Chapter 16

Chapter 17Spontaneity, 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?

ThermodynamicsThe science that deals with heat and energy effects.First Law 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 versaCH4+2O2(g) CO2(g)+ 2H2O(g) + energy Here potential energy has been converted to thermal energy.

Figure 16.1 Methane and Oxygen React

Spontaneous ProcessesA 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.

Figure 16.2 Rate of Reaction

EntropyWhat 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 universeWhat is Entropy?Entropy is a measure of randomness or disorderState property depends upon the state of a systemThermodynamic function that describe the number of arrangements S = Sfinal - Sinitial

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

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

ExamplesChoose 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)

Second Law of ThermodynamicsIn any spontaneous process there is always an increase in the entropy of the universe.UniverseSystem: Portion of the universe in which we are interestedSurroundings: everything else in the universe besides the system.

Change in entropy of the universe

Suniv = Ssys+ SsurWhere 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.

The Effect of Temperature on SpontaneityConsider, 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 Ssurr depends on the direction of the heat flowThe magnitude of Ssurr depends on the temperatureSsurr depends directly on the quantity of heat transferred and inversely on temperature.

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 = HEndothermic H positive and exothermic H negative

Ssurr = - H/T (at constant temp. and pressure)

ExampleCalculate Ssurr for the following reactions at 25 oC and 1 atm.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2221kJSsurr = - 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 kJSsurr = - H/T = -112 kJ/298 K = -0.376 kJ/K = -376 J/K

Free EnergyFree 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 entropyFor a process that occurs at constant temperature, the changes in free energy (G) is given by the equation, G = H TS

G = H TS

Lets see how this equation relates tospontaneity,-G/T = - H/T TS/-(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 PG negative process is spontaneous at constant T and P

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 - TSAt the boiling point, G = 0, so TS = 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

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 = TST=H/S=58.51X103J/mol92.92 J/K.mol = 629.7 K

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

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.

Entropy Changes in Chemical ReactionsDoes positional entropy increases or decreases for the following reaction4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)9 gaseous molecule 10 gaseous molecule Positional entropy increasesIf 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.

Figure 16.5 Entropy

Predicting the Sign of SoPredict the sign of So for each of the following changesa. 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 (-)

Third Law of ThermodynamicsThe entropy of a perfect crystal at 0 K is zeroEntropy is a state function of the systemEntropy 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- nrSoreactantsEntropy is an extensive property (depends on amount)

ExampleCalculate So at 25oC for the reaction2NiS(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/KWe would expect So to be negative because the number of gaseous molecules decreases.

Free Energy and chemical ReactionsGo, 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 - TSo

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 - TSo = -3.935 x 105J (298)(3.05 J/K) = -3.944 x 105J = -394.4 KJ(per mol of CO2)

Free Energy Change and Chemical ReactionsGo = 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 = npGof (products) nrGof (reactants)

The standard free energy of formation of an element in its standard state is zero.

Free Energy and PressureG = Go + RTln(P) where,Go = the free energy of the gas at a pressure of 1 atmG = the free energy of the gas at a pressure of P atmR = the universal gas constantT = the Kelvin temperatureG = npG products nrG reactantsWhere, Gproducts = Go products + RTln(Pproducts) Greactants= Go reactants + RTln(Preactants)G = Go + RTln(Q) where, Q = reaction quotient from the law of mass action

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

Figure 16.8 The Dependence of Free Energy on Partial Pressure

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 = 1Case 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.

Temperature Dependence of KGo = -RTln(K) = Ho - TSoln(K) = - Ho/RT + So/Rln(K) = -[Ho/R][1/T] + So/R This is a linear equation of the formy = 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)

SummaryFirst law of thermodynamicsSpontaneous processEntropyS = Sfinal Sinitial MicrostateSecond law of thermodynamicsEffect of temperatureSystemSurroundings

Summary

Suniv = Ssys + SsurrFree energy G = H TSEntropy change in chemical reactionsSoreaction = npSoproducts- nrSoreactantsGo = npGof(products) = nrGof(reactants)Go = Go + RTln(p) = Go + RTln(Q)Go = -RTln(K)ln(K) = -[Ho/R][1/T] + So/R

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