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Chapter 18 - Entropy, Free Energy, and Equilibrium

Spontaneous Physical and Chemical Processes• A waterfall runs downhill• A lump of sugar dissolves in a cup of coffee• At 1 atm, water freezes below 00C and ice melts above 00C• Heat flows from a hotter object to a colder object• A gas expands in an evacuated bulb• Iron exposed to oxygen and water forms rust

Does a decrease in enthalpy mean a reaction proceeds spontaneously?

Spontaneous reactionsCH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ΔH0 = -890.4 kJ/mol

H+ (aq) + OH- (aq) H2O (l) ΔH0 = -56.2 kJ/mol

H2O (s) H2O (l) ΔH0 = 6.01 kJ/mol

NH4NO3 (s) NH4+(aq) + NO3

- (aq) ΔH0 = 25 kJ/mol

Entropy (S) is a measure of the randomness or disorder of a system.Order S Disorder S

ΔS = Sf - Si

If the change from initial to final results in an increase in randomness

Sf > Si ΔS > 0

For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state

Ssolid < Sliquid << Sgas Processes that lead to an increase in entropy (ΔS > 0)

H2O (s) H2O (l) ΔS > 0

AP Chemistry Page 2 of 10Mr. MarkicEntropyState functions are properties that are determined by the state of the system, regardless of how that condition was achieved.energy, enthalpy, pressure, volume, temperature, entropy

Sample ExercisePredict whether the entropy change is greater or less than zero for each of the following processes:(a) Freezing ethanol

(b) Evaporating a beaker of liquid bromine at room temperature

(c) Dissolving glucose in water

(d) Cooling nitrogen gas from 80°C to 20°C

How does the entropy of a system change for each of the following processes?(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed.

Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

Spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0Equilibrium process: ΔSuniv = ΔSsys + ΔSsurr = 0

Entropy Changes in the System (ΔSsys)The standard entropy of reaction (ΔS0) is the entropy change for a reaction carried out at 1 atm and 250C.aA + bB cC + dD

ΔS0rxn = [cS0(C) + dS0(D)] – [aS0(A) + bS0(B)]

ΔS0rxn = ΣnS0(products) - ΣmS0(reactants)

Sample ExerciseFrom the following standard entropy values in Appendix 3, calculate the standard entropy changes for the following reactions 25°C

(a) CaCO3(s) CaO(s) + CO2(g)[CaCO3(s) = 92.9 J/Kmol, CaO(s) = 39.8 J/Kmol, CO2(g) = 213.6 J/Kmol]

(b) N2(g) + 3H2(g) 2NH3(g)

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[N2(g) = 192 J/Kmol, H2(g) = 131 J/Kmol, NH3(g)= 193 J/Kmol]

(c) H2(g) + Cl2(g) 2HCl(g)[H2(g) = 131 J/Kmol, Cl2(g) = 223 J/Kmol, HCl(g) = 187 J/Kmol]

(d) 2CO (g) + O2 (g) 2CO2 (g)[2CO (g) =197.9 J/Kmol, O2 (g) = 205.0 J/Kmol, CO2 (g) = 213.6 J/Kmol]

Entropy Changes in the System (ΔSsys)When gases are produced (or consumed)• If a reaction produces more gas molecules than it consumes, ΔS0 > 0.• If the total number of gas molecules diminishes, ΔS0 < 0.• If there is no net change in the total number of gas molecules, then ΔS0 may be positive or

negative BUT ΔS0 will be a small number.

Sample ExercisePredict whether the entropy change of the system in each of the following reactions is positive or negative.(a) 2H2(g) + O2(g) 2H2O(l) (b) NH4Cl(s) NH3(g) + HCl(g) (c) H2(g) + Br2(g) 2HBr(g)

Practice ExerciseDiscuss qualitatively the sign of the entropy change expected for each of the following processes:(a) I2(s) 2I(g) (b) 2Zn(s) + O2(g) 2ZnO(s) (c) N2(g) + O2(g) 2NO(g)

Review of ConceptsConsider the gas-phase reaction of A2 (blue) and B2 (orange) to form AB3.

(a) Write a balanced equation for the reaction. (b) What is the sign of S for the reaction?

aA + bB cC + dD

G0rxn dG0 (D)f

cG0 (C)f= [ + ] - bG0 (B)

faG0 (A)

f[ + ]

G0rxn nG0 (products)f= mG0 (reactants)f

-

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Entropy Changes in the Surroundings (ΔSsurr)

Exothermic Process Endothermic ProcessΔSsurr > 0 ΔSsurr < 0

Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

Gibbs Free Energy

Spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0

Equilibrium process: ΔSuniv = ΔSsys + ΔSsurr = 0

For a constant-temperature process:

Gibbs free energy (G) ΔG = ΔHsys -TΔSsys

ΔG < 0 The reaction is spontaneous in the forward direction.

ΔG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.

ΔG = 0 The reaction is at equilibrium.

The standard free-energy of reaction (ΔG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

Standard free energy of formation (ΔG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

ΔG0 of any element in its stable form is zero.

S = k ln WW = 1

S = 0

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Sample ExerciseCalculate the standard free-energy changes for the following reactions at 25°C

(a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)[CH4(g) = -50.8 kJ/mol, O2(g) = 0 kJ/mol, CO2(g) = -394.4 kJ/mol H2O(l) = -237.2 kJ/mol]

(b) 2MgO(s) 2Mg(s) + O2(g)[MgO(s) = -569.6 kJ/mol, Mg(s) = 0 kJ/mol, O2(g) = 0 kJ/mol]

Practice ExerciseWhat is the standard free-energy change for the following reaction at 25 0C? Is the reaction spontaneous at 25 0C? 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)[C6H6 (l) = 124.5 kJ/mol, O2 (g) = 0 kJ/mol, CO2 (g) = -394.4 kJ/mol, H2O (l) = -237.2 kJ/mol]

ΔG = ΔH - TΔS

Review of Concepts(a) Under what circumstances will an endothermic reaction proceed spontaneously?

(b) Explain why, in many reactions in which both the reactant and product species are in the solution phase, ΔH often gives a good hint about the spontaneity of a reaction at 298 K.

Temperature and Spontaneity of Chemical ReactionsCaCO3 (s) CaO (s) + CO2 (g)

Equilibrium Pressure of CO2

ΔH0 = 177.8 kJ/mol

ΔS0 = 160.5 J/K·mol

ΔG0 = ΔH0 – TΔS0

At 25 oC, ΔG0 = 130.0 kJ/mol

ΔG0 = 0 at 835 oC

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Gibbs Free Energy and Phase Transitions

ΔG0 = 0 = ΔH0 – TΔS0

H2O (l) H2O (g)

ΔS = ΔHT =

40.79 kJ /mol373 K

= 1.09 x 10-1 kJ/K·mol = 109 J/K·mol

Sample ExerciseThe molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid liquid and liquid vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.

Practice ExericeThe molar heats of fusion and vaporization of argon are1.3 kJ/mol and 6.3 kJ/mol, and argon’s melting and boiling points are -190°C and -186°C respectively. Calculate the entropy changes for fusion and vaporization.

Review of ConceptsConsider the sublimation of iodine (I2) at 45°C in a close flask shown here. If the enthalpy of sublimation is 62.4 kJ/mol, what is the ΔS for sublimation?

Gibbs Free Energy and Chemical EquilibriumΔG = ΔG0 + RT lnQ

R is the gas constant (8.314 J/K•mol)T is the absolute temperature (K)Q is the reaction quotient

Free Energy Versus Extent of Reaction

At EquilibriumΔG = 0 Q = K

0 = ΔG0 + RT lnKΔG0 = - RT lnK

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ΔG0 < 0 ΔG0 > 0ΔG0 = - RT lnK

Sample ExerciseUsing data listed in Apx 3, calculate the equilibrium constant (KP) for the following reaction at 25°C

2H2O(l) 2H2(g) + O2(g) ΔG°rxn = 474.4 kJ/mol ,

Practice ExerciseCalculate the equilibrium constant (KP) for the reaction at 25°C

2O3(g) 3O2(g) ΔG° = -326.8 kJ/mol

In chapter 16 we discussed the solubility product of slightly soluble substances. Using the solubility product of silver chloride at 25°C (1.6 x 10-10), calculate G° for the process:

AgCl(s) Ag+(aq) + Cl-(aq)

Calculate G° for the following process at 25°C, the Ksp of BaF2 is 1.7 x 10-6:BaF2(s) Ba2+(aq) + 2F-(aq)

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Sample ExerciseThe equilibrium constant (KP) for the reaction

N2O4(g) 2NO2(g)is 0.113 at 298 K, which corresponds to a standard free energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2 = 0.122 atm and PN2O4 = 0.453 atm. Calculate G for the reaction at these pressures and predict the direction of the net reaction.

Practice ExerciseThe ΔG° for the reaction: H2(g) + I2(g) 2HI(g)Is 2.60 kJ/mol at 25C. In one experiment, the initial pressures are PH2 = 4.26 atm, PI2 = 0.024 atm, and PHI = 0.23 atm. Calculate ΔG for the reaction and predict the direction of the net reaction

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Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matterDuration: Early AprilTextbook Chapter: 17 – Entropy, Free Energy and EquilibriumEnduring Understanding Essential Knowledge5.E: Chemical or physical processes are driven by a decrease in enthalpy or an increase in entropy, or both.

5.E.1: Entropy is a measure of the dispersal of matter and energy.5.E.2: Some physical or chemical processes involve both a decrease in the internal energy of the components (ΔH° < 0) under consideration and an increase in the entropy of those components (ΔS°> 0). These processes are necessarily “thermodynamically favored” (ΔG°< 0).5.E.3: If a chemical or physical process is not driven by both entropy and enthalpy changes, then the Gibbs free energy change can be used to determine whether the process is thermodynamically favored.5.E.4: External sources of energy can be used to drive change in cases where the Gibbs free energy change is positive.5.E.5: A thermodynamically favored process may not occur due to kinetic constraints (kinetic vs. thermodynamic control).

6.D: The equilibrium constant is related to temperature and the difference in Gibbs free energy between reactants and products.

6.D.1: When the difference in Gibbs free energy between reactants and products (ΔG°) is much larger than the thermal energy (RT), the equilibrium constant is either very small (for ΔG° > 0) or very large (for ΔG° < 0). When ΔG° is comparable to the thermal energy (RT), the equilibrium constant is near 1.

Learning Objectives3.11 The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or process to generate a relevant symbolic and/or graphical representation of the energy changes.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.5.4 The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow.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.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 ΔH° and ΔS°, and calculation or estimation of ΔG° when needed5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.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.5.16 The student can use LeChatelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product.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.6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.6.24 The student can analyze the enthalpic and entropic changes associated with the dissolution of a salt, using particulate level interactions and representations.6.25 The student is able to express the equilibrium constant in terms of ΔG° and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process.

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