Chapter 19 Lecture- Thermodynamics

Chapter 19- Chemical Thermodynamics

Sections 19.1 - 19.3 Spontaneous processes & entropy

ChemicalThermodynamics

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.




• 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• 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




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

A process is either spontaneous, non-spontaneous, or at equilibrium.



Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.


Melting a cube of water ice at −10°C and atmospheric pressure is a ________ process.

1. Spontaneous

2. Nonspontaneous


Correct Answer:

1. Spontaneous

2. Nonspontaneous

Melting of an ice cube at 1 atm pressure will only be spontaneous above the freezing point (0°C).


Spontaneous Processes• Processes that are spontaneous at one

temperature may be nonspontaneous at other temperatures.




• Above 0°C it is spontaneous for ice to melt.




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


No. Nonspontaneous processes can occur with some external assistance.

SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes

Predict whether these processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150°C is added to water at 40°C, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and O2(g). (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1°C.

SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes

Predict whether these processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150°C is added to water at 40°C, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and O2(g). (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1°C.

Analyze: We are asked to judge whether each process will proceed spontaneously in the direction indicated, in the reverse direction, or in neither direction.Plan: We need to think about whether each process is consistent with our experience about the natural direction of events or whether it is the reverse process that we expect to occur.

Solve: (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. In this instance heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gas bubbling up out of water! Rather, the reverse process—the reaction of H2 and O2 to form H2O—is spontaneous once initiated by a spark or flame. (c) By definition, the normal boiling point is the temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. Neither the condensation of benzene vapor nor the reverse process is spontaneous. If the temperature were below 80.1°C, condensation would be spontaneous.


PRACTICE EXERCISEUnder 1 atm pressure CO2(s) sublimes at –78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100ºC and 1 atm pressure?


PRACTICE EXERCISEUnder 1 atm pressure CO2(s) sublimes at –78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100ºC and 1 atm pressure?

Answer: No, the reverse process is spontaneous at this temperature.


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.



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

• Spontaneous processes are irreversible.


No


Entropy


Entropy

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


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

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


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


Entropy

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


Entropy


• Therefore,


Entropy


• Therefore, ΔS = 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:


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:

ΔS =qrevT


• ΔS depends not merely on q but on qrev. There is only one reversible isothermal path between two states regardless of the number of possible paths.


SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change

The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?


SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change

The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?

SolutionAnalyze: We first recognize that freezing is an exothermic process; heat is transferred from the system to the surroundings when a liquid freezes (q < 0). The enthalpy of fusion is ΔΗ for the melting process. Because freezing is the reverse of melting, the enthalpy change that accompanies the freezing of 1 mol of Hg is –ΔΗfusion = 2.29 kJ/mol.

Plan: We can use –ΔΗfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg:

We can use this value of q as qrev in Equation 19.2. We must first, however, convert the temperature to K:

Check: The entropy change is negative because heat flows from the system making qrev negative.Comment: The procedure we have used here can be used to calculate ΔS for other isothermal phase changes, such as the vaporization of a liquid at its boiling point.

Solve: We can now calculate the value of ΔSsys:


PRACTICE EXERCISEThe normal boiling point of ethanol, C2H5OH, is 78.3°C (Figure 11.24), and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point?

SAMPLE EXERCISE 19.2 continued


PRACTICE EXERCISEThe normal boiling point of ethanol, C2H5OH, is 78.3°C (Figure 11.24), and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point?


Answer: –163 J/K


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:

ΔSuniv = ΔSsystem + ΔSsurroundings = 0

For irreversible processes:ΔSuniv = ΔSsystem + ΔSsurroundings > 0



These last truths mean that as a result of all spontaneous processes

the entropy of the universe increases.


The entropy of the universe for the process must increase by a greater amount than the entropy of the system decreases.


Entropy on the Molecular Scale



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



• 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 σ bonds.


A molecule can vibrate (atoms moving relative to one another) and rotate (tumble); a single atom can only rotate.


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.



microstates, W, associated with it.• Entropy is




S = k lnW




S = k lnW where k is the Boltzmann constant, 1.38 × 10−23 J/K.


S = k lnW


S = 0

S = k lnW



• The change in entropy for a process, then, is

ΔS = k lnWfinal − k lnWinitial

lnWfinallnWinitial

ΔS = k ln



• The change in entropy for a process, then, is

ΔS = k lnWfinal − k lnWinitial

lnWfinallnWinitial

ΔS = k ln

• Entropy increases with the number of microstates in the system.



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

molecules.


Entropy and Physical States



• Entropy increases with the freedom of motion of molecules.




• Therefore,




• Therefore, S(g) > S(l) > S(s)


Solutions

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


Entropy Changes


Entropy Changes

• In general, entropy increases whenGases are formed from

liquids and solids.Liquids or solutions are

formed from solids.The number of gas

molecules increases.The number of moles

increases.


The sign of entropy change, ΔS, associated with the boiling of water is _______.

1. Positive

2. Negative

3. Zero


Correct Answer:

Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; ΔS must be positive.

1. Positive

2. Negative

3. Zero


SAMPLE EXERCISE 19.3 Predicting the Sign of ΔS

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

SolutionAnalyze: 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.

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

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 22.4 L as a gas at STP. 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.


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


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

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


SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy

Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.




Solution Analyze: We need to select the system in each pair that has the greater entropy.Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.





Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms cannot.





PRACTICE EXERCISEChoose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.








Answers: (a) 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP, (d) 2 mol NO2(g) of at STP

PRACTICE EXERCISEChoose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.Answers: (a) 1 mol of H2(g) at 100°C and 0.5 atm,

(b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP, (d) 2 mol NO2(g) of at STP


Which of the following would be expected to have the highest entropy?

1. 1 mole N2(l)2. 1 mole N2(g)3. 1 mole Ar(g)4. 1 mole Ar(s)


Correct Answer:

Given the same amounts, gases have higher entropy than liquids, which have higher entropy than solids. Of the gases, N2 has greater entropy than Ar because N2 is a molecule that can store energy in ways not possible for Ar.

1. 1 mole N2(l)2. 1 mole N2(g)3. 1 mole Ar(g)4. 1 mole Ar(s)


Third Law of Thermodynamics

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


It must be a perfect crystal at 0K (absolute zero).


Sections 19.4 - 19.6All about Free Energy


Standard Entropies


Standard Entropies

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


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.


1.Butane2.Isobutane3.Pentane4.Isopentane5.Neopentane

Which substance will have the highest

entropy?


Predict whether entropy increases, decreases, or stays constant for: 2 NO2(g) → 2 NO(g) + O2(g)

1. Increases2. Decreases3. Stays constant


1. Increases2. Decreases3. Stays constant

The entropy increases because two molecules of gas react to form three molecules of gas, leading to more disorder.

Predict whether entropy increases, decreases, or stays constant for: 2 NO2(g) → 2 NO(g) + O2(g)


Which process will have a negative ΔS?

1.Dissolving a substance2.Melting3.Falling leaves4.Oxidation of a metal 5.Decomposition of water

into H2 and O2 Dissolution of NaCl


Which process will have a negative ΔS?

1.Dissolving a substance2.Melting3.Falling leaves4.Oxidation of a metal 5.Decomposition of water

into H2 and O2Dissolution of NaCl


Entropy Changes


Entropy Changes

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

ΔS° = ΣnΔS°(products) - ΣmΔS°(reactants)

where n and m are the coefficients in the

balanced chemical equation.


SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies

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


SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies

SolutionAnalyze: 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.

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

Solve: Using Equation 19.8, we have

Substituting the appropriate S° values from Table 19.2 yields

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.


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

ChemicalThermodynamicsAnswer: 180.39 J/K

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


Calculate ΔS° for the equation below using the standard entropy data given:

2 NO(g) + O2(g) → 2 NO2(g)ΔS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205.

1. +176 J/mol2. +147 J/mol

3. −147 J/mol4. −76 J/mol


Correct Answer:

ΔS° = 2(240) − [2(211) + 205]ΔS° = 480 − [627]

ΔS° = − 147

( ) ( )ΣΣ °° Δ−Δ=°Δ reactantsproducts SSS

1. +176 J/mol2. +147 J/mol

3. −147 J/mol4. −76 J/mol


Entropy Changes in Surroundings



• Heat that flows into or out of the system changes the entropy of the surroundings.




• For an isothermal process:

ΔSsurr =−qsys

T




• For an isothermal process:

ΔSsurr =−qsys

T

• At constant pressure, qsys is simply ΔH° for the system.


ΔSsurr =−qsys

T



always increase

ΔSsurr =−qsys

T



Entropy Change in the Universe



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




• Therefore,




• Therefore,ΔSuniverse = ΔSsystem + ΔSsurroundings





• For spontaneous processes





• For spontaneous processes ΔSuniverse > 0



• This becomes: ΔSuniverse = ΔSsystem + Multiplying both sides by −T, −TΔSuniverse = ΔHsystem − TΔSsystem

−ΔHsystemT


Gibbs Free Energy


Gibbs Free Energy

• −TΔSuniverse is defined as the Gibbs free energy, ΔG.


Gibbs Free Energy


• When ΔSuniverse is positive, ΔG is negative.


Gibbs Free Energy


• When ΔSuniverse is positive, ΔG is negative.

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


Entropy of universe increases and free energy of the system decreases.


Gibbs Free Energy


Gibbs Free Energy1. If ΔG is negative, the

forward reaction is spontaneous.




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




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

3. If ΔG 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, ΔG°.

f


Standard Free Energy Changes

Analogous to standard enthalpies of formation are standard free energies of formation, ΔG°.

f

ΔG° = ΣnΔG°(products) − ΣmΔG°(reactants)f f

where n and m are the stoichiometric coefficients.


Calculate ΔG° for the equation below using the Gibbs free energy data given:

2 SO2(g) + O2(g) 2 SO3(g)ΔG° values (kJ): SO2(g) = −300.4, SO3(g) = −370.4

1. +70 kJ2. +140 kJ3. −140 kJ4. −70 kJ


Correct Answer:

( ) ( )∑ ∑ °° −=° reactantsproducts ff GGG ΔΔΔ

ΔG° = (2 × −370.4) − [(2 × −300.4) + 0]ΔG° = −740.8 − [−600.8] = −140

1. +70 kJ2. +140 kJ3. −140 kJ4. −70 kJ


It indicates the process has taken place under standard conditions.



298K = 25°C this is NOT STP (273K)



298K = 25°C this is NOT STP (273K)1 atm for gases



298K = 25°C this is NOT STP (273K)1 atm for gases1M for solutions


Calculate ΔG° for the equation below using the thermodynamic data given:

N2(g) + 3 H2(g) → 2 NH3(g)

ΔH° (NH3) = −46 kJ; ΔS° values (J/mol-K) are

NH3 = 192.5, N2 = 191.5, H2 = 130.6.

1. +33 kJ2. +66 kJ

3. −66 kJ4. −33 kJ


Correct Answer:

ΔH° = −92 kJΔS° = −198.7J/mol − KΔG° = ΔH° − TΔS°

ΔG° = (−92) − (298)(−0.1987)ΔG° = (−92) + (59.2) = −33

1. +33 kJ2. +66 kJ3. −66 kJ4. −33 kJ


SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change from Free Energies of Formation

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

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


SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change

from Free Energies of Formation

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

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

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 19.13: We multiply the molar quantities by the coefficients in the balanced equation, and subtract the total for the reactants from that for the products.

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.

Solve: (a) Cl2(g) is in its standard state, so is zero for this reactant. P4(g), however, is not in its standard state, so is not zero for this reactant. From the balanced equation and using Appendix C, we have:

(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):


PRACTICE EXERCISEBy using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane:


Answer: –800.7 kJ

PRACTICE EXERCISEBy using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane:

SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG°

(a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct?

In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K:


SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG°

(a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct?

In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K:

Solution Analyze: In part (a) we must predict the value for ΔG° relative to that for ΔH° on the basis of the balanced equation for the reaction. In part (b) we must calculate the value for ΔG° and compare with our qualitative prediction.Plan: The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction (Equation 19.11), so under standard conditions:

ΔG° = ΔH° – TΔS°

To determine whether ΔG° is more negative or less negative than ΔH° we need to determine the sign of the term TΔS°. T is the absolute temperature, 298 K, so it is a positive number. We can predict the sign of ΔS° by looking at the reaction.Solve: (a) We see that the reactants consist of six molecules of gas, and the products consist of three molecules of gas and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By using the general rules we discussed in Section 19.3, we would expect a decrease in the number of gas molecules to lead to a decrease in the entropy of the system—the products have fewer accessible microstates than the reactants. We therefore expect ΔS° and TΔS° to be negative numbers. Because we are subtracting TΔS°, which is a negative number, we would predict that ΔG° is less negative than ΔH°.

(b) Using Equation 19.13 and values from Appendix C, we can calculate the value of ΔG°:

Notice that we have been careful to use the value of as in the calculation of ΔH values, the phases of the reactants and products are important. As we predicted, ΔG° is less negative than ΔH° because of the decrease in entropy during the reaction.


PRACTICE EXERCISEConsider the combustion of propane to form CO2(g) and H2O(g) at 298 K:

Would you expect ΔG° to be more

negative or less negative than ΔH°?


Answer: more negative

PRACTICE EXERCISEConsider the combustion of propane to form CO2(g) and H2O(g) at 298 K:

Would you expect ΔG° to be more

negative or less negative than ΔH°?


Free Energy Changes


Free Energy Changes

At temperatures other than 25°C,

ΔG° = ΔH° − TΔS°

How does ΔG° change with temperature?


Free Energy and Temperature



• There are two parts to the free energy equation:ΔH°— the enthalpy termTΔS° — the entropy term

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


Upon heating, limestone (CaCO3) decomposes to CaO and CO2. Speculate on the sign of ΔH and ΔS for this process.

CaCO3(s) CaO(s) + CO2(g)

Limestone CaveCarlsbad Caverns, New Mexico


ΔH < TΔS


ΔH < TΔS

ΔG(Negative/spontaneous)=


ΔH < TΔS


ΔH (Positive/vaporization) - TΔS (positive/liquid to gas)


ΔH < TΔS


ΔH (Positive/vaporization) - TΔS (positive/liquid to gas) ↑must be larger


Both the entropy, ΔS° and the enthalpy ΔH° have positive values. Near absolute zero (0 K), will the process be spontaneous or nonspontaneous?

1. Spontaneous

2. Nonspontaneous


Correct Answer:

At low temperatures, the enthalpy term dominates the entropy term, leading to a positive ΔG and a nonspontaneous process.

1. Spontaneous

2. Nonspontaneous


A reaction will be spontaneous at lower temperatures when the sign

of ΔH and ΔS is

SAMPLE EXERCISE 19.8 Determining the Effect of Temperature on Spontaneity

Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values of ΔG° for the reaction at 25°C and 500°C.

The Haber process for the production of ammonia involves the equilibrium


(b) We calculated the value of ΔH° in Sample Exercise 15.14, and the value of ΔS° was determined in Sample Exercise 19.5: ΔH° = –92.38 kJ and ΔS° = –198.4 J/K. If we assume that these values don’t change with temperature, we can calculate ΔG° at any temperature by using Equation 19.20. At T = 298 K we have:

Notice that we have been careful to convert –TΔS° into units of kJ so that it can be added to ΔH°, which has units of kJ.

Comment: Increasing the temperature from 298 K to 773 K changes ΔG° from –33.3 kJ to +61 kJ. Of course, the result at 773 K depends on the assumption that ΔH° and ΔS° do not change with temperature. In fact, these values do change slightly with temperature. Nevertheless, the result at 773 K should be a reasonable approximation. The positive increase in ΔG° with increasing T agrees with our prediction in part (a) of this exercise. Our result indicates that a mixture of N2(g), H2(g), and NH3(g), each present at a partial pressure of 1 atm, will react spontaneously at 298 K to form more NH3(g). In contrast, at 773 K the positive value of ΔG° tells us that the reverse reaction is spontaneous. Thus, when the mixture of three gases, each at a partial pressure of 1 atm, is heated to 773 K, some of the NH3(g) spontaneously decomposes into N2(g) and H2(g).

Solve: (a) Equation 19.20 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –TΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negative, the term –TΔS° is positive and grows larger with increasing temperature. As a result, ΔG° becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature.

Solution Analyze: In part (a) we are asked to predict the direction in which ΔG° for the ammonia synthesis reaction changes as temperature increases. In part (b) we need to determine ΔG° for the reaction at two different temperatures.Plan: In part (a) we can make this prediction by determining the sign of ΔS° for the reaction and then using that information to analyze Equation 19.20. In part (b) we need to calculate ΔH° and ΔS° for the reaction by using the data in Appendix C. We can then use Equation 19.20 to calculate ΔG°.

PRACTICE EXERCISE(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the following reaction: (b) Using the values obtained in part (a), estimate ΔG° at 400 K.

Answer: (a) ΔH° = –196.6 kJ, ΔS° = –189.6 J/K; (b) ΔG° = –120.8 kJ

PRACTICE EXERCISE(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the following reaction: (b) Using the values obtained in part (a), estimate ΔG° at 400 K.


Section 19.7

Free Energy & Keq


Free Energy and Equilibrium



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

ΔG = ΔG° + RT lnQ

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

SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium

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 19.20 to estimate the normal boiling point of CCl4.


SAMPLE EXERCISE 19.9 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 19.20 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 19.21 at equilibrium (ΔG = 0). (c) We can use Equation 19.20 to calculate Twhen ΔG = 0.

(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). As a consequence, Q = 1, ln 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.

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:

(c) Combining Equation 19.20 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:

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

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.

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):


PRACTICE EXERCISEUse 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.)

ChemicalThermodynamicsAnswer: 330 K

PRACTICE EXERCISEUse 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.)


SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard Conditions

We will continue to explore the Haber process for the synthesis of ammonia:

Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.

SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard ConditionsWe will continue to explore the Haber process for the synthesis of ammonia:Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.

SolutionAnalyze: We are asked to calculate ΔG under nonstandard conditions.Plan: We can use Equation 19.21 to calculate ΔG. Doing so requires that we calculate the value of the reaction quotient Q for the specified partial pressures of the gases and evaluate ΔG°, using a table of standard free energies of formation.

In Sample Exercise 19.8 we calculated ΔG° = 33.3 kJ for this reaction. We will have to change the units of this quantity in applying Equation 19.21, however. In order for the units in Equation 19.21 to work out, we will use kJ/mol as our units for ΔG°, where “per mole” means “per mole of the reaction as written.” Thus,ΔG° = –33.3kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3.

Solve: Solving for the reaction quotient gives:

We can now use Equation 19.21 to calculate ΔG for these nonstandard conditions:

Comment: We see that ΔG becomes more negative, changing from –33.3 kJ/mol to –44.9 kJ/mol as the pressures of N2, H2, and NH3 are changed from 1.0 atm each (standard conditions, ΔG° ) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger negative value for ΔG indicates a larger “driving force” to produce NH3. We would have made the same prediction on the basis of Le Châtelier’s principle. • (Section 15.6) Relative to standard conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Châtelier’s principle predicts that both of these changes should shift the reaction more to the product side, thereby forming more NH3.


PRACTICE EXERCISECalculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.


PRACTICE EXERCISECalculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.

Answer: –26.0 kJ/mol


Decomposition of CaCO3 has a ΔH° = 178.3 kJ/mol and ΔS° = 159 J/mol • K. At what temperature does this become spontaneous?

1. 121°C2. 395°C3. 848°C4. 1121°C


Correct Answer:

1. 121°C2. 395°C3. 848°C4. 1121°C

T = ΔH°/ΔS° T = 178.3 kJ/mol/0.159 kJ/mol • K

T = 1121 KT (°C) = 1121 – 273 = 848



• At equilibrium, Q = K, and ΔG = 0.



• At equilibrium, Q = K, and ΔG = 0.• The equation becomes




0 = ΔG° + RT lnK




0 = ΔG° + RT lnK• Rearranging, this becomes





ΔG° = −RT lnK





ΔG° = −RT lnK or,





ΔG° = −RT lnK or,

K = e−ΔG°/RT


Suppose ΔG° is a large, positive value. What then will be the value of the equilibrium constant, K?

1. K = 02. K = 13. 0 < K < 14. K > 1


Correct Answer:

ΔG° = −RTlnK

Thus, large positive values of ΔG° lead to large negative values of lnK. The value of K itself, then, is very small, approaching zero.

1. K = 02. K = 13. 0 < K < 14. K > 1


If the temperature is decreased, what is the effect on the equilibrium constant and ΔG?

1. K increases, ΔG increases2. K decreases, ΔG decreases3. K increases, ΔG decreases4. K decreases, ΔG increases5. Temperature does not affect

equilibrium or ΔG

N2(g) + 3 H2(g) 2 NH3(g) ΔH° = -92.4 kJ/mol

Josiah Willard Gibbs (1839-1903)

SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°

Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process:

The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.

SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process:


Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.



SolutionAnalyze: We are asked to calculate K for a reaction, given ΔG°.Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form




In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG° when using Equations 19.21, 19.22, or 19.23).





Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have









We insert this value into Equation 19.23 to obtain K:







We insert this value into Equation 19.23 to obtain K:


Thermodynamics can tell us the direction and extent of a reaction, but tells us nothing about the rate at which it occurs. If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be needed to force the equilibrium toward NH3.


PRACTICE EXERCISEUse data from Appendix C to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at 298 K for the reaction


Answer: ΔG° = –106.4 kJ/mol, K = 5 × 1018

PRACTICE EXERCISEUse data from Appendix C to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at 298 K for the reaction

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

(a) Calculate the value of ΔG° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? (c) Use the values of ΔG° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will ΔG° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?

Consider the equilibria in which these salts dissolve in water to form aqueous solutions of ions:


SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

(a) Calculate the value of ΔG° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? (c) Use the values of ΔG° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will ΔG° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?

Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions:

Solution (a) We will use Equation 19.13 along with values from Appendix C to calculate the values for each equilibrium. (As we did in Section 13.1, we use the subscript “soln” to indicate that these are thermodynamic quantities for the formation of a solution.) We find


SAMPLE INTEGRATIVE EXERCISE continued

The entropy terms for the solution of the two salts are very similar. That seems sensible because each solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions. • (Section 13.1) In contrast, we see a very large difference in the enthalpy term for the solution of the two salts. The difference in the values of is dominated by the difference in the values of

(c) The solubility product, Ksp, is the equilibrium constant for the solution process. • (Section 17.4) As such, we can relate Ksp directly to by using Equation 19.23:

The value calculated for the Ksp of AgCl is very close to that listed in Appendix D.

We can calculate the Ksp values in the same way we applied Equation 19.23 in Sample Exercise 19.13. We use the values we obtained in part (a), remembering to convert them from to J/mol:

(b) We can write as the sum of an enthalpy term, and an entropy term, We can calculate the values of and by using Equations 5.31 and 19.8.

We can then calculate at T = 298 K. All these calculations are now familiar to us. The results are summarized in the following table:


(d) A soluble salt is one that dissolves appreciably in water. • (Section 4.2) The Ksp value for NaCl is greater than 1, indicating that NaCl dissolves to a great extent. The Ksp value for AgCl is very small, indicating that very little dissolves in water. Silver chloride should indeed be considered an insoluble salt.

SAMPLE INTEGRATIVE EXERCISE continued

(e) As we expect, the solution process has a positive value of ΔS for both salts (see the table in part b). As such, the entropy term of the free-energy change, is negative. If we assume that and do not change much with temperature, then an increase in T will serve to make more negative. Thus, the driving force for dissolution of the salts will increase with increasing T, and we therefore expect the solubility of the salts to increase with increasing T. In Figure 13.17 we see that the solubility of NaCl (and that of nearly any salt) increases with increasing temperature. • (Section 13.3)

Chapter 19 Lecture- Thermodynamics

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Transcript of Chapter 19 Lecture- Thermodynamics