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 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. 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

The Expansion of An Ideal Gas Into an Evacuated Bulb Copyright © Cengage Learning. All rights reserved

Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. A measure of molecular randomness or disorder. A measure of how many micro states can exist. 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)

Positional Entropy A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system. Therefore: Ssolid < Sliquid << Sgas

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

ΔSsurr ΔSsurr = +; entropy of the universe increases ΔSsurr = -; process is spontaneous in opposite direction ΔSsurr = 0; process has no tendency to occur Copyright © Cengage Learning. All rights reserved

Entropy Changes in the Surroundings (ΔSsurr) ΔSsurr is determined by flow of energy as heat Exothermic process increases ΔSsurr Important driving force for spontaneity Endothermic process decreases ΔSsurr Impact of transfer of energy as heat to or from the surroundings is greater at lower temperatures Copyright © Cengage Learning. All rights reserved 13 13

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

ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow. The magnitude of ΔSsurr depends on the temperature. $50 to a millionaire or a high school student Copyright © Cengage Learning. All rights reserved

ΔSsurr Copyright © Cengage Learning. All rights reserved

ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH If the reaction is exothermic: ΔH has a negative sign ΔSsurr is positive since heat flows into the surroundings Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

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

All quantities refer to the system Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) All quantities refer to the system Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔSsurr ΔG 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

Effect of ΔH and ΔS on Spontaneity Copyright © Cengage Learning. All rights reserved

Interactive Example 17.5 - Free Energy and Spontaneity At what temperatures is the following process spontaneous at 1 atm? What is the normal boiling point of liquid Br2?

Interactive Example 17.5 - Solution The vaporization process will be spontaneous at all temperatures where ΔG°is negative Note that ΔS°favors the vaporization process because of the increase in positional entropy, and ΔH°favors the opposite process, which is exothermic These opposite tendencies will exactly balance at the boiling point of liquid Br2, since at this temperature liquid and gaseous Br2 are in equilibrium (ΔG°= 0)

Interactive Example 17.5 - Solution (Continued 1) We can find this temperature by setting ΔG°= 0 in the following equation:

Interactive Example 17.5 - Solution (Continued 2) At temperatures above 333 K, TΔS°has a larger magnitude than ΔH°, and ΔG°is negative Above 333 K, the vaporization process is spontaneous The opposite process occurs spontaneously below this temperature At 333 K, liquid and gaseous Br2 coexist in equilibrium

Interactive Example 17.5 - Solution (Continued 3) Summary of observations (the pressure is 1 atm in each case) T > 333 K The term ΔS°controls, and the increase in entropy when liquid Br2 is vaporized is dominant T < 333 K The process is spontaneous in the direction in which it is exothermic, and the term ΔH°controls

Interactive Example 17.5 - Solution (Continued 4) T = 333 K The opposing driving forces are just balanced (ΔH°= 0), and the liquid and gaseous phases of bromine coexist This is the normal boiling point

Entropy Changes and Chemical Reactions Positional probability determines the changes that occur in a chemical system Fewer the molecules, fewer the possible configurations Copyright © Cengage Learning. All rights reserved 29 29

Figure 17.6 - Entropy of Water H2O molecule can vibrate and rotate in several ways Freedom of motion leads to a higher entropy for water 30

Entropy Changes in Reactions That Involve Gaseous Molecules Change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products If the number of product molecules is greater than the number of reactant molecules: Positional entropy increases ΔS is positive Copyright © Cengage Learning. All rights reserved 31 31

Interactive Example 17.6 - Predicting the Sign of ΔS° Predict the sign of ΔS°for each of the following reactions Thermal decomposition of solid calcium carbonate Oxidation of SO2 in air

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

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: EXERCISE! 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/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

CONCEPT CHECK! A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS° ΔG° 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) 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. 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

Interactive Example 17.11 - Calculating ΔG° Methanol is a high-octane fuel used in high-performance racing engines Calculate ΔG °for the following reaction: The following energies of formation are provided: 39

Interactive Example 17.11 - Solution Use the following equation:

Interactive Example 17.11 - Solution (Continued) The large magnitude and the negative sign of ΔG°indicate that this reaction is very favourable thermodynamically

Free Energy and Pressure System under constant P and T proceeds spontaneously in the direction that lowers its free energy Free energy of a reaction system changes as the reaction proceeds Dependent on the pressure of a gas or on the concentration of species in solution Equilibrium - Point where free energy value is at its lowest Copyright © Cengage Learning. All rights reserved 42 42

Free Energy and Pressure (Continued 1) For ideal gases: Enthalpy is not pressure-dependent Entropy depends on pressure due to its dependence on volume At a given temperature for 1 mole of ideal gas: Slarge volume > Ssmall volume Or, Slow pressure > Shigh pressure Copyright © Cengage Learning. All rights reserved 43 43

Free Energy and Pressure (Continued 2) G° - Free energy of a gas at 1 atm G - Free energy of the gas at a pressure of P atm R - Universal gas constant T - Temperature in Kelvin Copyright © Cengage Learning. All rights reserved 44 44

Free Energy and Pressure (Continued 3) Q - Reaction quotient T - Temperature in Kelvin R - Universal gas constant (8.3145 J/K·mol) ΔG°- Free energy change at 1 atm ΔG - Free energy change at specified pressures Copyright © Cengage Learning. All rights reserved 45 45

Interactive Example 17.13 - Calculating ΔG° One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol

Interactive Example 17.13 - Solution To calculate ΔG for this process, use the following equation: First compute ΔGₒ from standard free energies of formation

Interactive Example 17.13 - Solution (Continued 1) One might call this the value of ΔG°for one round of the reaction or for 1 mole of the reaction Thus, the ΔG°value might better be written as –2.9×104 J/mol of reaction, or –2.9×104 J/mol rxn Use this value to calculate the value of ΔG

Interactive Example 17.13 - Solution (Continued 2) ΔG°= – 2.9×104 J/mol rxn R = 8.3145 J/K·mol T = 273 + 25 = 298 K Note that the pure liquid methanol is not included in the calculation of Q

Interactive Example 17.13 - Solution (Continued 3)

Interactive Example 17.13 - Solution (Continued 4) Note that ΔG is significantly more negative than ΔG°, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm This result can be expected from Le Châtelier’s principle

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

Copyright © Cengage Learning. All rights reserved

Interactive Example 17.15 - Free Energy and Equilibrium II   56

Interactive Example 17.15 - Solution Calculate ΔG° from ΔG° = ΔH°– TΔS°

Interactive Example 17.15 - Solution (Continued 1) Therefore, ΔG° = ΔH°– TΔS° = (– 1.652×106 J) – (298 K) (– 543 J/K) = – 1.490 ×106 J

Interactive Example 17.15 - Solution (Continued 2) Therefore, K = e601 This is a very large equilibrium constant The rusting of iron is clearly very favourable from a thermodynamic point of view

Reversible and Irreversible Processes Reversible process: Universe is exactly the same as it was before a cyclic process Irreversible process: Universe is different after a cyclic process All real processes are irreversible Characteristics of a real cyclic process Work is changed to heat Entropy of the universe increases Copyright © Cengage Learning. All rights reserved 60 60