Chapter 17 Spontaneity, Entropy, and Free Energy AP*

Chapter 17

Spontaneity, Entropy, and Free Energy

AP*

AP Learning Objectives

LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. (Sec 17.1)

LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. (Sec 17.3)

LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes. (Sec 17.1-17.3, 17.5)

LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed. (Sec 17.4, 17.6)

LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. (Sec 17.4, 17.6)


LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. (Sec 17.6, 17.9)

LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. (Sec 17.6)

LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. (Sec 17.6)

LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. (Sec 17.1, 17.6-17.8)


LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. (Sec 17.8)

Section 17.1Spontaneous Processes and Entropy

AP Learning Objectives, Margin Notes and References

Learning Objectives LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water

and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the

entropy change associated with chemical or physical processes. LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large

amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.

Additional AP References LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)


Copyright © Cengage Learning. All rights reserved 6

Thermodynamics vs. Kinetics

Domain of Kinetics Rate of a reaction depends

on the pathway from reactants to products.

Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products (do not need knowledge of pathway b/n R and P)



Spontaneous Processes and Entropy Thermodynamics lets us predict the direction in

which a process will occur but gives no information about the speed of the process.

A spontaneous process is one that occurs without outside intervention. (Spontaneous does not mean fast)


First Law of Thermodynamics You will recall from Chapter 6 that energy

cannot be created or destroyed. (E = q + w) 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.

The first law only guarantees that energy is conserved, but it says nothing about the preferred direction of the process.


Enthalpy/Entropy

Enthalpy is the heat absorbed by a system during a constant-pressure process.

Entropy is a measure of the randomness in a system. It can also be thought of as dispersion or spreading out of energy.

Both play a role in determining whether a process is spontaneous.


Spontaneous processes proceed without any outside assistance.

The gas in vessel A will spontaneously effuse into vessel B, but it will not spontaneously return to vessel A.

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


Experimental Factors Affect Spontaneous Processes Temperature and pressure can affect spontaneity. An example of how temperature affects spontaneity is ice

melting or freezing.


Reversible and Irreversible Processes

Reversible process: The system changes so that the system and surroundings can be returned to the original state by exactly reversing the process. This maximizes work done by a system on the surroundings.

Irreversible processes cannot be undone by exactly reversing the change to the system or cannot have the process exactly followed in reverse. Also, any spontaneous process is irreversible!



Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.

a) What should happen to the gas when you open the valve?

CONCEPT CHECK!




b) Calculate ΔH, ΔE, q, and w for the process you described above.

All are equal to zero.

CONCEPT CHECK!




c) Given your answer to part b, what is thedriving force for the process?

Entropy

CONCEPT CHECK!



The Expansion of An Ideal Gas Into an Evacuated Bulb



Entropy The driving force for a spontaneous process is an

increase in the entropy of the universe. A measure of molecular randomness or disorder. It

can also be thought of as dispersion or spreading out of energy.



Entropy

Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state.

Nature spontaneously proceeds toward the states that have the highest probabilities of existing.


Microstate: A single possible arrangement of position and kinetic energy of molecules


Entropy on the Molecular Scale

Boltzmann described entropy on the molecular level. Gas molecule expansion: Two molecules are in the apparatus

above; both start in one side. What is the likelihood they both will end up there? (1/2)2

If one mole is used? (1/2)6.02×1023! (No chance!) Gases spontaneously expand to fill the volume given. Most probable arrangement of molecules: approximately

equal molecules in each side


Effect of Volume and Temperature Change on the System

If we increase volume, there are more positions possible for the molecules. This results in more microstates, so increased entropy.

If we increase temperature, the average kinetic energy increases. This results in a greater distribution of molecular speeds. Therefore, there are more possible kinetic energy values, resulting in more microstates, increasing entropy.


Molecular Motions 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 about an axis Note: More atoms means more microstates (more

possible molecular motions).


Entropy on the Molecular Scale

The number of microstates and, therefore, the entropy tend to increase with increases in

temperature. volume. the number of independently moving molecules. Increase with the formation of more complex

molecules.


Entropy and Physical States

Entropy increases with the freedom of motion of molecules. S(g) > S(l) > S(s) Entropy of a system increases for processes where gases form from either solids or liquids. liquids or solutions form from solids (Carbonates are an

exception) the number of gas molecules increases

during a chemical reaction.



Predict the sign of ΔS for each of the following, and explain:

a) The evaporation of alcoholb) The freezing of waterc) Compressing an ideal gas at constant

temperatured) Heating an ideal gas at constant

pressuree) Dissolving NaCl in water

+––

+

+

CONCEPT CHECK!

Section 17.2Entropy and the Second Law of Thermodynamics


Learning Objectives LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the

entropy change associated with chemical or physical processes.



Second Law of Thermodynamics

In any spontaneous process there is always an increase in the entropy of the universe.

The entropy of the universe is increasing. The total energy of the universe is constant, but the

entropy is increasing.

Δ Suniverse = ΔSsystem + ΔSsurroundings



ΔSuniv

ΔSuniv = +; entropy of the universe increases; spontaneous

ΔSuniv = -; process is spontaneous in opposite direction ΔSuniv = 0; process has no tendency to occur


Section 17.3The Effect of Temperature on Spontaneity


Learning Objectives LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between

systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions.

LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes.

Additional AP References LO 5.3 (see Appendix 7.2, “Thermal Equilibrium, the Kinetic Molecular Theory, and the Process of Heat”) LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)


For the process A(l) A(s), which direction involves an increase in energy randomness (exothermic)? Positional randomness? Explain your answer.

As temperature increases/decreases (answer for both), which takes precedence? Why?

At what temperature is there a balance between energy randomness and positional randomness?


CONCEPT CHECK!


Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is:

a) Exothermic and becomes more positionally randomSpontaneous

b) Exothermic and becomes less positionally randomCannot tell

c) Endothermic and becomes more positionally randomCannot tell

d) Endothermic and becomes less positionally randomNot spontaneous

Explain how temperature affects your answers.

CONCEPT CHECK!


ΔSsurr

The sign of ΔSsurr depends on the direction of the heat flow.

The magnitude of ΔSsurr depends on the temperature.



ΔSsurr



ΔSsurr



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

Entropy Changes in Surroundings Heat that flows into or out of the system

changes the entropy of the surroundings. For an isothermal process




Entropy Change in the Universe The universe is composed of the system and

the surroundings. Therefore,

Suniverse = Ssystem + Ssurroundings

For spontaneous processes

Suniverse > 0

Section 17.4Free Energy


Learning Objectives LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored

by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed.

LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.


Total Entropy and Spontaneity

ΔSuniverse = ΔSsystem + ΔSsurroundings

Substitute for the entropy of the surroundings:ΔSuniverse = ΔSsystem – ΔHsystem/T

Multiply by −T:−TΔSuniverse = −TΔSsystem + ΔHsystem

Rearrange:−TΔSuniverse = ΔHsystem − TΔSsystem

Call −TΔSuniverse the Gibbs Free Energy (ΔG):

• ΔG = ΔH − T ΔS


Free Energy (G)

A process (at constant T and P) is spontaneous in the direction in which the free energy decreases. Negative ΔG means positive ΔSuniv.


univ = (at constant and ) G

S T PT


Gibbs Free Energy1. 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.

More convenient to use DG than DSuniv to determine spontaneity because DG relates to the system alone (no need to examine the surroundings.


A liquid is vaporized at its boiling point. Predict the signs of:wqΔHΔSΔSsurr

ΔG

Explain your answers.


–+++–0

CONCEPT CHECK!


The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.

Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.

ΔS = 132 J/K·molΔSsurr = -132 J/K·mol

ΔG = 0 kJ/molCopyright © Cengage Learning. All rights reserved 43

EXERCISE!



Comparison of ΔSuniv and ΔG◦ to Predict Spontaneity



Effect of ΔH and ΔS on Spontaneity



Section 17.5Entropy Changes in Chemical Reactions


Learning Objectives LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the

entropy change associated with chemical or physical processes.



Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant.

Predict the signs of: ΔH ΔSsurr ΔS ΔSuniv

Explain.


– + 0 +

CONCEPT CHECK!


Third Law of Thermodynamics

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

Consider all atoms or molecules in the perfect lattice at 0 K; there will only be one microstate.

The entropy of a substance increases with temperature.


Standard Entropies

The reference for entropy is 0 K, so the values for elements are not 0 J/mol K at 298 K.

Standard molar entropy for gases are generally greater than liquids and solids. (Be careful of size!)

Standard entropies increase with molar mass.

Standard entropies increase with number of atoms in a formula.


Standard MolarEntropy Values (S°)

Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.



Entropy Changes (a state function)Entropy changes for a reaction can be calculated in a manner analogous to that by which H is calculated:

S° = nS°(products) – mS°(reactants)

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


Calculate ΔS° for the following reaction:2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Given the following information: S° (J/K·mol)

Na(s) 51 H2O(l) 70

NaOH(aq) 50 H2(g) 131

ΔS°= –11 J/KCopyright © Cengage Learning. All rights reserved 55

EXERCISE!

Section 17.6Free Energy and Chemical Reactions


Learning Objectives LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored

by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed.

LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.

LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.

LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product.

LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction.

LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.



Additional AP References LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 5.16 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 5.17 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)


Standard Free Energy Change (ΔG°)

The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.

We use ΔG°for various reactions to compare the relative tendency of the reactions to occur under the same pressure and concentration conditions. The more neg ΔG°, the further the reaction will go right to reach equilibrium.

ΔG° = ΔH° – TΔS°



Standard Free Energy Change (ΔG°)

There are several ways to calculate ΔG°(this is 3 out of 4)

1) ΔG° = ΔH° – TΔS°2) Hess’s Law (additions of reaction ΔG°terms)

3) ΔG°reaction = ΣnpGf°products – ΣnrGf°reactants

ΔGf°is defined as the change in free energy that accompanies the formation of one mole of that substance from its constituent elements with all reactants and products in their standard states.




CONCEPT CHECK!



CONCEPT CHECK!



CONCEPT CHECK!


A stable diatomic molecule spontaneously forms from its atoms.

Predict the signs of:ΔH° ΔS° ΔG°

Explain.


– – –

CONCEPT CHECK!


Consider the following system at equilibrium at 25°C.

PCl3(g) + Cl2(g) PCl5(g)

ΔG° = −92.50 kJ

What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain. The ratio will decrease.


CONCEPT CHECK!

Section 17.8Free Energy and Equilibrium


Learning Objectives LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large

amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.

LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process.

Additional AP References LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 6.25 (see Appendix 7.11, “Non-Spontaneous Reactions”)


Free Energy and Equilibrium

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

R= 8.314 J/mol KG = G° + RT ln Q

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


Solution

Plan We need to calculate the value of the reaction quotient Q for the specified partial pressures, for which we use the partial-pressures form: Q = [D]d[E]e/[A]a[B]b. We then use a table of standard free energies of formation to evaluate ΔG°.

We calculated ΔG° = −33.3 kJ for this reaction. We will have to change the units of this, however, for the units 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° = −33.3 kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3.

Calculate ΔG at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process:

Calculating the Free-Energy Change under Nonstandard Conditions


We now use Equation 19.19 to calculate ∆G for these nonstandard conditions:

ΔG = ΔG° + RT ln Q = (−33.3 kJ/mol) + (8.314 J/mol-K)(298 K)(1 kJ/1000 J) ln(9.3 × 10–3) = (−33.3 kJ/mol) + (−11.6 kJ/mol) = −44.9 kJ/mol

Comment We see that ΔG becomes more negative as the pressures of N2, H2, and NH3 are changed from 1.0 atm (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 make the same prediction based on Le Châtelier’s principle. 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 changes shift the reaction to the product side, thereby forming more NH3.

Continued

Calculating the Free-Energy Change under Nonstandard Conditions


The Meaning of ΔG for a Chemical Reaction

A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.



The Meaning of ΔG for a Chemical Reaction

A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.

For reaction A(g) ↔ B(g)



The equilibrium point occurs at the lowest value of free energy available to the reaction system.

ΔG = 0 = ΔG° + RT ln(K)ΔG° = –RT ln(K)


Fourth way to calculate ΔG°


Free Energy and Equilibrium

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

0 = G° + RT ln K Rearranging, this becomes

G° = RT ln Kor

K = eG/RT


Solution

Remembering to use the absolute temperature for T in and the form of R that matches our units, we have

K = e−ΔG°/RT = e−(−33,300 J ⁄ mol)/(8.314 J/mol-K)(298 K) = e13.4 = 7 × 105

The standard free-energy change for the Haber process at 25 °C was obtained for the Haber reaction:

N2(g) + 3 H2(g) 2 NH3(g) ∆G° = −33.3 kJ/mol = –33,300 J/mol

Use this value of ∆G° to calculate the equilibrium constant for the process at 25 °C.

Calculating an Equilibrium Constant from ΔG°



Section 17.9Free Energy and Work


Learning Objectives LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable

with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.



Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.

wmax = ΔG



Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.

All real processes are irreversible. First law: You can’t win, you can only break even. Second law: You can’t break even. As we use energy, we degrade its usefulness because we

spread it out.


Documents

Chapter 17 Spontaneity, Entropy, and Free Energy AP*