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Chapter 17 Spontaneity, Entropy, and Free Energy

Chapter 17 Spontaneity, Entropy, and Free Energy The goal of this chapter is to answer a basic question: will a given reaction occur “ by itself ” at a particular temperature and pressure, without the exertion of any outside force?. Thermodynamics.

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Chapter 17 Spontaneity, Entropy, and Free Energy

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  1. Chapter 17 Spontaneity, Entropy, and Free Energy • The goal of this chapter is to answer a basic question: will a given reaction occur “by itself” at a particular temperature and pressure, without the exertion of any outside force?

  2. Thermodynamics • The science that deals with heat and energy effects. • FirstLaw of thermodynamics / The Law of Conservation of Energy: • Energy can be neither created nor destroyed by a process, it is only transformed from one form to another. • Any energy lost by a system must be gained by the surroundings and vice versa • CH4+2O2(g) CO2(g)+ 2H2O(g) + energy Here potential energy has been converted to thermal energy.

  3. Figure 16.1 Methane and Oxygen React

  4. Spontaneous Processes • A process is said to be spontaneous if it occurs without outside intervention. • Thermodynamics lets us predict whether a process will occur or not. • Spontaneous processes may be fast or slow. • Thermodynamics can tell the direction in which a process will occur but can say nothing about the speed of the process.

  5. Figure 16.2 Rate of Reaction

  6. Entropy • What common characteristic cause the processes to be spontaneous? • The driving force for the spontaneous process is an increase in the Entropy (denoted by s) of the universe • What is Entropy? • Entropy is a measure of randomness or disorder • State property depends upon the state of a system • Thermodynamic function that describe the number of arrangements S = Sfinal - Sinitial

  7. Microstate • Each configuration that gives a particular arrangement is called a microstate. • Which arrangement is most likely to occur? • One with greatest number of microstate. • Positional probability (microstates) which depends on the number of configurations in space. • Positional probability increases (entropy increases) in going from solid to liquid to gas. Ssolid < Sliquid << Sgas

  8. Figure 16.3 The Expansion of an Ideal Gas into an Evacuated Bulb

  9. Figure 16.4 Three Possible Arrangements (states) of Four Molecules in a Two-Bulbed Flask

  10. Examples • Choose the compound with the greatest positional entropy in each case. a. 1 mol H2 (at STP) or 1 mol H2 (at 100 oC, 0.5 atm) H2 at 100 oC and 0.5 atm; higher temperature and lower pressure means greater volume and hence, greater positional entropy. b. 1 mol N2 (at STP) or 1 mol H2 (at 100 K, 2.0 atm) N2 at STP has the greater volume. c. 1 mol H2O(s) (at 0 oC) or 1 mol H2O(l) (at 20 oC) H2O(l) is more disordered than H2O(s)

  11. Second Law of Thermodynamics • In any spontaneous process there is always an increase in the entropy of the universe. • Universe • System: Portion of the universe in which we are interested • Surroundings: everything else in the universe besides the system.

  12. Change in entropy of the universe Suniv = Ssys+ Ssur • Where Ssys and Ssurr represent the changes in entropy • Suniv - positive the process is spontaneous in the direction written. • Suniv - negative the process is spontaneous in the opposite direction. • Suniv - zero the system is at equilibrium.

  13. The Effect of Temperature on Spontaneity Consider, H2O(l) H2O(g) 1 mole, 18 grams, 18 mL  1 mole, 18 grams, 31 litters (1 atm, 100 oC) Positional probability increases – entropy of the system increases,Ssys  Positive • The sign of Ssurrdepends on the direction of the heat flow • The magnitude of Ssurrdepends on the temperature • Ssurrdepends directly on the quantity of heat transferred and inversely on temperature.

  14. Continued.... Exothermic process: Ssurr = + quantity of heat (J) / temperature (K) Endothermic process: Ssurr = - quantity of heat (J) / temperature (K) Heat flow (constant P) = Change in enthalpy = H Endothermic H positive and exothermic H negative Ssurr = - H/T (at constant temp. and pressure)

  15. Example Calculate Ssurr for the following reactions at 25 oC and 1 atm. • C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Ho = -2221kJ Ssurr = - H/T = -(-2221 kJ) / (25+273)K = 7.45 kJ/K = 7.45 X 103 J/K • 2NO2(g)  2NO(g) + O2(g) H = 112 kJ Ssurr = - H/T = -112 kJ/298 K = -0.376 kJ/K = -376 J/K

  16. Free Energy • Free energy is a thermodynamic function related to spontaneity and is useful in dealing with temperature dependence of spontaneity, defined by the relationship: G = H – TS where, G is free energy, H is enthalpy, T is Kelvin temperature, S is entropy For a process that occurs at constant temperature, the changes in free energy (G) is given by the equation, G = H – TS

  17. G = H – TS Lets see how this equation relates to spontaneity, • -G/T = - H/T – TS/-(T) [Divide both side by –T] -G/T = - H/T + S = Ssurr+ S = Suniv [Recall, Ssurr = - H/T] • Suniv = - G/T at constant T and P • G negative – process is spontaneous at constant T and P

  18. Example • Example: The boiling point of chloroform (CHCl3) is 61.7 oC. The enthalpy of vaporization is 31.4 kJ/mol. Calculate the entropy of vaporization. G = H - TS At the boiling point, G = 0, so TS = H S = H/ T = 31.4 kJ/mol ÷ (273.2+61.7)K =9.38 X 10-2 kJ/K.mol = 93.8 J/K.mol

  19. Example: For mercury, the enthalpy of vaporization is 58.51 kJ/mole and the entropy of vaporization is 92.92 J/K.mole. What is the normal boiling point of mercury? At the boiling point , G = 0, so H = TS T=H/S=58.51X103J/mol92.92 J/K.mol = 629.7 K

  20. Example: For ammonia (NH3), the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 J/K.mol a. Will NH3(s) spontaneously melt at 200 K? b. What is the approximate melting point of ammonia? a. NH3(s)  NH3(l) G = H - TS = 5650 J/mol – 200 K.(28.9 J/K.mol) = 5650 J/mol - 5780 J/mol = - 130 J/mol Yes, NH3 will melt since G < 0 (negative) at this temperature b. At the melting point G = 0 H = TS T = H / S = 5650 J/mol  28.9 J/K.mol = 196 K

  21. Entropy Changes in Chemical Reactions • The entropy changes in the system (the reactants and the products of the reaction) are determined by positional probability. eg. In the ammonia synthesis reaction N2(g) + 3H2(g) 2NH3(g) • Four reactant molecules becomes two product molecules. • Fewer molecules mean fewer possible configuration.

  22. Entropy Changes in Chemical Reactions • Does positional entropy increases or decreases for the following reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) 9 gaseous molecule  10 gaseous molecule  Positional entropy increases If the number of molecules of the gaseous products is greater than the number of molecules of the gaseous reactants, positional entropy increases, and S will be positive for the reaction.

  23. Figure 16.5 Entropy

  24. Predicting the Sign of So • Predict the sign of So for each of the following changes a. AgCl(s)  Ag+(aq) + Cl-(aq) Increase in disorder; So(+) b. 2H2(g) + O2(g) 2H2O(l) Decrease in disorder; n < 0; So(-) c. H2O(l)  H2O(g) Increase in disorder; n > 0; So(+) d. Na(s) + 1/2Cl2(g)  NaCl(s) Decrease in disorder; n < 0; So(-) e. 2SO2(g) + O2(g)  2SO3(g) Decrease in disorder; n < 0; So (-)

  25. Third Law of Thermodynamics • The entropy of a perfect crystal at 0 K is zero • Entropy is a state function of the system • Entropy change for a given chemical reaction can be calculated by taking the difference between the standard entropy values of products and those of the reactants: Soreaction = npSoproducts- nrSoreactants • Entropy is an extensive property (depends on amount)

  26. Example Calculate So at 25oC for the reaction 2NiS(s) + 3O2(g)  2SO2(g) + 2NiO(s) Soreaction = npSoproducts- nrSoreactants = 2SoSO2(g) + 2SoNiO(s) – (2SoNiS(s) + 3SoO2(g)) = 2 mol(248 J/K.mol) + 2mol(38 J/K.mol) - 2mol(53 J/K.mol) - 3mol(205 J/K.mol) = 496 J/K + 76 J/K - 106 J/K - 615 J/K = -149 J/K We would expect So to be negative because the number of gaseous molecules decreases.

  27. Free Energy and chemical Reactions • Go, the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. N2(g) + 3H2(g)  2NH3(g)Go = -33.3 KJ Go = Ho - TSo

  28. Example: C(s) + O2(g) CO2(g) The values of Ho and So are -393.5 KJ and 3.05 J/K, calculate Go at 298 K. Go = H - TSo = -3.935 x 105J – (298)(3.05 J/K) = -3.944 x 105J = -394.4 KJ (per mol of CO2)

  29. Free Energy Change and Chemical Reactions • Go = standard free energy change that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states. Go = npGof (products) – nrGof (reactants) • The standard free energy of formation of an element in its standard state is zero.

  30. Free Energy and Pressure • G = Go + RTln(P) where, Go = the free energy of the gas at a pressure of 1 atm G = the free energy of the gas at a pressure of P atm R = the universal gas constant T = the Kelvin temperature G = npG products – nrG reactants Where, Gproducts = Go products + RTln(Pproducts) Greactants= Go reactants + RTln(Preactants) G = Go + RTln(Q) where, Q = reaction quotient from the law of mass action

  31. Free Energy and Equilibrium G = Go + RTln(Q) At equilibrium, G = 0 (Gproducts = Greactants) and Q = K (equilibrium constant) So, G = 0 = Go + RTln(K) Go = -RTln(K)

  32. Figure 16.8 The Dependence of Free Energy on Partial Pressure

  33. Continued…. Case 1: Go = 0, the system is at equilibrium when the pressures of all reactants and products are 1 atm, which means that K = 1 Case 2: Go < 0 , in this case Go products < Go reactants. The system will adjust to the right to reach equilibrium, K will be greater than 1, since the pressure of the products at equilibrium will be greater than 1 atm and the pressure of the reactants will be less than 1 atm. Case 3: Go > O, in this case Go reactants < Go products. The system will adjust to the left to reach equilibrium. The value of K will be less than 1.

  34. Temperature Dependence of K Go = -RTln(K) = Ho - TSo ln(K) = - Ho/RT + So/R ln(K) = -[Ho/R][1/T] + So/R This is a linear equation of the form y = mx + b where, y = ln(K), m = -Ho/R = slope, x = 1/T, and b = So/R = intercept (Ho and So independent of temperature over a small temperature range)

  35. Summary • First law of thermodynamics • Spontaneous process • Entropy • S = Sfinal – Sinitial • Microstate • Second law of thermodynamics • Effect of temperature • System • Surroundings

  36. Summary • Suniv = Ssys + Ssurr • Free energy G = H – TS • Entropy change in chemical reactions • Soreaction = npSoproducts- nrSoreactants • Go = npGof(products) = nrGof(reactants) • Go = Go + RTln(p) = Go + RTln(Q) • Go = -RTln(K) • ln(K) = -[Ho/R][1/T] + So/R

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