Chapter 16 Spontaneity, Entropy, and Free Energy Why do some things happen and others don’t?

16.1 Spontaneous Processes and Entropy 16.2 Entropy and the Second Law of Thermodynamics 16.3 The Effect of Temperature on Spontaneity 16.4 Free Energy 16.5 Entropy Changes in Chemical Reactions 16.6 Free Energy and Chemical Reactions 16.7 The Dependence of Free Energy on Pressure 16.8 Free Energy and Equilibrium 16.9 Free Energy and Work Copyright © Cengage Learning. All rights reserved

To compare kinetics with thermodynamics To define spontaneity Objectives To compare kinetics with thermodynamics To define spontaneity To answer why some things happen and others don’t. To define Entropy To assign positive or negative sign depending on the process To understand why a diverging diamond interchange is safer than other designs Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved

Spontaneous Processes and Entropy Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process. A spontaneous process is one that occurs without outside intervention. Examples?? Copyright © Cengage Learning. All rights reserved

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What should happen to the gas when you open the valve? 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. What should happen to the gas when you open the valve? The gas should spread evenly throughout the two bulbs. Copyright © Cengage Learning. All rights reserved

The Expansion of An Ideal Gas Into an Evacuated Bulb What is the driving force for the process? Why does it happen? Copyright © Cengage Learning. All rights reserved

Entropy (S) is a measure of molecular randomness or disorder. The driving force for a spontaneous process is an increase in the entropy (S) of the universe. Entropy (S) is a measure of molecular randomness or disorder. Copyright © Cengage Learning. All rights reserved

Is a deck of cards usually ordered or disordered? 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. Is a deck of cards usually ordered or disordered? There are so many ways for a deck of cards to be disordered, but Only one way for it to be ordered… Copyright © Cengage Learning. All rights reserved

The Microstates That Give a Particular Arrangement (State) Copyright © Cengage Learning. All rights reserved

Therefore: Ssolid < Sliquid << Sgas Why do solutions form? Positional Entropy A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system. Therefore: Ssolid < Sliquid << Sgas Why do solutions form? Why do ice cubes get smaller, even though they are in the freezer? If a process moves in a direction of greater disorder or higher entropy, the sign is positive. Copyright © Cengage Learning. All rights reserved

Predict the sign of S for each of the following, and explain: Concept Check Predict the sign of S for each of the following, and explain: The evaporation of alcohol The freezing of water Compressing an ideal gas at constant temperature Heating an ideal gas at constant pressure Dissolving NaCl in water + – a) + (a liquid is turning into a gas) b) - (more order in a solid than a liquid) c) - (the volume of the container is decreasing) d) + (the volume of the container is increasing) e) + (there is less order as the salt dissociates and spreads throughout the water) Copyright © Cengage Learning. All rights reserved

To compare kinetics with thermodynamics To define spontaneity Objectives Review To compare kinetics with thermodynamics To define spontaneity To answer why some things happen and others don’t. To define Entropy To assign positive or negative sign depending on the process Why is a diverging diamond safer? #12 Work Session Page 782 # 1, 19, 23 (think IGL and volume), 25, 33, 35 Copyright © Cengage Learning. All rights reserved

To define the Second Law of Thermodynamics Objectives To define the Second Law of Thermodynamics Wait, what was the first? I’m constantly forgetting it… Copyright © Cengage Learning. All rights reserved

Second Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. What is the sign of a spontaneous process? The entropy of the universe is increasing. The total energy of the universe is constant, but the entropy is increasing. Suniverse = ΔSsystem + ΔSsurroundings Copyright © Cengage Learning. All rights reserved

Second Law of Thermodynamics Suniverse = ΔSsystem + ΔSsurroundings How can life exist if it requires cells to take small molecules to make bigger ones? Is this a spontaneous process? Does S increase? Is this consistent with the 2nd Law of Thermodynamics? For a spontaneous process, how are ΔSsys & ΔSsurr related? For a spontaneous process, how are ΔSsys & ΔSsurr related? They have to combine such that entropy of the universe is increased. Copyright © Cengage Learning. All rights reserved

To define the Second Law of Thermodynamics 1st Law of Thermodynamics: Objectives Review To define the Second Law of Thermodynamics 1st Law of Thermodynamics: The amount of energy in the Universe is constant Work Session #8 Copyright © Cengage Learning. All rights reserved

To predict spontaneity of a process based on ΔS and ΔH Objectives To predict spontaneity of a process based on ΔS and ΔH To understand the effect of temperature on Spontaneity Copyright © Cengage Learning. All rights reserved

Which side is favored by Entropy? H2O (l) H2O (g) water is the system… 18 ml 31 L @1atm, 100C Positional Probability? What is the sign of ΔSsystem? What direction is the heat flow? ΔHsystem? …what about the surroundings? What is the sign of ΔSsurr? Heat flow? ΔHsurr? What is the overall ΔS of the universe? ΔSuniv = ΔSsys + ΔSsurr Copyright © Cengage Learning. All rights reserved

Concept Check For the process A(l) A(s), which direction involves an increase in energy randomness(-ΔH)? Positional randomness(+ΔS)? Explain your answer. As temperature increases/decreases , does energy or positional randomness take precedence? Why? At what temperature is there a balance between energy randomness and positional randomness? Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid). As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored. There is a balance at the melting point. Copyright © Cengage Learning. All rights reserved

The magnitude of ΔSsurr depends on the temperature. ΔSsurr = - ΔH T The sign of ΔSsurr depends on the direction of the heat flow. –ΔH, +ΔSsurr The magnitude of ΔSsurr depends on the temperature. Why does ice freeze, but only at lower temperatures? Balance between energy randomness and positional randomness. Liquid  Solid At high temps, ΔSsurr = - ΔH is very small…as T gets smaller, exothermicity becomes a stronger driving force. T Copyright © Cengage Learning. All rights reserved

ΔSsurr = - ΔH enthalpy determines the sign T Copyright © Cengage Learning. All rights reserved

ΔSsurr = - ΔH temperature determines the magnitude T Heat flow (constant P) = change in enthalpy = ΔH Copyright © Cengage Learning. All rights reserved

Interplay of Ssys and Ssurr in Determining the Sign of Suniv ΔSuniv = ΔSsys + ΔSsurr Copyright © Cengage Learning. All rights reserved

To predict spontaneity of a process based on ΔS and ΔH Objectives Review To predict spontaneity of a process based on ΔS and ΔH To understand the effect of temperature on Spontaneity Copyright © Cengage Learning. All rights reserved

To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH Objectives To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH To predict spontaneity based on Gibbs Free Energy To conceptually and mathematically use Gibbs Equation Copyright © Cengage Learning. All rights reserved

ΔG = ΔH – TΔS (at constant T and P) Negative ΔG means positive ΔSuniv Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) Negative ΔG means positive ΔSuniv positive ΔSuniv means Spontaneous! Negative ΔG means Spontaneous! ΔG < 0 means spontaneous Copyright © Cengage Learning. All rights reserved

How to determine spontaneity based on ΔS and ΔH … For a process to be considered spontaneous, it must be both : exothermic (favor energy randomness (-ΔH) and Favor positional randomness (entropy (+ΔS)) ΔG = ΔH – TΔS If the opposite of the above conditions exist, we can conclude that the process is not spontaneous. Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid). As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored. There is a balance at the melting point. Spontaneous Not Spontaneous Don’t Know ΔH - + ΔS Copyright © Cengage Learning. All rights reserved

Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is: Exothermic and becomes more positionally random Spontaneous Exothermic and becomes less positionally random Cannot tell Endothermic and becomes more positionally random Endothermic and becomes less positionally random Not spontaneous Explain how temperature affects your answers. a) Spontaneous (both driving forces are favorable). An example is the combustion of a hydrocarbon. b) Cannot tell (exothermic is favorable, positional randomness is not). An example is the freezing of water, which becomes spontaneous as the temperature of water is decreased. c) Cannot tell (positional randomness is favorable, endothermic is not). An example is the vaporization of water, which becomes spontaneous as the temperature of water is increased.. d) Not spontaneous (both driving forces are unfavorable). Questions "a" and "d" are not affected by temperature. Choices "b" and "c" are explained above. Copyright © Cengage Learning. All rights reserved

Effect of ΔH and Δ S on Spontaneity ΔG = ΔH – TΔS Copyright © Cengage Learning. All rights reserved

A liquid is vaporized at its boiling point. Predict the signs of: w q Concept Check A liquid is vaporized at its boiling point. Predict the signs of: w q H S Ssurr G Explain your answers. – + As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium. Copyright © Cengage Learning. All rights reserved

Predict the spontaneity of: H2O(s)  H2O(l) Exercise Predict the spontaneity of: H2O(s)  H2O(l) Given: ΔH = 6.03 X 103 J/mol ΔS = 22.1 J/K*mol 5-Step with Gibbs! ΔG = ΔH – TΔS Try at -10 oC, 0 oC, and 10 oC -10 oC, ΔG = +2.2 X 102 J/mol 0 oC, ΔG = 0 J/mol 10 oC, ΔG = -2.2 X 102 J/mol S = 132 J/K·mol Ssurr = -132 J/k·mol G = 0 kJ/mol Copyright © Cengage Learning. All rights reserved

Exercise The value of Hvaporization of substance X is 45.7 J/mol, and its normal boiling point is 72.5°C. Calculate S, Ssurr, and G for the vaporization of one mole of this substance at 72.5°C and 1 atm. ΔSsurr = - ΔH T S = 132 J/K·mol Ssurr = -132 J/K·mol G = 0 kJ/mol S = 132 J/K·mol @ point, no other S is occuring Ssurr = -132 J/k·mol G = 0 kJ/mol Copyright © Cengage Learning. All rights reserved

Find the normal boiling point of liquid Br2. Br2(l)  Br2(g) Exercise Find the normal boiling point of liquid Br2. Br2(l)  Br2(g) ΔH = 31.0 KJ/mol ΔS = 93.0 J/K*mol 5-Step with Gibbs! ΔG = ΔH – TΔS S = 132 J/K·mol Ssurr = -132 J/k·mol G = 0 kJ/mol Copyright © Cengage Learning. All rights reserved

Spontaneous Reactions Copyright © Cengage Learning. All rights reserved

To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH Objectives Review To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH To predict spontaneity based on Gibbs Free Energy To conceptually and mathematically use Gibbs Equation Work Session: 27, 29, 30, 31 Copyright © Cengage Learning. All rights reserved

To predict the sign of ΔS for a given chemical reaction. Objectives To predict the sign of ΔS for a given chemical reaction. To calculate ΔS°reaction given ΔS° values for products and reactants To calculate ΔS°reaction , ΔH°reaction , ΔG°reaction Copyright © Cengage Learning. All rights reserved

ΔS changes in Chemical Reactions Entropy changes in chemical reactions are determined by positional probability. This means that the more molecules, the more possible configurations, the more disorder. Predict the sign of ΔS for the following reaction: H2  2H N2(g) + 3H2(g)  2NH3(g) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) CaCO3(s)  CaO(s) + CO2(g) 2SOS(g) + O2(g)  2SO3(g) Copyright © Cengage Learning. All rights reserved

Concept Check Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The potential energy of AB is much greater than that of either reactant. Predict the signs of: H Ssurr Suniv Explain. Since the POTENTIAL energy of the products is greater than the average bond energies of the reactants, the reaction is endothermic as written. Thus, the sign of H is positive; Ssurr = -H/T, so S is negative, but close to zero (cannot tell for sure); and Suniv is positive. + - + Copyright © Cengage Learning. All rights reserved

Third Law of Thermodynamics…there is a 3rd Law?? So far, we’ve discussed the relative change in entropy, ΔS. Can we assign an absolute S? The entropy of a perfect crystal at 0 K is zero. A perfect crystal represents the lowest possible entropy. Or in other words, everything is perfectly ordered, in its place, just plain ducky…. Therefore, from 0 K, as the temperature increases, so does the randomness of vibrational motion and, therefore, the entropy of the substance. Copyright © Cengage Learning. All rights reserved

Third Law of Thermodynamics…there is a 3rd Law?? Oh, yeah, … Since a perfect crystal at absolute zero is an unattainable ideal, we take the Third Law as a standard although we have never actually observed it……. You are welcome…. Copyright © Cengage Learning. All rights reserved

Standard Enthalpy Values (H°) Represent the increase in enthalpy of formation of a compound. ΔH°reaction = ΣnpH°products – ΣnrH°reactants n = moles (balanced coefficient) H° = standard enthalpy (from a given list in problem # or on pages A20-A21) See formula on AP reference sheet. Copyright © Cengage Learning. All rights reserved

Standard Entropy Values (S°) Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. ΔS°reaction = ΣnpS°products – ΣnrS°reactants n = moles (balanced coefficient) S° = standard entropy (from a given list in problem # or on pages A20-A21) See formula on AP reference sheet. Copyright © Cengage Learning. All rights reserved

Calculate Δ S° for the following reaction: Exercise Calculate Δ S° 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 Δ S°= –11 J/K [2(50) + 131] – [2(51) + 2(70)] = –11 J/K ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved

Calculate Δ S° for the following reaction: Exercise Calculate Δ S° for the following reaction: 2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s) Given the following information: S° (J/K·mol) SO2(g) 248 NiO(s) 38 O2(g) 205 NiS(s) 53 Δ S°= –149 J/K [2(50) + 131] – [2(51) + 2(70)] = –11 J/K ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved

Calculate Δ S° for the following reaction: Exercise Calculate Δ S° for the following reaction: Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g) Given the following information: S° (J/K·mol) Al2O3(s) 51 H2(g) 131 Al(s) 28 H2O(g) 189 Δ S°= 179 J/K [2(50) + 131] – [2(51) + 2(70)] = –11 J/K ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved

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. ΔG° = ΔH° – TΔS° ΔG°reaction = ΣnpG°products – ΣnrG°reactants ΔH°reaction = ΣnpH°products – ΣnrH°reactants ΔS°reaction = ΣnpS°products – ΣnrS°reactants A20-A21 for ΔS°, ΔH°, ΔG° data… Copyright © Cengage Learning. All rights reserved

A stable diatomic molecule spontaneously forms from its atoms. Concept Check A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: H° S° G° Explain. The reaction is exothermic, more ordered, and spontaneous. Thus, the sign of H is negative; S is negative; and G is negative. – – – Copyright © Cengage Learning. All rights reserved

H° S° G° Predict the sign of: Explain Copyright © Cengage Learning. All rights reserved

To predict the sign of ΔS for a given chemical reaction. Objectives To predict the sign of ΔS for a given chemical reaction. To calculate ΔS°reaction given ΔS°values for products and reactants To calculate ΔS°reaction , ΔH°reaction , ΔG°reaction Work Session # 33, 34, 37, 39, 40, 41, 43, 45, 47, 51 (like Hess’s Law!) A20-A21 for ΔS°, ΔH°, ΔG° data… Copyright © Cengage Learning. All rights reserved

Consider the following system at equilibrium at 25°C. Concept Check Consider the following system at equilibrium at 25°C. PCl3(g) + Cl2(g) PCl5(g) G° = −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. S is negative (unfavorable) yet the reaction is spontaneous (G is negative). Thus, H must be negative (exothermic, favorable). Thus, as the temperature is increased, the reaction proceeds to the left, decreasing the ratio of partial pressure of PCl5 to the partial pressure of PCl3. Copyright © Cengage Learning. All rights reserved

Free Energy and Pressure G = G° + RT ln(P) or ΔG = ΔG° + RT ln(Q) Copyright © Cengage Learning. All rights reserved

ln(K) vs. 1/T (for both endothermic and exothermic cases) Concept Check Sketch graphs of: G vs. P H vs. P ln(K) vs. 1/T (for both endothermic and exothermic cases) G vs. P: A natural log graph (levels off as P increases). H vs. P: no relationship (slope of zero). lnK vs 1/T: endothermic - straight line, negative slope exothermic - straight line, positive slope Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved

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) Copyright © Cengage Learning. All rights reserved

Change in Free Energy to Reach Equilibrium Copyright © Cengage Learning. All rights reserved

Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction Copyright © Cengage Learning. All rights reserved

Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. wmax = ΔG Copyright © Cengage Learning. All rights reserved

All real processes are irreversible. 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. Copyright © Cengage Learning. All rights reserved

Calculate H, E, q, and w for the process you described above. Concept Check 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. Calculate H, E, q, and w for the process you described above. All are equal to zero. All are equal to zero. Since it is a constant temperature process, H = 0 and E = 0. The gas is working against zero pressure (evacuated bulb) so w = 0. E = q + w, so q = 0. Copyright © Cengage Learning. All rights reserved

Concept Check 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. b) Given your answer to part a, what is the driving force for the process? Entropy Some students are probably aware of the concept of entropy. This is a good introduction to the concept. Copyright © Cengage Learning. All rights reserved

The Microstates That Give a Particular Arrangement (State) Copyright © Cengage Learning. All rights reserved

Negative ΔG means positive ΔSuniv. 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. Copyright © Cengage Learning. All rights reserved