LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. (Sec 17.1) LO 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. (Sec 17.3) LO 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. (Sec 17.1-17.3, 17.5) LO 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 Ho and So, and calculation or estimation of Go when needed. (Sec 17.4, 17.6) LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. (Sec 17.4, 17.6)

LO 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. (Sec 17.6, 17.9) LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. (Sec 17.6) LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. (Sec 17.6) LO 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. (Sec 17.1, 17.6-17.8)

LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. (Sec 17.8)

AP Learning Objectives, Margin Notes and References LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. LO 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. LO 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. Additional AP References LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)

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 (do not need knowledge of pathway b/n R and P) 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. (Spontaneous does not mean fast) Copyright © Cengage Learning. All rights reserved

First Law of Thermodynamics You will recall from Chapter 6 that energy cannot be created or destroyed. (E = q + w) Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. The first law only guarantees that energy is conserved, but it says nothing about the preferred direction of the process.

Both play a role in determining whether a process is spontaneous. Enthalpy/Entropy Enthalpy is the heat absorbed by a system during a constant-pressure process. Entropy is a measure of the randomness in a system. It can also be thought of as dispersion or spreading out of energy. Both play a role in determining whether a process is spontaneous.

Spontaneous processes proceed without any outside assistance. The gas in vessel A will spontaneously effuse into vessel B, but it will not spontaneously return to vessel A. Processes that are spontaneous in one direction and are nonspontaneous in the reverse direction. Both processes are directional spontaneous even though energy is conserved in both directions. First Law not enough to predict direction of a chemical reaction.

Experimental Factors Affect Spontaneous Processes Temperature and pressure can affect spontaneity. An example of how temperature affects spontaneity is ice melting or freezing. Unless at normal melting point T=0C, then freezing and melting are in equilibrium and there is no preferred direction.

Reversible and Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system or cannot have the process exactly followed in reverse. Also, any spontaneous process is irreversible! Reversible process: The system changes so that the system and surroundings can be returned to the original state by exactly reversing the process. This maximizes work done by a system on the surroundings.

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 ΔH, ΔE, 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 ΔH, ΔE, 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

Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. A measure of molecular randomness or disorder. It can also be thought of as dispersion or spreading out of energy. 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

Microstate: A single possible arrangement of position and kinetic energy of molecules Each configuration that gives a particular arrangement is called a microstate.

Entropy on the Molecular Scale Boltzmann described entropy on the molecular level. Gas molecule expansion: Two molecules are in the apparatus above; both start in one side. What is the likelihood they both will end up there? (1/2)2 If one mole is used? (1/2)6.02×1023! (No chance!) Gases spontaneously expand to fill the volume given. Most probable arrangement of molecules: approximately equal molecules in each side

Effect of Volume and Temperature Change on the System If we increase volume, there are more positions possible for the molecules. This results in more microstates, so increased entropy. If we increase temperature, the average kinetic energy increases. This results in a greater distribution of molecular speeds. Therefore, there are more possible kinetic energy values, resulting in more microstates, increasing entropy.

Molecular Motions Molecules exhibit several types of motion. Translational: Movement of the entire molecule from one place to another Vibrational: Periodic motion of atoms within a molecule Rotational: Rotation of the molecule about an axis Note: More atoms means more microstates (more possible molecular motions). 

Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tend to increase with increases in temperature. volume. the number of independently moving molecules. Increase with the formation of more complex molecules. Tendency to mix is due to increased volume available to particles so more potential positions.

Entropy and Physical States Entropy increases with the freedom of motion of molecules. S(g) > S(l) > S(s) Entropy of a system increases for processes where gases form from either solids or liquids. liquids or solutions form from solids (Carbonates are an exception) the number of gas molecules increases during a chemical reaction. Increase, increase, decrease microstates so decrease entropy Disordering of ions compensates for the ordering of the water molecules in hydration.

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

AP Learning Objectives, Margin Notes and References LO 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. Additional AP References LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)

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. Δ Suniverse = ΔSsystem + ΔSsurroundings System S primarily driven by positional entropy. Surrounding S primarily driven by heat transfer (endo vs exo). Copyright © Cengage Learning. All rights reserved

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

AP Learning Objectives, Margin Notes and References LO 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. LO 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. Additional AP References LO 5.3 (see Appendix 7.2, “Thermal Equilibrium, the Kinetic Molecular Theory, and the Process of Heat”) LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)

CONCEPT CHECK! For the process A(l) A(s), which direction involves an increase in energy randomness (exothermic)? 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.

ΔSsurr 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

Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process At constant pressure, qsys is simply H° for the system.

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Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore, Suniverse = Ssystem + Ssurroundings For spontaneous processes Suniverse > 0

AP Learning Objectives, Margin Notes and References LO 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 Ho and So, and calculation or estimation of Go when needed. LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.

Total Entropy and Spontaneity ΔSuniverse = ΔSsystem + ΔSsurroundings Substitute for the entropy of the surroundings: ΔSuniverse = ΔSsystem – ΔHsystem/T Multiply by −T: −TΔSuniverse = −TΔSsystem + ΔHsystem Rearrange: −TΔSuniverse = ΔHsystem − TΔSsystem Call −TΔSuniverse the Gibbs Free Energy (ΔG): ΔG = ΔH − T ΔS ΔSuniverse predict spontaneity for all processes. ΔG (constant T and P) predict spontaneity for chemical processes. More practical. Helps to understand the temperature dependence of spontaneity. Accounts for heat and entropy changes. Involves measuring system only.

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

Gibbs Free Energy If DG is negative, the forward reaction is spontaneous. If DG is 0, the system is at equilibrium. If G is positive, the reaction is spontaneous in the reverse direction. More convenient to use DG than DSuniv to determine spontaneity because DG relates to the system alone (no need to examine the surroundings.

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

EXERCISE! The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm. ΔS = 132 J/K·mol ΔSsurr = -132 J/K·mol ΔG = 0 kJ/mol S = 132 J/K·mol Ssurr = -132 J/k·mol G = 0 kJ/mol Copyright © Cengage Learning. All rights reserved

Comparison of ΔSuniv and ΔG◦ to Predict Spontaneity Copyright © Cengage Learning. All rights reserved

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

Set ΔG◦ = 0 and solve for T. T= 333K, so spontaneous at >333K.

AP Learning Objectives, Margin Notes and References LO 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.

a) Increases b) decreases

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: ΔH ΔSsurr ΔS ΔSuniv 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 pure crystalline substance at absolute zero is 0. Consider all atoms or molecules in the perfect lattice at 0 K; there will only be one microstate. The entropy of a substance increases with temperature. Can assign absolute values for entropy because of 3rd Law. One of few thermodynamic characteristics that can have an absolute value. Usually measure change.

Standard Entropies The reference for entropy is 0 K, so the values for elements are not 0 J/mol K at 298 K. Standard molar entropy for gases are generally greater than liquids and solids. (Be careful of size!) Standard entropies increase with molar mass. Standard entropies increase with number of atoms in a formula.

Standard MolarEntropy Values (S°) Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. Copyright © Cengage Learning. All rights reserved

Entropy Changes (a state function) Entropy changes for a reaction can be calculated in a manner analogous to that by which H is calculated: S° = nS°(products) – mS°(reactants) where n and m are the coefficients in the balanced chemical equation.

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 Hydrogen is less complex that water so entropy decreases. ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 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 Ho and So, and calculation or estimation of Go when needed. LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. LO 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. LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. LO 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.

AP Learning Objectives, Margin Notes and References Additional AP References LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 5.16 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 5.17 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)

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. We use ΔG°for various reactions to compare the relative tendency of the reactions to occur under the same pressure and concentration conditions. The more neg ΔG°, the further the reaction will go right to reach equilibrium. ΔG° = ΔH° – TΔS° Can not measure ΔG° because rxn goes to equilibrium not completion and no instrument to measure ΔG° directly Copyright © Cengage Learning. All rights reserved

Standard Free Energy Change (ΔG°) There are several ways to calculate ΔG°(this is 3 out of 4) ΔG° = ΔH° – TΔS° Hess’s Law (additions of reaction ΔG°terms) ΔG°reaction = ΣnpGf°products – ΣnrGf°reactants ΔGf°is defined as the change in free energy that accompanies the formation of one mole of that substance from its constituent elements with all reactants and products in their standard states. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! H= -198kJ, s= -187J/K G= -142kJ Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! -3kJ Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! G= -6kJ Thermodynamically favored at 25C but may be too slow. Must study kinetics. If raise temp to speed up, need to recalculate G at that temp because may not be favored. 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

AP Learning Objectives, Margin Notes and References LO 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. LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. Additional AP References LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 6.25 (see Appendix 7.11, “Non-Spontaneous Reactions”)

Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way: R= 8.314 J/mol K G = G° + RT ln Q (Under standard conditions, concentrations are 1 M, so Q = 1 and ln Q = 0; the last term drops out.) Q in pressures or concentrations

Calculating the Free-Energy Change under Nonstandard Conditions Calculate ΔG at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process: Solution Plan We need to calculate the value of the reaction quotient Q for the specified partial pressures, for which we use the partial-pressures form: Q = [D]d[E]e/[A]a[B]b. We then use a table of standard free energies of formation to evaluate ΔG°. We calculated ΔG° = −33.3 kJ for this reaction. We will have to change the units of this, however, for the units to work out, we will use kJ/mol as our units for ΔG°, where “per mole” means “per mole of the reaction as written.” Thus, ΔG° = −33.3 kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3.

Calculating the Free-Energy Change under Nonstandard Conditions Continued We now use Equation 19.19 to calculate ∆G for these nonstandard conditions: ΔG = ΔG° + RT ln Q = (−33.3 kJ/mol) + (8.314 J/mol-K)(298 K)(1 kJ/1000 J) ln(9.3 × 10–3) = (−33.3 kJ/mol) + (−11.6 kJ/mol) = −44.9 kJ/mol Comment We see that ΔG becomes more negative as the pressures of N2, H2, and NH3 are changed from 1.0 atm (standard conditions, ΔG°) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger negative value for ΔG indicates a larger “driving force” to produce NH3. We would make the same prediction based on Le Châtelier’s principle. Relative to standard conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Châtelier’s principle predicts that both changes shift the reaction to the product side, thereby forming more NH3.

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. a) Phase change b) chemical reactions do not go to completion . point c is equilibrium Copyright © Cengage Learning. All rights reserved 69

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. For reaction A(g) ↔ B(g) Figure 17.8 | (a) The initial free energies of A and B. (b) As A(g) changes to B(g), the free energy of A decreases and that of B increases. (c) Eventually, pressures of A and B are achieved such that GA = GB, the equilibrium position. Copyright © Cengage Learning. All rights reserved 70

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) Fourth way to calculate ΔG° Copyright © Cengage Learning. All rights reserved

Free Energy and Equilibrium At equilibrium, Q = K, and G = 0. The equation becomes 0 = G° + RT ln K Rearranging, this becomes G° = RT ln K or K = e G/RT

N2(g) + 3 H2(g) 2 NH3(g) ∆G° = −33.3 kJ/mol = –33,300 J/mol Calculating an Equilibrium Constant from ΔG° The standard free-energy change for the Haber process at 25 °C was obtained for the Haber reaction: N2(g) + 3 H2(g) 2 NH3(g) ∆G° = −33.3 kJ/mol = –33,300 J/mol Use this value of ∆G° to calculate the equilibrium constant for the process at 25 °C. Solution Remembering to use the absolute temperature for T in and the form of R that matches our units, we have K = e−ΔG°/RT = e−(−33,300 J ⁄ mol)/(8.314 J/mol-K)(298 K) = e13.4 = 7 × 105

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AP Learning Objectives, Margin Notes and References LO 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. Additional AP References LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”)

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. As we use energy, we degrade its usefulness because we spread it out. Henry Bendt Copyright © Cengage Learning. All rights reserved