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Copyright  2011 Pearson Education, Inc. Chapter 17 Free Energy and Thermodynamics Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts.

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Presentation on theme: "Copyright  2011 Pearson Education, Inc. Chapter 17 Free Energy and Thermodynamics Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts."— Presentation transcript:

1 Copyright  2011 Pearson Education, Inc. Chapter 17 Free Energy and Thermodynamics Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2 Copyright  2011 Pearson Education, Inc. First Law of Thermodynamics You can’t win! First Law of Thermodynamics: Energy cannot be created or destroyed the total energy of the universe cannot change though you can transfer it from one place to another   E universe = 0 =  E system +  E surroundings 2Tro: Chemistry: A Molecular Approach, 2/e

3 Copyright  2011 Pearson Education, Inc. First Law of Thermodynamics Conservation of Energy For an exothermic reaction, “lost” heat from the system goes into the surroundings Two ways energy is “lost” from a system converted to heat, q used to do work, w Energy conservation requires that the energy change in the system equal the heat released + work done  E = q + w  E =  H + P  V  E is a state function internal energy change independent of how done 3Tro: Chemistry: A Molecular Approach, 2/e

4 Copyright  2011 Pearson Education, Inc. The Energy Tax You can’t break even! To recharge a battery with 100 kJ of useful energy will require more than 100 kJ because of the Second Law of Thermodynamics Every energy transition results in a “loss” of energy an “Energy Tax” demanded by nature and conversion of energy to heat which is “lost” by heating up the surroundings 4Tro: Chemistry: A Molecular Approach, 2/e

5 Copyright  2011 Pearson Education, Inc. Heat Tax fewer steps generally results in a lower total heat tax 5Tro: Chemistry: A Molecular Approach, 2/e

6 Copyright  2011 Pearson Education, Inc. Thermodynamics and Spontaneity Thermodynamics predicts whether a process will occur under the given conditions processes that will occur are called spontaneous  nonspontaneous processes require energy input to go Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction if the system after reaction has less potential energy than before the reaction, the reaction is thermodynamically favorable. Spontaneity ≠ fast or slow 6Tro: Chemistry: A Molecular Approach, 2/e

7 Copyright  2011 Pearson Education, Inc. Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end 7Tro: Chemistry: A Molecular Approach, 2/e

8 Copyright  2011 Pearson Education, Inc. Reversibility of Process Any spontaneous process is irreversible because there is a net release of energy when it proceeds in that direction it will proceed in only one direction A reversible process will proceed back and forth between the two end conditions any reversible process is at equilibrium results in no change in free energy If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction 8Tro: Chemistry: A Molecular Approach, 2/e

9 Copyright  2011 Pearson Education, Inc. Thermodynamics vs. Kinetics 9Tro: Chemistry: A Molecular Approach, 2/e

10 Copyright  2011 Pearson Education, Inc. Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations) 10Tro: Chemistry: A Molecular Approach, 2/e

11 Copyright  2011 Pearson Education, Inc. Spontaneous Processes Spontaneous processes occur because they release energy from the system Most spontaneous processes proceed from a system of higher potential energy to a system at lower potential energy exothermic But there are some spontaneous processes that proceed from a system of lower potential energy to a system at higher potential energy endothermic How can something absorb potential energy, yet have a net release of energy? 11Tro: Chemistry: A Molecular Approach, 2/e

12 Copyright  2011 Pearson Education, Inc. Melting Ice 12 Melting is an Endothermic process, yet ice will spontaneously melt above 0 °C. When a solid melts, the particles have more freedom of movement. More freedom of motion increases the randomness of the system. When systems become more random, energy is released. We call this energy, entropy Tro: Chemistry: A Molecular Approach, 2/e

13 Copyright  2011 Pearson Education, Inc. Factors Affecting Whether a Reaction Is Spontaneous There are two factors that determine whether a reaction is spontaneous. They are the enthalpy change and the entropy change of the system The enthalpy change,  H, is the difference in the sum of the internal energy and PV work energy of the reactants to the products The entropy change,  S, is the difference in randomness of the reactants compared to the products 13Tro: Chemistry: A Molecular Approach, 2/e

14 Copyright  2011 Pearson Education, Inc. Enthalpy Change   H generally measured in kJ/mol Stronger bonds = more stable molecules A reaction is generally exothermic if the bonds in the products are stronger than the bonds in the reactants exothermic = energy released,  H is negative A reaction is generally endothermic if the bonds in the products are weaker than the bonds in the reactants endothermic = energy absorbed,  H is positive The enthalpy change is favorable for exothermic reactions and unfavorable for endothermic reactions 14Tro: Chemistry: A Molecular Approach, 2/e

15 Copyright  2011 Pearson Education, Inc. Entropy Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S S generally J/mol S = k ln W k = Boltzmann Constant = 1.38 x 10 −23 J/K W is the number of energetically equivalent ways a system can exist  unitless Random systems require less energy than ordered systems 15Tro: Chemistry: A Molecular Approach, 2/e

16 Copyright  2011 Pearson Education, Inc. W These are energetically equivalent states for the expansion of a gas. 16 It doesn’t matter, in terms of potential energy, whether the molecules are all in one flask, or evenly distributed But one of these states is more probable than the other two Tro: Chemistry: A Molecular Approach, 2/e

17 Copyright  2011 Pearson Education, Inc. This macrostate can be achieved through several different arrangements of the particles Macrostates → Microstates 17 These microstates all have the same macrostate So there are six different particle arrangements that result in the same macrostate Tro: Chemistry: A Molecular Approach, 2/e

18 Copyright  2011 Pearson Education, Inc. Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State B There are six possible arrangements that give State C Therefore State C has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy There is six times the probability of having the State C macrostate than either State A or State B 18Tro: Chemistry: A Molecular Approach, 2/e

19 Copyright  2011 Pearson Education, Inc. Changes in Entropy,  S   S = S final − S initial Entropy change is favorable when the result is a more random system  S is positive Some changes that increase the entropy are reactions whose products are in a more random state  solid more ordered than liquid more ordered than gas reactions that have larger numbers of product molecules than reactant molecules increase in temperature solids dissociating into ions upon dissolving 19Tro: Chemistry: A Molecular Approach, 2/e

20 Copyright  2011 Pearson Education, Inc. Increases in Entropy 20Tro: Chemistry: A Molecular Approach, 2/e

21 Copyright  2011 Pearson Education, Inc. SS For a process where the final condition is more random than the initial condition,  S system is positive and the entropy change is favorable for the process to be spontaneous For a process where the final condition is more orderly than the initial condition,  S system is negative and the entropy change is unfavorable for the process to be spontaneous   S system  S reaction =  n(S° products ) −  n(S° reactants ) 21Tro: Chemistry: A Molecular Approach, 2/e

22 Copyright  2011 Pearson Education, Inc. Entropy Change in State Change When materials change state, the number of macrostates it can have changes as well the more degrees of freedom the molecules have, the more macrostates are possible solids have fewer macrostates than liquids, which have fewer macrostates than gases 22Tro: Chemistry: A Molecular Approach, 2/e

23 Copyright  2011 Pearson Education, Inc. Entropy Change and State Change 23Tro: Chemistry: A Molecular Approach, 2/e

24 Copyright  2011 Pearson Education, Inc. Practice – Predict whether  S system is + or − for each of the following A hot beaker burning your fingers Water vapor condensing Separation of oil and vinegar salad dressing Dissolving sugar in tea 2 PbO 2 (s)  2 PbO(s) + O 2 (g) 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) Ag + (aq) + Cl − (aq)  AgCl(s)  S is +  S is −  S is +  S is − 24Tro: Chemistry: A Molecular Approach, 2/e

25 Copyright  2011 Pearson Education, Inc. The 2 nd Law of Thermodynamics The 2 nd Law of Thermodynamics says that the total entropy change of the universe must be positive for a process to be spontaneous for reversible process  S univ = 0 for irreversible (spontaneous) process  S univ > 0  S universe =  S system +  S surroundings If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when  S system is negative,  S surroundings must be positive and big for a spontaneous process 25Tro: Chemistry: A Molecular Approach, 2/e

26 Copyright  2011 Pearson Education, Inc. Heat Flow, Entropy, and the 2 nd Law According to the 2 nd Law, heat must flow from water to ice because it results in more dispersal of heat. The entropy of the universe increases. 26 When ice is placed in water, heat flows from the water into the ice Tro: Chemistry: A Molecular Approach, 2/e

27 Copyright  2011 Pearson Education, Inc. Heat Transfer and Changes in Entropy of the Surroundings The 2 nd Law demands that the entropy of the universe increase for a spontaneous process Yet processes like water vapor condensing are spontaneous, even though the water vapor is more random than the liquid water If a process is spontaneous, yet the entropy change of the process is unfavorable, there must have been a large increase in the entropy of the surroundings The entropy increase must come from heat released by the system – the process must be exothermic! 27Tro: Chemistry: A Molecular Approach, 2/e

28 Copyright  2011 Pearson Education, Inc. Entropy Change in the System and Surroundings 28 When the entropy change in system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive), and large to allow the process to be spontaneous Tro: Chemistry: A Molecular Approach, 2/e

29 Copyright  2011 Pearson Education, Inc. Heat Exchange and  S surroundings When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings The amount the entropy of the surroundings changes depends on its original temperature the higher the original temperature, the less effect addition or removal of heat has 29Tro: Chemistry: A Molecular Approach, 2/e

30 Copyright  2011 Pearson Education, Inc. Temperature Dependence of  S surroundings When heat is added to surroundings that are cool it has more of an effect on the entropy than it would have if the surroundings were already hot Water freezes spontaneously below 0 °C because the heat released on freezing increases the entropy of the surroundings enough to make  S positive above 0 °C the increase in entropy of the surroundings is insufficient to make  S positive 30Tro: Chemistry: A Molecular Approach, 2/e

31 Copyright  2011 Pearson Education, Inc. Quantifying Entropy Changes in Surroundings The entropy change in the surroundings is proportional to the amount of heat gained or lost q surroundings = −q system The entropy change in the surroundings is also inversely proportional to its temperature At constant pressure and temperature, the overall relationship is 31Tro: Chemistry: A Molecular Approach, 2/e

32 Copyright  2011 Pearson Education, Inc. Example 17.2a: Calculate the entropy change of the surroundings at 25ºC for the reaction below C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  H rxn = −2044 kJ 32 combustion is largely exothermic, so the entropy of the surroundings should increase significantly  H system = −2044 kJ, T = 25 ºC = 298 K  S surroundings, J/K Check: Solution: Conceptual Plan: Relationships: Given: Find: SST,  H Tro: Chemistry: A Molecular Approach, 2/e

33 Copyright  2011 Pearson Education, Inc. Practice – The reaction below has  H rxn = kJ at 25 °C. (a) Determine the  s surroundings, (b) the sign of  S system, and (c) whether the process is spontaneous 2 O 2 (g) + N 2 (g)  2 NO 2 (g) 33Tro: Chemistry: A Molecular Approach, 2/e

34 Copyright  2011 Pearson Education, Inc. Practice – The reaction 2 O 2 (g) + N 2 (g)  2 NO 2 (g) has  H rxn = kJ at 25 °C. Calculate the entropy change of the surroundings. Determine the sign of the entropy change in the system and whether the reaction is spontaneous.  H sys = kJ, T = 25 ºC = 298 K  S surr, J/K,  S react + or −,  S universe + or − Solution: Conceptual Plan: Relationships: Given: Find: SST,  H the major difference is that there are fewer product molecules than reactant molecules, so the  S reaction is unfavorable and (−) because  S surroundings is (−) it is unfavorable both  S surroundings and  S reaction are (−),  S universe is (−) and the process is nonspontaneous 34Tro: Chemistry: A Molecular Approach, 2/e

35 Copyright  2011 Pearson Education, Inc. Gibbs Free Energy and Spontaneity It can be shown that −T  S univ =  H sys −T  S sys The Gibbs Free Energy, G, is the maximum amount of work energy that can be released to the surroundings by a system for a constant temperature and pressure system the Gibbs Free Energy is often called the Chemical Potential because it is analogous to the storing of energy in a mechanical system   G sys =  H sys −T  S sys Because  S univ determines if a process is spontaneous,  G also determines spontaneity  S univ is + when spontaneous, so  G is − 35Tro: Chemistry: A Molecular Approach, 2/e

36 Copyright  2011 Pearson Education, Inc. Gibbs Free Energy,  G A process will be spontaneous when  G is negative   G will be negative when  H is negative and  S is positive  exothermic and more random  H is negative and large and  S is negative but small  H is positive but small and  S is positive and large  or high temperature  G will be positive when  H is + and  S is − never spontaneous at any temperature When  G = 0 the reaction is at equilibrium 36Tro: Chemistry: A Molecular Approach, 2/e

37 Copyright  2011 Pearson Education, Inc.  G,  H, and  S 37Tro: Chemistry: A Molecular Approach, 2/e

38 Copyright  2011 Pearson Education, Inc. Free Energy Change and Spontaneity 38Tro: Chemistry: A Molecular Approach, 2/e

39 Copyright  2011 Pearson Education, Inc. Example 17.3a: The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = kJ and  S = J/K at 25 °C. Calculate  G and determine if it is spontaneous. 39 Because  G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.  H = kJ,  S = J/K, T = 298 K  G, kJ Answer: Solution: Conceptual Plan: Relationships: Given: Find: GGT,  H,  S Tro: Chemistry: A Molecular Approach, 2/e

40 Copyright  2011 Pearson Education, Inc. Practice – The reaction Al (s) + Fe 2 O 3(s)  Fe (s) + Al 2 O 3(s) has  H = −847.6 kJ and  S = −41.3 J/K at 25 °C. Calculate  G and determine if it is spontaneous. 40Tro: Chemistry: A Molecular Approach, 2/e

41 Copyright  2011 Pearson Education, Inc. Practice – The reaction Al (s) + Fe 2 O 3(s)  Fe (s) + Al 2 O 3(s) has  H = −847.6 kJ and  S = −41.3 J/K at 25 °C. Calculate  G and determine if it is spontaneous. Because  G is −, the reaction is spontaneous at this temperature. To make it nonspontaneous, we need to increase the temperature.  H = −847.6 kJ and  S = −41.3 J/K, T = 298 K  G, kJ Answer: Solution: Conceptual Plan: Relationships: Given: Find: GGT,  H,  S 41Tro: Chemistry: A Molecular Approach, 2/e

42 Copyright  2011 Pearson Education, Inc. Example 17.3b: The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = kJ and  S = J/K. Calculate the minimum temperature it will be spontaneous. 42 the temperature must be higher than 673K for the reaction to be spontaneous  H = kJ,  S = J/K,  G < 0 T, K Answer: Solution: Conceptual Plan: Relationships: Given: Find: T  G,  H,  S Tro: Chemistry: A Molecular Approach, 2/e

43 Copyright  2011 Pearson Education, Inc. Practice – The reaction Al (s) + Fe 2 O 3(s)  Fe (s) + Al 2 O 3(s) has  H = −847.6 kJ and  S = −41.3 J/K. Calculate the maximum temperature it will be spontaneous. 43Tro: Chemistry: A Molecular Approach, 2/e

44 Copyright  2011 Pearson Education, Inc. Practice – The reaction Al (s) + Fe 2 O 3(s)  Fe (s) + Al 2 O 3(s) has  H = −847.6 kJ and  S = −41.3 J/K at 25 °C. Determine the maximum temperature it is spontaneous. any temperature above 1775  C will make the reaction nonspontaneous  H = −847.6 kJ and  S = −41.3 J/K,  G > 0 T,  C Answer: Solution: Conceptual Plan: Relationships: Given: Find: T  G,  H,  S 2048 − 273 = 1775  C 44Tro: Chemistry: A Molecular Approach, 2/e

45 Copyright  2011 Pearson Education, Inc. Standard Conditions The standard state is the state of a material at a defined set of conditions Gas = pure gas at exactly 1 atm pressure Solid or Liquid = pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest usually 25 °C Solution = substance in a solution with concentration 1 M 45Tro: Chemistry: A Molecular Approach, 2/e

46 Copyright  2011 Pearson Education, Inc. The 3 rd Law of Thermodynamics: Absolute Entropy The absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles The 3 rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy therefore, the absolute entropy of substances is always + 46Tro: Chemistry: A Molecular Approach, 2/e

47 Copyright  2011 Pearson Education, Inc. Standard Absolute Entropies S° Extensive Entropies for 1 mole of a substance at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution 47Tro: Chemistry: A Molecular Approach, 2/e

48 Copyright  2011 Pearson Education, Inc. Standard Absolute Entropies 48Tro: Chemistry: A Molecular Approach, 2/e

49 Copyright  2011 Pearson Education, Inc. Relative Standard Entropies: States The gas state has a larger entropy than the liquid state at a particular temperature The liquid state has a larger entropy than the solid state at a particular temperature Substance S°, (J/mol∙K) H 2 O (l)70.0 H 2 O (g) Tro: Chemistry: A Molecular Approach, 2/e

50 Copyright  2011 Pearson Education, Inc. Relative Standard Entropies: Molar Mass The larger the molar mass, the larger the entropy Available energy states more closely spaced, allowing more dispersal of energy through the states 50Tro: Chemistry: A Molecular Approach, 2/e

51 Copyright  2011 Pearson Education, Inc. Relative Standard Entropies: Allotropes The less constrained the structure of an allotrope is, the larger its entropy The fact that the layers in graphite are not bonded together makes it less constrained 51Tro: Chemistry: A Molecular Approach, 2/e

52 Copyright  2011 Pearson Education, Inc. Relative Standard Entropies: Molecular Complexity Larger, more complex molecules generally have larger entropy More available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) Ar (g) NO (g) Substance Molar Mass S°, (J/mol∙K) CO (g) C 2 H 4 (g) Tro: Chemistry: A Molecular Approach, 2/e

53 Copyright  2011 Pearson Education, Inc. Relative Standard Entropies Dissolution Dissolved solids generally have larger entropy Distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO 3 (s)143.1 KClO 3 (aq) Tro: Chemistry: A Molecular Approach, 2/e

54 Copyright  2011 Pearson Education, Inc. The Standard Entropy Change,  S The standard entropy change is the difference in absolute entropy between the reactants and products under standard conditions  Sº reaction = (∑n p Sº products ) − (∑n r Sº reactants ) remember: though the standard enthalpy of formation,  H f °, of an element is 0 kJ/mol, the absolute entropy at 25 °C, S°, is always positive 54Tro: Chemistry: A Molecular Approach, 2/e

55 Copyright  2011 Pearson Education, Inc. Example 17.4: Calculate  S  for the reaction 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (g) 55  S is +, as you would expect for a reaction with more gas product molecules than reactant molecules standard entropies from Appendix IIB  S, J/K Check: Sol’n: Conceptual Plan: Relationships: Given: Find: SSS  NH3, S  O2, S  NO, S  H2O Substance S , J/mol  K NH 3 (g)192.8 O2(g)O2(g)205.2 NO(g)210.8 H 2 O(g)188.8 Tro: Chemistry: A Molecular Approach, 2/e

56 Copyright  2011 Pearson Education, Inc. Practice – Calculate the  S  for the reaction 2 H 2(g) + O 2(g)  2 H 2 O (g) Substance S , J/mol  K H2(g)H2(g)130.7 O2(g)O2(g)205.2 H 2 O(g) Tro: Chemistry: A Molecular Approach, 2/e

57 Copyright  2011 Pearson Education, Inc. Example 17.4: Calculate  S  for the reaction 2 H 2(g) + O 2(g)  2 H 2 O (g) 57  S is −, as you would expect for a reaction with more gas reactant molecules than product molecules standard entropies from Appendix IIB  S, J/K Check: Solution: Conceptual Plan: Relationships: Given: Find: SSS  H2, S  O2, S  H2O Substance S , J/mol  K H2(g)H2(g)130.6 O2(g)O2(g)205.2 H 2 O(g)188.8 Tro: Chemistry: A Molecular Approach, 2/e

58 Copyright  2011 Pearson Education, Inc. Calculating  G  At 25  C  G o reaction =  n  G o f (products) -  n  G o f (reactants) At temperatures other than 25  C assuming the change in  H o reaction and  S o reaction is negligible  G  reaction =  H  reaction – T  S  reaction or  G  total =  G  reaction 1 +  G  reaction Tro: Chemistry: A Molecular Approach, 2/e

59 Copyright  2011 Pearson Education, Inc. Example 17.6: The reaction SO 2(g) + ½ O 2(g)  SO 3(g) has  H  = −98.9 kJ and  S  = −94.0 J/K at 25 °C. Calculate  G  at 125  C and determine if it is more or less spontaneous than at 25 °C with  G° = −70.9 kJ/mol SO because  G is −, the reaction is spontaneous at this temperature, though less so than at 25  C  H  = −98.9 kJ,  S  = −94.0 J/K, T = 398 K  G , kJ Answer: Solution: Conceptual Plan: Relationships: Given: Find: GG T,  H ,  S  Tro: Chemistry: A Molecular Approach, 2/e

60 Copyright  2011 Pearson Education, Inc. Practice – Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) Substance  H , kJ/molS , J/mol H2O2(g)H2O2(g)− O2(g)O2(g) H 2 O(g)− Tro: Chemistry: A Molecular Approach, 2/e

61 Copyright  2011 Pearson Education, Inc. Practice – Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) 61 standard energies from Appendix IIB, T = 298 K  G , kJ Solution: Conceptual Plan: Relationships: Given: Find: GG  H f ° Sº of prod & react  H ,  Sº  H  = kJ,  S  = −117.4 J/K, T = 298 K  G , kJ Tro: Chemistry: A Molecular Approach, 2/e

62 Copyright  2011 Pearson Education, Inc. Standard Free Energies of Formation The free energy of formation (  G f °) is the change in free energy when 1 mol of a compound forms from its constituent elements in their standard states The free energy of formation of pure elements in their standard states is zero 62Tro: Chemistry: A Molecular Approach, 2/e

63 Copyright  2011 Pearson Education, Inc. Example 17.7: Calculate  G  at 25  C for the reaction CH 4(g) + 8 O 2(g)  CO 2(g) + 2 H 2 O (g) + 4 O 3(g) 63 standard free energies of formation from Appendix IIB  G , kJ Solution: Conceptual Plan: Relationships: Given: Find: GG G f ° of prod & react Substance  G f °, kJ/mol CH 4 (g)−50.5 O2(g)O2(g)0.0 CO 2 (g)−394.4 H 2 O(g)−228.6 O3(g)O3(g) Tro: Chemistry: A Molecular Approach, 2/e

64 Copyright  2011 Pearson Education, Inc. Practice – Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) Substance  G , kJ/mol H2O2(g)H2O2(g)−105.6 O2(g)O2(g)0 H 2 O(g)− Tro: Chemistry: A Molecular Approach, 2/e

65 Copyright  2011 Pearson Education, Inc. Practice – Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) 65 standard free energies of formation from Appendix IIB  G , kJ Solution: Conceptual Plan: Relationships: Given: Find: GG  G f ° of prod & react Tro: Chemistry: A Molecular Approach, 2/e

66 Copyright  2011 Pearson Education, Inc.  G Relationships If a reaction can be expressed as a series of reactions, the sum of the  G values of the individual reaction is the  G of the total reaction  G is a state function If a reaction is reversed, the sign of its  G value reverses If the amount of materials is multiplied by a factor, the value of the  G is multiplied by the same factor the value of  G of a reaction is extensive 66Tro: Chemistry: A Molecular Approach, 2/e

67 Copyright  2011 Pearson Education, Inc. Example 17.8: Find  Gº rxn for the following reaction using the given equations 3 C (s) + 4 H 2(g)  C 3 H 8(g) Given: Find: C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  Gº = −2074 kJ C (s) + O 2(g)  CO 2(g)  Gº = −394.4 kJ 2 H 2(g) + O 2(g)  2 H 2 O (g)  Gº = −457.1 kJ  Gº of 3 C (s) + 4 H 2(g)  C 3 H 8(g) Rel:Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the  Gº’s is the  Gº of the reaction you want to find. Solution: 3 CO 2(g) + 4 H 2 O (g)  C 3 H 8(g) + 5 O 2(g)  Gº = kJ 3 x [C (s) + O 2(g)  CO 2(g) ]  Gº = 3(−394.4 kJ) 2 x [2 H 2(g) + O 2(g)  2 H 2 O (g) ]  Gº = 2(−457.1 kJ) Given: Find: C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  Gº = −2074 kJ C (s) + O 2(g)  CO 2(g)  Gº = −394.4 kJ 2 H 2(g) + O 2(g)  2 H 2 O (g)  Gº = −457.1 kJ  Gº of 3 C (s) + 4 H 2(g)  C 3 H 8(g) Rel:Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the  Gº’s is the  Gº of the reaction you want to find. Solution: 3 CO 2(g) + 4 H 2 O (g)  C 3 H 8(g) + 5 O 2(g)  Gº = kJ 3 C (s) + 3 O 2(g)  3 CO 2(g)  Gº = − kJ 4 H 2(g) + 2 O 2(g)  4 H 2 O (g)  Gº = −914.2 kJ 3 C (s) + 4 H 2(g)  C 3 H 8(g)  Gº = −23 kJ 67Tro: Chemistry: A Molecular Approach, 2/e

68 Copyright  2011 Pearson Education, Inc. Practice – Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) Substance  G , kJ/mol H 2 (g) + O 2 (g)  H 2 O 2 (g) − H 2 (g) + O 2 (g)  2 H 2 O(g) − Tro: Chemistry: A Molecular Approach, 2/e

69 Copyright  2011 Pearson Education, Inc. Practice – Determine the free energy change in the following reaction at 298 K 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) Given: Find: H 2(g) + O 2(g)  H 2 O 2(g)  Gº = −105.6 kJ 2 H 2(g) + O 2(g)  2 H 2 O (g)  Gº = −457.2 kJ  Gº of 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g) Rel:Manipulate the given reactions so they add up to the reaction you wish to find. The sum of the  Gº’s is the  Gº of the reaction you want to find. Solution: 2 H 2(s) + 2 O 2(g)  2 H 2 O 2(g)  Gº = −211.2 kJ 2 H 2 O (g)  2 H 2(g) + O 2(g)  Gº = kJ 2 H 2 O (g) + O 2(g)  2 H 2 O 2(g)  Gº = kJ 69Tro: Chemistry: A Molecular Approach, 2/e

70 Copyright  2011 Pearson Education, Inc. What’s “Free” About Free Energy? The free energy is the maximum amount of energy released from a system that is available to do work on the surroundings For many exothermic reactions, some of the heat released due to the enthalpy change goes into increasing the entropy of the surroundings, so it is not available to do work And even some of this free energy is generally lost to heating up the surroundings 70Tro: Chemistry: A Molecular Approach, 2/e

71 Copyright  2011 Pearson Education, Inc. Free Energy of an Exothermic Reaction C (s, graphite) + 2 H 2(g) → CH 4(g)  H° rxn = −74.6 kJ = exothermic  S° rxn = −80.8 J/K = unfavorable  G° rxn = −50.5 kJ = spontaneous 71  G° is less than  H° because some of the released heat energy is lost to increase the entropy of the surroundings Tro: Chemistry: A Molecular Approach, 2/e

72 Copyright  2011 Pearson Education, Inc. Free Energy and Reversible Reactions The change in free energy is a theoretical limit as to the amount of work that can be done If the reaction achieves its theoretical limit, it is a reversible reaction 72Tro: Chemistry: A Molecular Approach, 2/e

73 Copyright  2011 Pearson Education, Inc. Real Reactions In a real reaction, some of the free energy is “lost” as heat if not most Therefore, real reactions are irreversible 73Tro: Chemistry: A Molecular Approach, 2/e

74 Copyright  2011 Pearson Education, Inc.  G under Nonstandard Conditions   G =  G  only when the reactants and products are in their standard states their normal state at that temperature partial pressure of gas = 1 atm concentration = 1 M  Under nonstandard conditions,  G =  G  + RTlnQ Q is the reaction quotient  At equilibrium  G = 0  G  = −RTlnK 74Tro: Chemistry: A Molecular Approach, 2/e

75 Copyright  2011 Pearson Education, Inc. 75Tro: Chemistry: A Molecular Approach, 2/e

76 Copyright  2011 Pearson Education, Inc. Example 17.9: Calculate  G at 298 K for the reaction under the given conditions 2 NO (g) + O 2(g)  2 NO 2(g)  Gº = −71.2 kJ 76 non-standard conditions,  Gº, T  G, kJ Solution: Conceptual Plan: Relationships: Given: Find: GG  G , P NO, P O2, P NO2 SubstanceP, atm NO(g)0.100 O2(g)O2(g) NO 2 (g)2.00 Tro: Chemistry: A Molecular Approach, 2/e

77 Copyright  2011 Pearson Education, Inc. Practice – Calculate  G rxn for the given reaction at 700 K under the given conditions N 2(g) + 3 H 2(g)  2 NH 3(g)  Gº = kJ SubstanceP, atm N2(g)N2(g)33 H2(g)H2(g)99 NH 3 (g)2.0 77Tro: Chemistry: A Molecular Approach, 2/e

78 Copyright  2011 Pearson Education, Inc. Practice – Calculate  G rxn for the given reaction at 700 K under the given conditions N 2(g) + 3 H 2(g)  2 NH 3(g)  Gº = kJ 78 non-standard conditions,  Gº, T  G, kJ Solution: Conceptual Plan: Relationships: Given: Find: GG  G , P NO, P O2, P NO2 SubstanceP, atm N2(g)N2(g)33 H2(g)H2(g)99 NH 3 (g)2.0 Tro: Chemistry: A Molecular Approach, 2/e

79 Copyright  2011 Pearson Education, Inc.  Gº and K Because  G rxn = 0 at equilibrium, then  Gº = −RTln(K) When K < 1,  Gº is + and the reaction is spontaneous in the reverse direction under standard conditions nothing will happen if there are no products yet! When K > 1,  Gº is − and the reaction is spontaneous in the forward direction under standard conditions When K = 1,  Gº is 0 and the reaction is at equilibrium under standard conditions 79Tro: Chemistry: A Molecular Approach, 2/e

80 Copyright  2011 Pearson Education, Inc. 80Tro: Chemistry: A Molecular Approach, 2/e

81 Copyright  2011 Pearson Education, Inc. Example 17.10: Calculate K at 298 K for the reaction N 2 O 4(g)  2 NO 2(g) 81 standard free energies of formation from Appendix IIB K Solution: Conceptual Plan: Relationships: Given: Find: Substance  G f °, kJ/mol N2O4(g)N2O4(g)+99.8 NO 2 (g)+51.3 GG G  f of prod & react K  Gº = −RTln(K) Tro: Chemistry: A Molecular Approach, 2/e

82 Copyright  2011 Pearson Education, Inc. Practice – Estimate the equilibrium constant for the given reaction at 700 K N 2(g) + 3 H 2(g)  2 NH 3(g)  Gº = kJ 82Tro: Chemistry: A Molecular Approach, 2/e

83 Copyright  2011 Pearson Education, Inc. Practice – Estimate the Equilibrium Constant for the given reaction at 700 K N 2(g) + 3 H 2(g)  2 NH 3(g)  Gº = kJ 83 GºGº K Solution: Conceptual Plan: Relationships: Given: Find: GG K  Gº = −RTln(K) Tro: Chemistry: A Molecular Approach, 2/e

84 Copyright  2011 Pearson Education, Inc. Combining these two equations  G° =  H° − T  S°  G° = −RTln(K) It can be shown that This equation is in the form y = mx + b The graph of ln(K) versus inverse T is a straight line with slope and y-intercept Why Is the Equilibrium Constant Temperature Dependent? 84Tro: Chemistry: A Molecular Approach, 2/e


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