver metal initially, the reaction must proceed to the right to form silver metal. Now, we enter the solution concentration data into the ICE format.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
58Cumulative Example continued
AssessmentIn choosing the correct root of the quadratic equation, we were guided by the fact that x < 0.200. That is, [Ag+] at equilibrium is (0.200 – x) M, and this value must be a positive quantity. The rejected root had x = 0.6. We can verify that the equilibrium concentrations are correct by showing that Qc based on equilibrium concentrations is equal to the value of Keq used in the calculation. Here, we obtain Qc = 0.08/(0.14)2 = 4. This matches Keq = 3.3 quite well, given the rather limited precision of the calculation. The limitation in precision was introduced in calculating ∆G°, where we had to work with the difference between two numbers of nearly equal magnitude.
Chapter SeventeenGeneral Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Prentice Hall © 2005Hall © 2005
59Cumulative Example continued
Solution continuedNext we substitute the equilibrium concentrations from the ICE table into the Keq expression and express the equation in the customary quadratic format.
We now solve the quadratic equation for x, keeping in mind that x must be less than 0.200.
Then we determine the equilibrium concentrations of the ions.
Finally, we can determine the number of moles of Ag(s) displaced from solutionand appearing as Ag(s) by taking the difference between initial and equilibrium concentrations and multiplying by the volume.
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