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Chapter 17 Spontaneity, Entropy, and Free Energy AP*

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AP Learning Objectives 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 H o and S o, and calculation or estimation of G o 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)

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AP Learning Objectives 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)

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

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Section 17.1 Spontaneous Processes and Entropy AP Learning Objectives, Margin Notes and References Learning Objectives 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”)

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 6 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)

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 7 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)

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Section 17.1 Spontaneous Processes and Entropy 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.

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Section 17.1 Spontaneous Processes and Entropy 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.

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Section 17.1 Spontaneous Processes and Entropy 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.

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Section 17.1 Spontaneous Processes and Entropy Experimental Factors Affect Spontaneous Processes Temperature and pressure can affect spontaneity. An example of how temperature affects spontaneity is ice melting or freezing.

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Section 17.1 Spontaneous Processes and Entropy Reversible and Irreversible Processes 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. 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!

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 13 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. a)What should happen to the gas when you open the valve? CONCEPT CHECK!

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 14 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)Calculate ΔH, ΔE, q, and w for the process you described above. All are equal to zero. CONCEPT CHECK!

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 15 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 CONCEPT CHECK!

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 16 The Expansion of An Ideal Gas Into an Evacuated Bulb

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 17 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.

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 18 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.

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Section 17.1 Spontaneous Processes and Entropy Microstate: A single possible arrangement of position and kinetic energy of molecules

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Section 17.1 Spontaneous Processes and Entropy 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) 6.02×10 23 ! (No chance!) Gases spontaneously expand to fill the volume given. Most probable arrangement of molecules: approximately equal molecules in each side

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Section 17.1 Spontaneous Processes and Entropy 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.

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Section 17.1 Spontaneous Processes and 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).

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Section 17.1 Spontaneous Processes and Entropy 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.

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Section 17.1 Spontaneous Processes and Entropy 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.

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Section 17.1 Spontaneous Processes and Entropy Copyright © Cengage Learning. All rights reserved 25 Predict the sign of ΔS for each of the following, and explain: a)The evaporation of alcohol b)The freezing of water c)Compressing an ideal gas at constant temperature d)Heating an ideal gas at constant pressure e)Dissolving NaCl in water + – – + + CONCEPT CHECK!

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Section 17.2 Entropy and the Second Law of Thermodynamics AP Learning Objectives, Margin Notes and References Learning Objectives 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”)

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Section 17.2 Entropy and the Second Law of Thermodynamics 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 S universe = ΔS system + ΔS surroundings Copyright © Cengage Learning. All rights reserved 27

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Section 17.2 Entropy and the Second Law of Thermodynamics ΔS univ ΔS univ = +; entropy of the universe increases; spontaneous ΔS univ = -; process is spontaneous in opposite direction ΔS univ = 0; process has no tendency to occur Copyright © Cengage Learning. All rights reserved 28

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Section 17.3 The Effect of Temperature on Spontaneity AP Learning Objectives, Margin Notes and References Learning Objectives 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”)

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Section 17.3 The Effect of Temperature on Spontaneity 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? Copyright © Cengage Learning. All rights reserved 30 CONCEPT CHECK!

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Section 17.3 The Effect of Temperature on Spontaneity Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is: a)Exothermic and becomes more positionally random Spontaneous b)Exothermic and becomes less positionally random Cannot tell a)Endothermic and becomes more positionally random Cannot tell a)Endothermic and becomes less positionally random Not spontaneous Explain how temperature affects your answers. CONCEPT CHECK!

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Section 17.3 The Effect of Temperature on Spontaneity ΔS surr The sign of ΔS surr depends on the direction of the heat flow. The magnitude of ΔS surr depends on the temperature. Copyright © Cengage Learning. All rights reserved 32

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Section 17.3 The Effect of Temperature on Spontaneity ΔS surr Copyright © Cengage Learning. All rights reserved 33

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Section 17.3 The Effect of Temperature on Spontaneity ΔS surr Copyright © Cengage Learning. All rights reserved 34

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Section 17.3 The Effect of Temperature on Spontaneity At constant pressure, q sys is simply H° for the system. Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process

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Section 17.3 The Effect of Temperature on Spontaneity Copyright © Cengage Learning. All rights reserved 36

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Section 17.3 The Effect of Temperature on Spontaneity Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore, S universe = S system + S surroundings For spontaneous processes S universe > 0

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Section 17.4 Free Energy AP Learning Objectives, Margin Notes and References Learning Objectives 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 H o and S o, and calculation or estimation of G o 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.

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Section 17.4 Free Energy Total Entropy and Spontaneity ΔS universe = ΔS system + ΔS surroundings Substitute for the entropy of the surroundings: ΔS universe = ΔS system – ΔH system /T Multiply by −T: −TΔS universe = −TΔS system + ΔH system Rearrange: −TΔS universe = ΔH system − TΔS system Call −TΔS universe the Gibbs Free Energy (ΔG): ΔG = ΔH − T ΔS

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Section 17.4 Free Energy 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 ΔS univ. Copyright © Cengage Learning. All rights reserved 40

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Section 17.4 Free Energy Gibbs Free Energy 1.If G is negative, the forward reaction is spontaneous. 2.If G is 0, the system is at equilibrium. 3.If G is positive, the reaction is spontaneous in the reverse direction. More convenient to use G than S univ to determine spontaneity because G relates to the system alone (no need to examine the surroundings.

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Section 17.4 Free Energy A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔS surr ΔG Explain your answers. Copyright © Cengage Learning. All rights reserved 42 – + + + – 0 CONCEPT CHECK!

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Section 17.4 Free Energy The value of ΔH vaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS, ΔS surr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm. ΔS = 132 J/K·mol ΔS surr = -132 J/K·mol ΔG = 0 kJ/mol Copyright © Cengage Learning. All rights reserved 43 EXERCISE!

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Section 17.4 Free Energy

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Section 17.4 Free Energy Comparison of ΔS univ and ΔG ◦ to Predict Spontaneity Copyright © Cengage Learning. All rights reserved 45

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Section 17.4 Free Energy Effect of ΔH and ΔS on Spontaneity Copyright © Cengage Learning. All rights reserved 46

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Section 17.4 Free Energy

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Section 17.5 Entropy Changes in Chemical Reactions AP Learning Objectives, Margin Notes and References Learning Objectives 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.

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Section 17.5 Entropy Changes in Chemical Reactions

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Section 17.5 Entropy Changes in Chemical Reactions Gas A 2 reacts with gas B 2 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: ΔHΔS surr ΔSΔS univ Explain. Copyright © Cengage Learning. All rights reserved 50 – + 0 + CONCEPT CHECK!

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Section 17.5 Entropy Changes in Chemical Reactions 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.

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Section 17.5 Entropy Changes in Chemical Reactions 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.

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Section 17.5 Entropy Changes in Chemical Reactions 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 53

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Section 17.5 Entropy Changes in Chemical Reactions 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.

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Section 17.5 Entropy Changes in Chemical Reactions Calculate ΔS° for the following reaction: 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) Given the following information: S° (J/K·mol) Na(s) 51 H 2 O(l) 70 NaOH(aq) 50 H 2 (g) 131 ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved 55 EXERCISE!

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Section 17.6 Free Energy and Chemical Reactions AP Learning Objectives, Margin Notes and References Learning Objectives 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 H o and S o, and calculation or estimation of G o 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.

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Section 17.6 Free Energy and Chemical Reactions 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”)

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Section 17.6 Free Energy and Chemical 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° Copyright © Cengage Learning. All rights reserved 58

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Section 17.6 Free Energy and Chemical Reactions Standard Free Energy Change (ΔG°) There are several ways to calculate ΔG°(this is 3 out of 4) 1)ΔG° = ΔH° – TΔS° 2)Hess’s Law (additions of reaction ΔG°terms) 3)ΔG°reaction = Σ n p G f ° products – Σ n r G f ° reactants ΔG f °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 59

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Section 17.6 Free Energy and Chemical Reactions Copyright © Cengage Learning. All rights reserved 60 CONCEPT CHECK!

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Section 17.6 Free Energy and Chemical Reactions Copyright © Cengage Learning. All rights reserved 61 CONCEPT CHECK!

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Section 17.6 Free Energy and Chemical Reactions Copyright © Cengage Learning. All rights reserved 62 CONCEPT CHECK!

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Section 17.6 Free Energy and Chemical Reactions A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS°ΔG° Explain. Copyright © Cengage Learning. All rights reserved 63 – – – CONCEPT CHECK!

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Section 17.6 Free Energy and Chemical Reactions Consider the following system at equilibrium at 25°C. PCl 3 (g) + Cl 2 (g) PCl 5 (g) ΔG° = −92.50 kJ What will happen to the ratio of partial pressure of PCl 5 to partial pressure of PCl 3 if the temperature is raised? Explain. The ratio will decrease. Copyright © Cengage Learning. All rights reserved 64 CONCEPT CHECK!

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Section 17.8 Free Energy and Equilibrium AP Learning Objectives, Margin Notes and References Learning Objectives 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 G o 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”)

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Section 17.8 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.)

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Section 17.8 Free Energy and Equilibrium 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 N 2, per 3 mol of H 2, and per 2 mol of NH 3. Calculate ΔG at 298 K for a mixture of 1.0 atm N 2, 3.0 atm H 2, and 0.50 atm NH 3 being used in the Haber process: Calculating the Free-Energy Change under Nonstandard Conditions

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Section 17.8 Free Energy and Equilibrium 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 N 2, H 2, and NH 3 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 NH 3. We would make the same prediction based on Le Châtelier’s principle. Relative to standard conditions, we have increased the pressure of a reactant (H 2 ) and decreased the pressure of the product (NH 3 ). Le Châtelier’s principle predicts that both changes shift the reaction to the product side, thereby forming more NH 3. Continued Calculating the Free-Energy Change under Nonstandard Conditions

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Section 17.8 Free Energy and Equilibrium 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 69

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Section 17.8 Free Energy and Equilibrium 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) Copyright © Cengage Learning. All rights reserved 70

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Section 17.8 Free Energy and Equilibrium 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 71 Fourth way to calculate ΔG°

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Section 17.8 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

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Section 17.8 Free Energy and Equilibrium 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) = e 13.4 = 7 × 10 5 The standard free-energy change for the Haber process at 25 °C was obtained for the Haber reaction: N 2 (g) + 3 H 2 (g) 2 NH 3 (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. Calculating an Equilibrium Constant from ΔG°

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Section 17.8 Free Energy and Equilibrium Copyright © Cengage Learning. All rights reserved 74

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Section 17.9 Free Energy and Work AP Learning Objectives, Margin Notes and References Learning Objectives 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”)

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Section 17.9 Free Energy and Work Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy. w max = ΔG Copyright © Cengage Learning. All rights reserved 76

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Section 17.9 Free Energy and Work 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. Copyright © Cengage Learning. All rights reserved 77

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