Spontaneity, Entropy, and Free Energy Chapter 17 Spontaneity, Entropy, and Free Energy

17.1 Spontaneous Processes and Entropy 17.2 Entropy and the Second Law of Thermodynamics 17.3 The Effect of Temperature on Spontaneity 17.4 Free Energy 17.5 Entropy Changes in Chemical Reactions 17.6 Free Energy and Chemical Reactions 17.7 The Dependence of Free Energy on Pressure 17.8 Free Energy and Equilibrium 17.9 Free Energy and Work 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. Copyright © Cengage Learning. All rights reserved

What should happen to the gas when you open the valve? 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. 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

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. c) Given your answer to part b, 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 Expansion of An Ideal Gas Into an Evacuated Bulb Copyright © Cengage Learning. All rights reserved

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

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. Copyright © Cengage Learning. All rights reserved

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

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

Therefore: Ssolid < Sliquid << Sgas 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 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

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. Suniverse = ΔSsystem + ΔSsurroundings Copyright © Cengage Learning. All rights reserved

Concept Check For the process A(l) A(s), which direction involves an increase in energy randomness? 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? 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

Concept Check 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

The sign of ΔSsurr depends on the direction of the heat flow. The magnitude of ΔSsurr depends on the temperature. Copyright © Cengage Learning. All rights reserved

ΔSsurr Copyright © Cengage Learning. All rights reserved

ΔSsurr Copyright © Cengage Learning. All rights reserved

Heat flow (constant P) = change in enthalpy = ΔH ΔSsurr 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 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

ΔG = ΔH – TΔS (at constant T and P) Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) 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

Exercise The value of Hvaporization of substance X is 45.7 kJ/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. S = 132 J/K·mol Ssurr = -132 J/K·mol G = 0 kJ/mol 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

Effect of H and S on Spontaneity Copyright © Cengage Learning. All rights reserved

Concept Check 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: H Ssurr S Suniv Explain. Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written. Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive. – + 0 + Copyright © Cengage Learning. All rights reserved

Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature. 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 Copyright © Cengage Learning. All rights reserved

Calculate S° for the following reaction: Exercise Calculate 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 S°= –11 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 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

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