Chemical Thermodynamics Chapter 19: Chemical Thermodynamics John D. Bookstaver, St. Charles Community College, St. Peters, MO, 2006, Prentice Hall, Inc.

ChemicalThermodynamics

Chapter 19: Chemical ThermodynamicsJohn D. Bookstaver, St. Charles Community College, St. Peters, MO,

2006, Prentice Hall, Inc.

(ppt modified for our requirements)

Chemistry, The Central Science, 10th edition AP editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten


Ch. 19 Chemical thermodynamics

• Spontaneous Reactions

• Entropy & Second Law of

Thermodynamics

• Molecular Interpretation of

Entropy

• Entropy Changes in Chemical

Reactions

• Gibb’s Free Energy

• Free energy and Temperature

• Free Energy and the

Equilibrium Constant

Resources and Activities

• Students review ch 5 notes• Textbook - chapter 19 & ppt file• Online practice quiz• POGIL activities • Chem Guy video lecture series on

thermodynamics (lectures 31-35)http://www.cosmolearning.com/courses/ap-chemistry-with-chemguy/video-lectures//

• Chemtour videos from Norton• http://www.wwnorton.com/college/

chemistry/chemistry3/ch/14/chemtours.aspx


Activities and Problem set for chapter 19 (due date_______)

POGILS (3) :• Entropy of the Universe and Free

Energy• Spontaneous Change and Entropy• Free energy and Chemical equilibrium

Study sample exercises and complete practice exercises

Chapter 19 reading guide and practice problems packet

Independent work - students to view animations & interactive activities (3 in total from Norton) and write summary notes on each. These summaries are to be included in your portfolio. Some of these may be previewed in class.

Norton Animations : Entropy, Dissolution of Ammonium Nitrate, Gibbs Free Energy

http://www.wwnorton.com/college/chemistry/chemistry3/ch/14/chemtours.aspx


© 2009, Prentice-Hall, Inc.

First Law of Thermodynamics

• You will recall from Chapter 5 that energy cannot be created nor destroyed.

• Therefore, the total energy of the universe is a constant.

• Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.



Spontaneous Processes

• Spontaneous processes are those that can proceed without any outside intervention.

• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.



Spontaneous or “thermodynamically favored” processes

Processes that are spontaneous (i.e. thermodynamically favored) in one direction are nonspontaneous in the reverse direction.



• Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.

• Above 0 C it is spontaneous for ice to melt.• Below 0 C the reverse process is spontaneous.

Spontaneous or “thermodynamically favored” processes



Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.



Irreversible Processes

• Irreversible processes cannot be undone by exactly reversing the change to the system.

• Spontaneous processes are irreversible.



Entropy

• Entropy (S) is a term coined by Rudolph Clausius in the 19th century.

• Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, .q

T



Entropy

• Entropy can be thought of as a measure of the randomness of a system.

• It is related to the various modes of motion in molecules.



Entropy

• Like total energy, E, and enthalpy, H, entropy is a state function.

• Therefore,

S = Sfinal Sinitial



Entropy

For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:

S =qrev

T



Second Law of Thermodynamics

The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.




In other words:

For reversible processes:

Suniv = Ssystem + Ssurroundings = 0

For irreversible processes:

Suniv = Ssystem + Ssurroundings > 0




These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.



Entropy on the Molecular Scale

• Ludwig Boltzmann described the concept of entropy on the molecular level.

• Temperature is a measure of the average kinetic energy of the molecules in a sample.



Entropy on the Molecular Scale• Molecules exhibit several types of motion:

– Translational: Movement of the entire molecule from one place to another.

– Vibrational: Periodic motion of atoms within a molecule.– Rotational: Rotation of the molecule on about an axis or

rotation about bonds.



Entropy on the Molecular Scale• Boltzmann envisioned the motions of a sample of

molecules at a particular instant in time.– This would be akin to taking a snapshot of all the

molecules.

• He referred to this sampling as a microstate of the thermodynamic system.



Entropy on the Molecular Scale• Each thermodynamic state has a specific number of

microstates, W, associated with it.• Entropy is

S = k lnW

where k is the Boltzmann constant, 1.38 1023 J/K.



Entropy on the Molecular Scale

• The number of microstates and, therefore, the entropy tends to increase with increases in– Temperature.– Volume.– The number of independently moving

molecules.



Entropy and Physical States

• Entropy increases with the freedom of motion of molecules.

• Therefore,

S(g) > S(l) > S(s)



Solutions

Generally, when a solid is dissolved in a solvent, entropy increases.



Entropy Changes

• In general, entropy increases when– Gases are formed from

liquids and solids;– Liquids or solutions are

formed from solids;– The number of gas

molecules increases;– The number of moles

increases.



Third Law of Thermodynamics

The entropy of a pure crystalline substance at absolute zero is 0.



Sample Exercise 19.3 Predicting the Sign of ΔS

Solution

Analyze: We are given four equations and asked to predict the sign of ΔS for each chemical reaction.Plan: The sign of ΔS will be positive if there is an increase in temperature, an increase in the volume in which the molecules move, or an increase in the number of gas particles in the reaction. The question states that the temperature is constant. Thus, we need to evaluate each equation with the other two factors in mind.Solve:(a) The evaporation of a liquid is accompanied by a large increase in volume. One mole of water (18 g) occupies about 18 mL as a liquid and if it could exist as a gas at STP it would occupy 22.4 L. Because the molecules are distributed throughout a much larger volume in the gaseous state than in the liquid state, an increase in motional freedom accompanies vaporization. Therefore, ΔS is positive.(b) In this process the ions, which are free to move throughout the volume of the solution, form a solid in which they are confined to a smaller volume and restricted to more highly constrained positions. Thus, ΔS is negative.(c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than do the molecules of a gas. Because O2 gas is converted into part of the solid product Fe2O3, ΔS is negative.(d) The number of moles of gases is the same on both sides of the equation, and so the entropy change will be small. The sign of ΔS is impossible to predict based on our discussions thus far, but we can predict that ΔS will be close to zero.

Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature:



Sample Exercise 19.3 Predicting the Sign of ΔS

Indicate whether each of the following processes produces an increase or decrease in the entropy of the system:

Answer: (a) increase, (b) decrease, (c) decrease, (d) decrease

Practice Exercise



Standard Entropies

• These are molar entropy values of substances in their standard states.

• Standard entropies tend to increase with increasing molar mass.



Standard Entropies

Larger and more complex molecules have greater entropies.



Entropy Changes

Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:

S = nS(products) — mS(reactants)

where n and m are the coefficients in the balanced chemical equation.



Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies

Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K:

N2(g) + 3 H2(g) → 2 NH3(g)

Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K:

Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g)Answers: 180.39 J/K

Practice Exercise

Solution

Analyze: We are asked to calculate the entropy change for the synthesis of NH3(g) from its constituent elements.Plan: We can make this calculation using Equation 19.8 and the standard molar entropy values for the reactants and the products that are given in Table 19.2 and in Appendix C.Solve: Using Equation 19.8, we have

Substituting the appropriate S° values from Table 19.2 yields

ΔS° = 2S°(NH3) - [S°(N2) + 3S°(H2)]

ΔS° = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K) + (3 mol)(130.6 J/mol-K)] = -198.3 J/K

Check: The value for ΔS° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction.



Entropy Change in the Universe

• The universe is composed of the system and the surroundings.

• Therefore,

Suniverse = Ssystem + Ssurroundings

• For spontaneous processes

Suniverse > 0



Gibbs Free Energy

• Energy available to do work TSuniverse is defined as the Gibbs free

energy, G.

• When Suniverse is positive, G is negative.

• Therefore, when G is negative, a process is spontaneous.



Gibbs Free Energy

1. If G is negative, the forward reaction is spontaneous.

2. If G is 0, the system is at equilibrium.

3. If G is positive, the reaction is spontaneous in the reverse direction.



Standard Free Energy Changes

Analogous to standard enthalpies of formation are standard free energies of formation, G.

f

G = nG (products) mG (reactants)f f

where n and m are the stoichiometric coefficients.



Sample Exercise 19.7 Calculating Standard Free-Energy Change from Free Energies of Formation

(a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K:

P4(g) + 6 Cl2(g) → 4 PCl3(g)

(b) What is ΔG° for the reverse of the above reaction?Solution

Analyze: We are asked to calculate the free-energy change for the indicated reaction and then to determine the free-energy change of its reverse.Plan: To accomplish our task, we look up the free-energy values for the products and reactants and use Equation 19.14: We multiply the molar quantities by the coefficients in the balanced equation, and subtract the total for the reactants from that for the products.Solve:(a) Cl2(g) is in its standard state, so ΔG°f is zero for this reactant. P4(g), however, is not in its standard state, so ΔG°f is not zero for this reactant. From the balanced equation and using Appendix C, we have:



Sample Exercise 19.7 Calculating Standard Free-Energy Change from Free Energies of Formation

By using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane:CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g).Answer: –800.7 kJ

Practice Exercise

Solution (continued)

The fact that ΔG° is negative tells us that a mixture of P4(g), Cl2(g), and PCl3(g) at 25 °C, each present at a partial pressure of 1 atm, would react spontaneously in the forward direction to form more PCl3. Remember, however, that the value of ΔG° tells us nothing about the rate at which the reaction occurs.(b) Remember that ΔG = G (products) – G (reactants). If we reverse the reaction, we reverse the roles of the reactants and products. Thus, reversing the reaction changes the sign of ΔG, just as reversing the reaction changes the sign of ΔH. (Section 5.4) Hence, using the result from part (a):

4 PCl3(g) → P4(g) + 6 Cl2(g) ΔG° = +1102.8 kJ



Free Energy Changes

At temperatures other than 25°C,

G° = H TS

How does G change with temperature?



Sample Exercise 19.6 Calculating Free-Energy Change from ΔH°, T, ΔS°

Calculate the standard free energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K:

N2(g) + O2(g) → 2 NO(g)

given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these circumstances?

A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG°.Is the reaction spontaneous under these conditions? Answers: ΔG° = –14.7 kJ; the reaction is spontaneous.

Practice Exercise

Solution

Analyze: We are asked to calculate ΔG° for the indicated reaction (given ΔH°, ΔS° and T) and to predict whether the reaction is spontaneous under standard conditions at 298 K.Plan: To calculate ΔG°, we use Equation 19.12, ΔG° = ΔH° – T ΔS°. To determine whether the reaction is spontaneous under standard conditions, we look at the sign of ΔG°.Solve:

Because ΔG° is positive, the reaction is not spontaneous under standard conditions at 298 K.Comment: Notice that we had to convert the units of the T ΔS° term to kJ so that they could be added to the ΔH° term, whose units are kJ.



Free Energy and Temperature

• There are two parts to the free energy equation:H— the enthalpy term– TS — the entropy term

• The temperature dependence of free energy, then comes from the entropy term.



Free Energy and Temperature



Free Energy and Equilibrium

Under any conditions, standard or nonstandard, the free energy change can be found this way:

G = G + RT lnQ

(Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)



Free Energy and Equilibrium

• At equilibrium, Q = K, and G = 0.

• The equation becomes

0 = G + RT lnK

• Rearranging, this becomes

G = RT lnK

or,

K = e-GRT



Sample Exercise 19.10 Relating ΔG to a Phase change at Equilibrium

As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation 19.12 to estimate the normal boiling point of CCl4.Solution

Analyze: (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous CCl4 at the normal boiling point. (b) We must determine the value of ΔG° for CCl4, in equilibrium with its vapor at the normal boiling point. (c) We must estimate the normal boiling point of CCl4, based on available thermodynamic data.Plan: (a) The chemical equation will merely show the change of state of CCl4 from liquid to solid. (b) We need to analyze Equation 19.16 at equilibrium (ΔG = 0). (c) We can use Equation 19.12 to calculate T when ΔG = 0.

Solve: (a) The normal boiling point of CCl4 is the temperature at which pure liquid CCl4 is in equilibrium with its vapor at a pressure of 1 atm:(b) At equilibrium ΔG = 0. In any normal boiling-point equilibrium both the liquid and the vapor are in their standard states (Table 19.2). Consequently, Q = 1, Q = 0, and ΔG = ΔG° for this process. Thus, we conclude that ΔG = 0 for the equilibrium involved in the normal boiling point of any liquid. We would also find that ΔG = 0 for the equilibria relevant to normal melting points and normal sublimation points of solids.




Solution (continued)(c) Combining Equation 19.12 with the result from part (b), we see that the equality at the normal boiling point, Tb, of CCl4(l) or any other pure liquid is

Solving the equation for Tb, we obtain

Strictly speaking, we would need the values of ΔH° and ΔS° for the equilibrium between CCl4(l) and CCl4(g) at the normal boiling point to do this calculation. However, we can estimate the boiling point by using the values of ΔH° and ΔS° for CCl4 at 298 K, which we can obtain from the data in Appendix C and Equations 5.31 and 19.8:

Notice that, as expected, the process is endothermic (ΔH > 0)and produces a gas in which energy can be more spread out (ΔS > 0). We can now use these values to estimate Tb for CCl4(l):

Note also that we have used the conversion factor between J and kJ to make sure that the units of ΔH° and ΔS° match.




Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3.)Answers: 330 K

Practice Exercise

Solution (continued)

Check: The experimental normal boiling point of CCl4(l) is 76.5 °C. The small deviation of our estimate from the experimental value is due to the assumption that ΔH° and ΔS° do not change with temperature.

Chemical Thermodynamics Chapter 19: Chemical Thermodynamics John D. Bookstaver, St. Charles Community College, St. Peters, MO, 2006, Prentice Hall, Inc.

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