Thermodynamics Entropy, Free Energy, and Equilibrium Chapter 18 3/e Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

_______________ 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 into an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

18.2

Refresh your memory: What is enthalpy? Enthalpy is a thermodynamic property that describes heat changes taking place at constant _____________. Enthalpy is an ___________property—its magnitude depends on the amount of the substance present. Enthalpy cannot be determined, so we measure H, the change in enthalpy. In earlier chapters, we examined Hrxn, Hsoln, and Hº Enthalpy is a _____________function… f

These are all reactions: Identify the following reactions as exothermic or endothermic. Does a decrease in enthalpy mean a reaction proceeds spontaneously? These are all reactions: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = - 890.4 kJ H+ (aq) + OH- (aq) H2O (l) DH0 = - 56.2 kJ H2O (s) H2O (l) DH0 = 6.01 kJ NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ H2O 18.2

So we see that a decrease in enthalpy does not mean a reaction proceeds spontaneously. What do we need to know in order to predict if a reaction is spontaneous? Two things (p.577). ______________________________________________ …AND… The greater the disorder or randomness of a system, the ___________________ _____ ________________. The more ordered a system is, the less its ___________.

Entropy (S) is a measure of the _____________________ of a system. disorder S order S DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the ____________ state is more ordered than the ____________ state, and the ____________ state is more ordered than ____________ state Ssolid < Sliquid << Sgas H2O (s) H2O (l) DS > 0 18.2

Thermodynamics __________________ _________________ are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, enthalpy, pressure, volume, temperature , entropy Standard Entropy Values (S º) for Some Substances at 25ºC Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 18.7

Processes that lead to an ______________ in entropy (DS > 0) 18.2

Look at page 580, Example 18.1. Let’s work through it together. Then, let’s look at the Practice Exercise below it.

(a) Condensing water vapor How does the entropy of a system change for each of the following processes? Ex.18.1, p.580 (a) Condensing water vapor Randomness _______________ Entropy _______________ (b) Forming sucrose crystals from a supersaturated solution Randomness _______________ Entropy _______________ (c) Heating hydrogen gas from 600C to 800C Randomness _______________ Entropy _______________ (d) Subliming dry ice Randomness _______________ Entropy _______________ 18.2

First Law of Thermodynamics Energy can be converted from one form to another, but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 18.3

Entropy Changes in the System (DSsys) The ____________ _____________ of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aA + bB cC + dD DS0 rxn dS0(D) cS0(C) = [ + ] — bS0(B) aS0(A) DS0 rxn nS0(products) = S mS0(reactants) — n = the sum of the stoichiometric coefficients of the products m = the sum of the stoichiometric coefficients of the reactants What does the superscripted zero mean? 18.3

Look at the bottom of page 581 and top of page 582, Example 18.2. Let’s work through it together. Then, let’s look at the Practice Exercise below it.

Entropy Changes in the System (DSsys) What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g) S0(CO) = 197.9 J/K•mol Where do we get these values? S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol DS0 rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol 18.3

You should work through the other parts of this Practice Exercise.

Entropy Changes in the System (DSsys) When gases are produced or consumed If a reaction produces more gas molecules than it consumes, DS0 > 0. If the total number of gas molecules diminishes, DS0 < 0. If there is no net change in the total number of gas molecules, then DS0 may be positive or negative—BUT DS0 will be a small number. 18.3

Look at page 582f, Example 18.3. Let’s work through it together. Then, let’s look at the Practice Exercise below it.

Entropy Changes in the System (DSsys) What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s) Practice Exercise 18.3, part (b), page 583 18.3

Since the total number of gas molecules goes down, DS is negative. Entropy Changes in the System (DSsys) What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s) Since the total number of gas molecules goes down, DS is negative. Practice Exercise 18.3, part (b), page 583 18.3

You should work through the other parts of this Practice Exercise.

Entropy Changes in the Surroundings (DSsurr) Exothermic Process DSsurr > 0 Endothermic Process DSsurr < 0 You should keep studying this until you understand it! 18.3

Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. 18.3

“How can we tell if a reaction is spontaneous?” You will soon have an answer to this question!

DSuniv = DSsys + DSsurr > 0 Spontaneous process: DSuniv = DSsys + DSsurr > 0 Substituting —DHsys/T for DSsurr DHsys DSuniv = DSsys — > 0 T Multiply by T TDSuniv = —DHsys + TDSsys > 0 Multiply by —1 —TDSuniv = DHsys — TDSsys < 0 This says that for a process carried out at constant P and T, if the changes in enthalpy (H) and entropy (S) of the system are such that DHsys — TDSsys is less than zero, then the process must be spontaneous. But we can predict spontaneity more directly…. p.586 18.4

Gibbs Free Energy (G) G = H — TS For a constant-temperature process: All quantities in the above equation refer to the system For a constant-temperature process: The change in Gibbs free energy (DG) ____________________________________ If DG is negative, there is a release of usable energy, and the reaction is spontaneous! If DG is positive, the reaction is not spontaneous! 18.4

For a constant-temperature process: DG = DHsys — TDSsys DG < 0 The reaction is ___________ in the forward direction. DG > 0 The reaction is ________________ as written. The reaction is ___________ in the reverse direction DG = 0 The reaction is _____________________ . 18.4

n = the sum of the stoichiometric coefficients of the products The _____________________________________ is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0 (A) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - n = the sum of the stoichiometric coefficients of the products m = the sum of the stoichiometric coefficients of the reactants The superscripted zero means ______________________ 18.4

DG0 of any element in ______________________ is zero. Conventions for Standard States ________________________ (DG0) is the free-energy change that occurs when: * 1 mole of the compound * is formed from its elements * in their standard states. f DG0 of any element in ______________________ is zero. f f 18.4

Look at page 587f, Example 18.4. Let’s work through it together. Then, let’s look at the Practice Exercise below it.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) What is the standard free-energy change for the following reaction at 25 0C? 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0 (CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Practice Exercise 18.4 (b), p.588 Is this reaction spontaneous at 25 0C? 18.4

This reaction is spontaneous What is the standard free-energy change for the following reaction at 25 0C? 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0 (CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is this reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 This reaction is spontaneous 18.4

DG = DH — TDS If both DH and DS are positive, then DG will be negative only when the TDS term is larger than DH. This occurs only when T is large. If DH is positive and DS is negative, DG will always be positive—regardless of the temperature. If DH is negative and DS is positive, then DG will always be negative regardless of temperature. If DH is negative and DS is negative, then DG will be negative only when TDS is smaller in magnitude than DH. This condition is met when T is small. p.588

DG = DH — TDS Factors Affecting the Sign of DG in the Relationship DG = DH — TDS p.589 18.4

Temperature and Spontaneity of Chemical Reactions CaCO3 (s) CaO (s) + CO2 (g) At what temperature will this reaction become spontaneous? Or…at what temperature does DGº become zero—or negative? 1. Use Eqn 6.16 to calculate DHº 2. Use Eqn 18.4 to calculate DSº 3. Substitute these into DGº = DHº — TDSº to calculate DGº (130.0 kJ) 4. Since this is a large positive number, we set DGº = 0 and solve the equation for T Follow along on pp.589f

Temperature and Spontaneity of Chemical Reactions CaCO3 (s) CaO (s) + CO2 (g) “Two points are worth making….” [Read quote p.590.] First, we used DHº and DSº values at 25ºC to calculate changes that occur at much higher temperatures. Both DHº and DSº change with temperature, so the DGº value will not be accurate—but it is a helpful, ballpark estimate. Second, don’t think nothing happens below 835ºC, and then suddenly, at 835ºC, CaCO3 begins to decompose. Look at the graph on the next frame to see what happens.

Temperature and Spontaneity of Chemical Reactions CaCO3 (s) CaO (s) + CO2 (g) DH0 = 177.8 kJ DS0 = 160.5 J/K DG0 = DH0 – TDS0 At 25 0C, DG0 = 130.0 kJ DG0 = 0 at 835 0C p.590 18.4

DG = DH — TDS 0 = DH — TDS DS = DH/ T The condition DG = 0 applies to any phase transition. At the temperature at which a phase transition occurs, the system is at equilibrium, so DG = 0. Substituting and rearranging, DG = DH — TDS 0 = DH — TDS DS = DH/ T

Gibbs Free Energy and Phase Transitions DG0 = 0 = DH0 – TDS0 H2O (l) H2O (g) DS = T DH = 40.79 kJ 373 K = 109 J/K 18.4

Gibbs Free Energy and Chemical Equilibrium DG0 describes a reaction in standard state conditions DG must be used when the reaction is not in standard state conditions So we derive this equation: DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) At ________________ Q is the reaction quotient DG = 0 Q = K 0 = DG0 + RT lnK Kp for gases Kc for reactions in solution DG0 = - RT lnK pp.592ff 18.4

DG0 < 0 DG0 > 0 p.593 18.4

DG0 = - RT lnK Relation between DG0 and K as Predicted by DG0 = - RT lnK p.593 18.4

18.5

Alanine + Glycine Alanylglycine DG0 = +29 kJ K < 1 ATP + H2O + Alanine + Glycine ADP + H3PO4 + Alanylglycine DG0 = -2 kJ K > 1 18.5