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# Thermodynamics STUDENT NOTES.pptx

these notes indicate or talk about the concept of thermodynamics which deals with hot and cool temperatures

these notes indicate or talk about the concept of thermodynamics which deals with hot and cool temperatures

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### Thermodynamics STUDENT NOTES.pptx

1. 1. CMY 127 THEME 4 9th Ed: Chapter 18 (P678 -711) Thermodynamics Deals with heat and temperature and their relation to energy and work 1
2. 2.  First law of thermodynamics: The law of conservation of energy; energy cannot be created or destroyed.  State Function: Quantity in which its determination is path independent.  U = q + w: The change in internal energy of a system is a function of heat and work done on or by the system.  H: Heat transferred at constant pressure.  Exothermic Process: H < 0  Endothermic Process: H > 0 A Review of Concepts of Thermodynamics 2
3. 3. Thermodynamics 1. Will there be a reaction, even if very slow? If so, to what extent? Questions: 2. If the reaction does take place, how fast will it be? THERMODYNAMICS KINETICS Thermodynamics (spontaneous reactions): • Most exothermic reactions (∆𝑯 < 𝟎) are spontaneous, BUT NOT ALL!! Enthalpy is a (one) “driving force”. • There must be another, what is it??? Spontaneous – change without intervention. Spontaneous changes occur ONLY in the direction that leads to equilibrium Nonspontaneous change - occurs only if driven e.g. coffee pot heating up only if placed on a stove 3
4. 4. What then is ALWAYS TRUE for SPONTANEOUS REACTIONS? Spontaneous processes contributes to the “distribution of energy” from more to less concentrated (more dispersed) • This happens without “external” influences. • Entropy (S) allows us to quantify the dispersal of energy Thermodynamics A few spontaneous, but endothermic (∆𝑯 > 𝟎) processes: • Ammonium nitrate dissolves easily in water, but the process is endothermic • Ice melts above 0°C • A gas moves into open space (diffusion) 4
5. 5. Measure of: Level of disorder in a system or Number of Microscopic Energy Levels Available to a Molecule (i.e. microstates) Second Law of Thermodynamics For a spontaneous process the ∆𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 > 𝟎 i.e. Spontaneous rxns cause an increase in the entropy of the universe Entropy (S) 5
6. 6. Entropy Spontaneous processes result in the dispersal * of energy * or of matter * or of energy & matter 6
7. 7. Entropy Increasing the number of microstates (and entropy) by an increase in: • number of particles – 2 molecules > entropy than 1 molecule • energy – substances at a higher temperature > entropy • phase – liquids have > entropy than solids and gasses > entropy than liquids • structure – larger molecules have > entropy than smaller ones (they have more ways to rotate etc.) e.g. C2H6 compared to CH4 A perfect crystal is the most ordered state of all, lowest entropy – at 0 Kelvin, it has an entropy of zero, S = 0. This is used as our entropy reference! For all others, S > 0 (Third Law of Thermodynamics) 7
8. 8. Because it is necessary to add energy as heat to raise the temperature, all substances have positive entropy values at temperatures above 0 Kelvin (usually at 298 K or 25°C) S = 𝒒𝒓𝒆𝒗 𝑻 • Heat forms part of the mathematical definition of S • Energy dispersed differs with temperature • q has a greater effect on S at low temp qrev heat exchange under reversible conditions 8
9. 9. Predicting Changes in Entropy, S • Entropy is higher for: –higher temperature –larger volume –more complex molecules –Increase in moles of gas –larger sample size –heavier atoms –gas relative to liquid or solid –liquid relative to solid 9
10. 10. Standard molar entropy: The standard molar entropy, S°, of a substance, is the entropy GAINED by converting 1 mole of it from a perfect crystal at 0 K to standard state conditions (pressure of 1 bar) at the specified temperature. See APPENDIX L 𝑺° of elements in the standard state is NOT ZERO!! Unit of S° At 298 K 10
11. 11. Predicting Relative Entropy • Which has higher S? – A. new deck of cards – B. shuffled deck of cards – A. water – B. ice – NO2(g) or N2O4(g) – I2(g) or I2(s) 11
12. 12. Entropy • When reactants are converted to products, matter and energy are redistributed…. • ∆rS° 𝐨𝒓 ∆sysS° (includes reactants and products) • S is a state function 3. ∆rS° = 𝛴𝒏𝑆𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝛴𝒏𝑆𝑜(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) Examples: Do the following changes represent an increase or decrease in entropy of the system? 1. NH4Cl(aq) → NH4Cl(s) 2. KClO3(s) → KClO(s) + O2(g) 3. NaCl(s) → NaCl(l) 4. 2HI(g) + F2(g) → 2HF(g) + I2(s) 5. Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) 12
13. 13. Entropy Question 1 Consider the following reaction: Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g) a) Consider qualitatively: do you expect the entropy of the system to increase or decrease? b) Consider quantitatively: calculate the entropy change for this reaction at 25°C. In future - See APPENDIX L S° (J/K.mol) Mg(s) 32.67 H2O(l) 69.95 Mg(OH)2(aq) 63.18 H2(g) 130.7 Question 2: And for 2Al(s) + 3Cl2(g) → 2AlCl3(s)? 13
14. 14. Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g) S° (J/K.mol) Mg(s) 32.67 H2O(l) 69.95 Mg(OH)2(aq) 63.18 H2(g) 130.7 3. ∆rS° = 𝛴𝒏𝑆𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝛴𝒏𝑆𝑜(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) Mg(OH)2(aq) + H2(g) Mg(s) + 2H2O(l) 14
15. 15. Entropy 18.5 Entropy changes and spontaneity (P691-696) • Is a reaction spontaneous or not? For all spontaneous reactions there will be an increase in entropy of the UNIVERSE. (2nd LAW of Thermodynamics revisited) • BUT, the universe consists of 2 parts: system and surroundings 4. ∆univS° = ∆sysS° + ∆surS° If Suniv > 0 If Suniv < 0 If Suniv = 0 The Question: Will system or surroundings control the situation??? Spontaneous Non-spontaneous System at equilibrium THIS DEPENDS ON THE TEMPERATURE 15
16. 16. At low temperatures water freezes spontaneously – exothermic process more important - Ssys < 0 Ssurr > 0 Suniverse? At higher temperatures water spontaneously evaporates – endothermic process more important - Ssys > 0 Ssurr < 0 Suniverse? Effect of temperature on spontaneity ∆univS° = ∆sysS° + ∆surS° So why is Suniverse > 0 ? At these Temps these processes are spontaneous Change in entropy depends on heat flow H2O(l) ↔ H2O(g) H2O(l) ↔ H2O(s) 16
17. 17. at constant pressure, qsystem = rH°system DS°system = qsystem T DS°surroundings = qsurroundings T system surroundings system surroundings q q 0 q q + = = - surroundings r system r system surroundings Therefore : q H H and S T          • Entropy change (S) in surroundings results from an enthalpy change (H) in system, affecting the surroundings…..(take notice of + and – signs) Determining whether a process is spontaneous 17
18. 18. Determining whether a process is spontaneous ∆𝒔𝒖𝒓𝒓S = − ∆𝒔𝒚𝒔𝑯 𝑻 For entropy changes of systems during 1) melting / freezing 2) evaporation (boiling)/ condensation ∆𝒔𝒚𝒔S = ∆𝒇𝒖𝒔𝑯 𝑻 ∆𝒔𝒚𝒔S = ∆𝒗𝒂𝒑𝑯 𝑻 NEVER use °C in S calculations only K NB Be careful not to make “– and +” and unit mistakes (combining  S (J/K.mol) with H (kJ/mol) – convert to J) ∆univS° = ∆sysS° + ∆surS° becomes ∆univS° = ∆sysS° + (− ∆𝒔𝒚𝒔 𝑯) 𝑻 18
19. 19. Question 3 Calculate the entropy change of the system for 1 g of ice melting (at 0.0°C). Given: Hfusion = 6.01  103 J/mol 19
20. 20. Question 4 Predict the normal boiling point of CCl4 in °C, given that the entropy of 1 mole of CCl4(ℓ) changes with + 95.00 J/K when forming CCl4 vapour at 1 bar pressure, and Hvap = +32.6 kJ/mol (Hint: at boiling point, the system is..???) 20
21. 21. Question 5 Calculate ∆𝒔𝒖𝒓𝒓S for the following reaction at 25°C and 1 bar: a) Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) H = –125 kJ 22
22. 22. Question 7 (a) Is the synthesis of ammonia from nitrogen and hydrogen spontaneous at 298.15 K and 1 bar? (b) Above which temperature (in °C) will it become non-spontaneous? 23
23. 23. Above which temperature (in C) will it become non- spontaneous? 24
24. 24. Question Under which of the following conditions will ∆univS° always be positive? a) ∆sysS° < 0; ∆surS° < 0 b) ∆sysS° > 0; ∆surS° > 0 c) ∆sysS° < 0; ∆surS° > 0 d) ∆sysS° > 0; ∆surS° < 0 Question When will ∆surS° be positive? a) When ∆sysH° is exothermic b) When ∆sysH° is endothermic 25
25. 25. 26 Question 6 Is the reaction of H2(g) and Cl2(g) to give HCl(g) predicted to be spontaneous under standard conditions at 298.15K? [+637.85 J/K mol, spontaneous] Question 8 Consider the reaction: CO(g) + 2H2(g)  CH3OH(ℓ) Is the synthesis of methanol spontaneous at 25°C or not? [-218.945 J/K mol, non-spontaneous] Homework
26. 26. 18.6 Gibbs free energy, G (P696) • So far: ∆sysS° and ∆surS° were combined to calculate ∆univS° (and predict spontaneity) • Problem: it is challenging to assess the surroundings. To have one thermodynamic function associated with the system only, would be convenient. • Introduced: Gibbs free energy (G) “Free” = available for work Definition: The change in free energy accompanying the complete conversion of reactants to products under standard conditions, is ∆G° multiplied with (-T) throughout Becomes -𝑻∆univS° = -T∆sysS° + sysH° ∆rG° = sysH° - T∆sysS° or (Gibbs-Helmholz equation) To calculate spontaneity under standard conditions: 1 molar for solutions, 1 bar (105 Pa) pressure for gasses, specified temperature, usually 25°C. ∆rG = rH - T∆rS 27
27. 27. What is free about “free energy”: • ∆rG° (for a reaction) represents the maximum useful (free) energy obtainable in the form of work from a given process at constant T and P. • Unit for ∆rG° = kJ/mole (like with ∆rH° ) Example 18.4 (P700): C(graphite) + 2H2(g)  CH4(g) Ho = –74.9 kJ So = –80.7 J/K Go = –50.0 kJ But… So<0……portion of energy is used to create more order (concentrate energy/ reverse dispersal) What is left, is “free” or available, to do work. Exothermic reaction: expect E transferred to surroundings available to do work ∆rG = rH - T∆rS heat organization 28
28. 28. Temperature and spontaneity (effect on Free Energy) ∆rG° becomes more negative if… • ∆rG° : used as indicator of spontaneity: H becomes more negative – process gives off more heat to surroundings) S becomes more positive – process results in greater disorder G Spontaneity (constant T, P) + Nonspontaneous 0 Equilibrium (e.g. at boiling point) – Spontaneous ∆rG° = rH° - T∆rS° Effect of temperature changes? 29
29. 29. H S G Reaction  +  Product Favored +  + Reactant Favored   ? Temperature dependent + + ? Temperature dependent Gibbs Free Energy, G G H T S         30
30. 30. Temperature and spontaneity (effect on Free Energy) (See Table 18.1) Spontaneous at high Temp Never spontaneous at any Temp Spontaneous at any Temp Spontaneous at low Temp 31 Type 2 Type 3 Type 1 Type 4
31. 31. 32 Gibbs Free Energy, ΔG Spontaneity: ΔG < 0 Spontaneous ΔG > 0 Nonspontaneous G H T S        
32. 32. Fig. 18.11 The variation of Gibbs free energy with Temperature If rH and T∆rS work in opposite directions, T determines whether the reaction will be spontaneous or not e.g. A) Type 2: If ∆rS < 0 (entropy disfavoured) and rH < 0 (enthalpy favoured) A) Type 3: If ∆rS > 0 (entropy favoured) and rH > 0 (enthalpy disfavoured) ∆rG° = rH° - T∆rS° 33
33. 33. Question 14 Consider the decomposition of calcium carbonate: CaCO3(s)→ CaO(s) + CO2(g) Go rxn = +131.4 kJ Ho rxn = +179.1 kJ So rxn = +160.2 J/K How high should the temperature for this reaction be to ensure that it will be spontaneous? 34
34. 34. Question 9 Will the following reaction products form at 350°C? 2HgO(s, red)  2Hg(l) + O2(g) Assume that ∆𝐻𝑜 and ∆𝑆𝑜 have the same values at this temperature as at 25°C. The boiling point of mercury is 356°C and the melting point of mercury(II) oxide is far above 500°C. From information pages – APPENDIX L: HgO(s, red) O2(g) Hg(l) Hf° (kJ/mol) -90.83 0 0 S° (J/Kmol) 70.29 205.14 76.02 Also see Example 18.6 (P704) 36
35. 35. Question 11 Calculate ∆rG° for the reaction using free energy of formation (APPENDIX L). 4NH3(g)+ 5O2(g) → 4NO(g) + 6H2O(l) 37
36. 36. Gibbs Free Energy, Spontaneity and Chemical equilibrium (P697-699) • Free energy at equilibrium (mixture of reactants and products) is always lower than free energy of pure reactants or pure products • All reactions proceed spontaneously towards minimum energy (equilibrium) Reactants Products Mixture 38
37. 37. Different values if not at equilibrium Gibbs Free Energy, Spontaneity and Chemical equilibrium (P697-699) A reaction for which ∆rG° < 0, will proceed to an equilibrium position at which point the products will dominate in the reaction mixture, because K > 1 Such a reaction is product favoured at equilibrium. Fig. 18.10 Free Energy changes in course of a reaction 𝑎𝐴 + 𝑏𝐵 ⇌ 𝑐𝐶 + 𝑑𝐷 𝐾 = 𝐶𝑐 𝐷𝑑 𝐴𝑎𝐵𝑏 Q = 𝐶𝑐𝐷𝑑 𝐴𝑎𝐵𝑏 Slope of graph indicates ∆rG 39 ∆rG° at standard conditions, ∆rGat non std conditions
38. 38. • A reaction of which ∆rG° > 0, will proceed towards an equilibrium position at which point the reactants will dominate in the reaction mixture, because K < 1. Such a reaction is reactant favoured at equilibrium. Gibbs Free Energy, Spontaneity and Chemical equilibrium (P697-699) Fig. 18.10 Free Energy changes in course of a reaction mM + 𝑛𝑁 ⇌ 𝑥𝑋 +yY Slope of graph indicates ∆rG 40
39. 39. Figure 18-10b p697 Chemical equilibrium and the interpretation of K and Q (P697-699) NB SUMMARY P699 • A reaction never goes beyond the equilibrium, unless disturbed • Reaction quotient (Q) and Equilibrium constant (K) tells us whether a reaction is at equilibrium or not…(is Q = K??) • A reaction with a very small K-value (reactant favoured) moves spontaneously towards the equilibrium, BUT the equilibrium is reached quickly, e.g. PbI2(s) dissolved in water. Only a few ions exist in the solution. For PbI2(s) the Ksp is 9.8 x 10-8 This means, for the reaction: PbI2(s) ↔Pb2+ (aq) + 2I- (aq) Ksp = Pb2+ I− 2 __ a very low ion concentration! Very small!! 41
40. 40. Chemical equilibrium and the interpretation of K and Q (P697-699) See SUMMARY P699 The more negative the value of rG°….. …the more product favoured the reaction.. …the higher the value of K….. Between pure reactant and pure product we do not have standard conditions, therefore rG: At any point, determine rG related to Go and then progresses towards equilibrium: rG = rGo + RT ln Q 8.314 J/Kmol Reaction quotient No ° Take notice of unit: J to kJ!!! 42
41. 41. • When rG < 0, (slope < 0) the reaction proceeds spontaneously towards equilibrium and Q < K • When rG > 0, (slope > 0) the reaction proceeds spontaneously towards equilibrium and Q > K G = Go + RT ln Q At equilibrium (any temperature): rG = 0 and Q = K , therefore: rG = rG° + RT ln K 0 = rG° + RT ln K rGo = - RT ln K OR Go = -2.303RTlogK For reaction-type with rG° < 0, ∴ forward reaction spontaneous: 43
42. 42. 44 rG = rGo + RT ln Q rGo = - RT ln K rG = - RT ln K + RT ln Q rG = RT (ln Q - ln K) rG = RT ln (Q/K)
43. 43. Page 710 rGo = - RT ln K G = Go + RT ln Q 45
44. 44. SUMMARY – under which circumstances /conditions can which formulas be used? 7. Gr = Gr o + RT ln Q 8. rGo = - RT ln K ) (reactants G n (products) G n G o f o f rxn        Δ 6. 𝟓. ∆rG = rH - T∆rS 46
45. 45. Question: Consider the synthesis of cyclohexane from benzene: 𝐶6𝐻6 𝑙 + 3𝐻2 𝑔 ⇌ 𝐶6𝐻12(𝑙) Assume there is 𝐶6𝐻6 𝑙 and 𝐶6𝐻12(𝑙) present at all times in this system. a) Calculate ∆𝑟𝐺° at 25℃. b) Is the reaction product-favoured at equilibrium at 25℃? c) What is the value of 𝐾 at 25℃? d) What was 𝑃𝐻2 at the start of the reaction? e) What will 𝑃𝐻2 be at equilibrium at 25 ℃? f) What will 𝑃𝐻2 be when ∆𝑟𝐺 is −50.0 kJ/mol at 25 ℃? g) Is the reaction product-favoured at equilibrium at 400.℃? h) Estimate the value of 𝐾 at 400.℃? i) What will 𝑃𝐻2 be at equilibrium at 400.℃? j) At what temperature does the reaction become reactant favoured? Given at 298.15 K: (Assume ∆𝒇𝑯° and 𝑺° values are valid at all temperatures) ∆𝑓𝐻°(𝑘𝐽/𝑚𝑜𝑙) 𝑆°(𝐽/𝐾 ∙ 𝑚𝑜𝑙) ∆𝑓𝐺°(𝑘𝐽/𝑚𝑜𝑙) 𝐶6𝐻6(𝑙) 48.95 173.26 124.21 𝐻2(𝑔) 0 130.7 0 𝐶6𝐻12 𝑙 -156 204.4 26.8 47
46. 46. Essential practice for understanding SnO2 (s)+2CO (g) →2CO2 (g)+Sn(s) 1) Is the reaction enthalpy- or entropy-driven? Support your answer with calculations. 2) Determine the temperature range in which this reaction will be spontaneous under standard conditions. 3) Tick the correct statement based on your answer in b): •At 298.15 K the value of the equilibrium constant is smaller than 0. •At 298.15 K the value of the equilibrium constant is smaller than 1. •At 298.15 K the value of the equilibrium constant is larger than 1. 4] Calculate the value of the equilibrium constant at 275.00℃. 5] What is the value of ∆rG at 275.00℃ when CO2=1.85 M and CO=1.59 M? ∆𝒓𝑯° 𝑺° SnO2 -580.7 52.3 Sn 0 51.55 CO2 -393,5 213,6 CO -110.525 197.67 48
47. 47. NB Also see Examples 18.7, 18.8, 18.9 Question 16 Consider the reaction SiCl4(g) + 2H2O(l) ↔ SiO2(s) + 4HCl(g). Assume H° and ∆S° stays constant even with temperature changes. Concentrations are all 1.0 M. a) Will this reaction be spontaneous at 25°C? b) Will this reaction be spontaneous at 30°C? [SiCl4(g] = 0.15 M, [HCl(g)] = 0.2M c) What will the Kc value be at 25°C? Question 17 Predict the boiling point of ethanol in a container where equilibrium is reached. C2H5OH(l) ↔ C2H5OH(g) 50
48. 48. Question 18 What is the value of Kp at 25℃ for the formation of SO3 in vehicle catalytic converters: 2SO2(g) + O2(g) ↔ 2SO3(g) At 25 ℃, rG° = –140. kJ/mol Question 19 The value of Ksp for AgCl(s) at 25 ℃ is 1.8 x 10-10. Determine rG° for the process Ag+(aq) + Cl-(aq) ↔ AgCl(s) at 298.15 K. Question 20 The standard Gibbs energy change for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is –33.2 kJ and the equilibrium constant, Kp, is 6.59 x 105 at 25° C. In a specific experiment, the initial pressures are PH2 = 0.250 bar, PN2 = 0.870 bar, and PNH3 = 12.9 bar. Calculate G for the reaction at these pressures, and predict the direction of the spontaneous reaction. 51
49. 49. Question 21 Consider the following reaction at 25℃ SnO2(s) + C(s, graphite) ↔ Sn(s, white) + CO2(g) a) Do calculations to determine whether the reaction is entropy-driven or enthalpy-driven at standard conditions. Motivate your answer. b) Which of the following statements are correct with respect to the spontaneity of the above reaction? Motivate your answer and calculate the value of x if your answer is C or D. A. Spontaneous at all temperatures C. Spontaneous above x℃ B. Never spontaneous D. Spontaneous under x℃ c) Calculate the value of the equilibrium constant, K, for the reaction at 25℃ 52
50. 50. Question 22 Consider the reaction: 2NO(g) + Cl2(g) ⇌ 2NOCl(g) Calculate the approximate equilibrium constant for this reaction at 431 K by using the following thermodynamic data: Substance S° (]/mol · K) ∆fH° (k]/mol) NO(g) 210.8 90.29 Cl2(g) 223.1 NOCl(g) 261.7 51.71 Question 23 Using values of ∆f H° and S°, calculate the standard molar free energy of formation, ∆f G°, for the following compound: NaOH(s) Do you predict that this formation reaction is product- favoured at equilibrium at 25 °C? 53
51. 51. Question 24 C2H4(g) +H2O (l) ⇌ C2H5OH(l) (a) Is the above reaction spontaneous at high or low temperatures given the following data? Do calculations and use the answers to motivate your answer. (b) What is the value of ∆rG for the reaction above if the pressure of ethene is 2.55 bar (non-standard conditions) at 25℃? Substance C H4(g) H2O(l) C2H5OH(l) S°(]/mol · K) 219.36 69.95 160.7 54
52. 52. Question 25 A crucial reaction for the production of synthetic fuels is the production of H2 by the reaction of coal with steam. The chemical reaction is C(s) + H2O(g) → CO(g) + H2(g) (a) Calculate ΔrG° for this reaction at 25 °C, assuming C(s) is graphite. (b) Calculate Kp for the reaction at 25°C. (c) Is the reaction predicted to be product-favoured at equilibrium at 25°C? If not, at what temperature will it become so? Question 26 Do the problem with integrated concepts (see ClickUP) 55
53. 53. 1. S = kBlnW 2. ∆S = 𝒒𝒓𝒆𝒗 𝑻 3. ∆rS° = 𝛴𝑛𝑆𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝛴𝑛𝑆𝑜(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) 4. ∆univS° = ∆sysS° + ∆surS° ∆𝒔𝒖𝒓𝒓S = − ∆𝒔𝒚𝒔 𝑯 𝑻 ∆univS° = ∆sysS° + (− ∆𝒔𝒚𝒔𝑯) 𝑻 𝟓. ∆rG = rH - T∆rS ∆rG° = rH° - T∆rS° ) (reactants n (products) n G Δ 6. o o rxn        G G f f 𝐾 = 𝐶𝑐𝐷𝑑 𝐴𝑎𝐵𝑏 Q = 𝐶𝑐𝐷𝑑 𝐴𝑎𝐵𝑏 7. rG = Go + RT ln Q 8. rGo = - RT ln K At Equilibrium FORMULAS 56