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Chapter 19 Chemical Thermodynamics 19.1 Chemical Thermodynamics Area of chemistry that studies energy relationships How fast is the reaction? Kinetics -- energy How far does the reaction proceed? Equilibrium – energy O Thermodynamics will not predict kinetics O Thermodynamics will predict equilibrium Spontaneous Process One that proceeds without any outside assistance; spontaneous in one direction but not the opposite Examples: O Na and Cl2 make NaCl O Brick falls off ledge O Gas expands into container O Rusting nail Spontaneous Process O Does not mean it occurs at a rate that you can see or notice O Rusting nail very slow O Ice melting pretty fast O Energy is lost in spontaneous process O Can be endothermic or exothermic O irreversible Reversible Process O Both the process and surroundings can be simultaneously returned to their initial state after the process has been completed O Ideal O Infinitesimal changes – small ΔT & heat flow O all real processes are irreversible 19.2 Entropy Associated with system randomness O S Phase changes O State function occur at constant T ΔHfus O If isothermal (constant T) ΔS = qrev Tconstant Crash course chemistry – entropy https://www.youtube.com/watch?v=ZsY4WcQOrfk ΔHvap 2nd Law of Thermodynamics entropy of the system and surroundings will increase for an irreversible process ΔSuniverse > 0 ΔSuniverse = 0 for spontaneous process for reversible process Example The normal boiling point of ethanol is 78.3ºC and the molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy of the system when 68.3 g of ethanol at 1 atm is condensed to liquid? -163 J/K Suggested Homework pp. 833-836 (1, 9, 13, 15, 23, 25) 19.3 Molecular Interpretation O KMT – ideal gas O Average KE direction proportional to absolute temperature (higher T = more KE) O More KE = more motion O Translational O Vibrational O Rotational Molecular Interpretation O Energy of ideal gas depends only on temperature Microstates snapshot of molecules at point in time; single possible state of arrangements and energies of molecules Molecular Interpretation Entropy is a measure of the number of microstates of the system. Entropy increases with the number of microstates of the system. Isothermally increase volume more arrangements Increase temperature (fixed volume) more kinetic energies Increase molecules (not ideal) more kinetic energies Qualitative Predictions To predict the change in entropy consider what is happening to: 1. Temperature 2. Volume 3. Number of moving particles Examples: O Melting O Evaporation O number of gas molecules increase 3rd Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is zero. Example Predict an increase or decrease in entropy. 1. CO2(s) CO2(g) 2. CaO(s) + CO2(g) CaCO3(s) 3. HCl(g) + NH3(g) NH4Cl(s) 4. 2 SO2(g) + O2(g) 2 SO3(g) Increase Decrease Decrease Decrease Example Which sample has the highest entropy? Assume 1 mole of sample unless otherwise specified. 1. H2(g) at STP or at 100°C and 0.5 atm 2. Water at freezing temperature or room temperature 3. H2(g) or SO2(g) at STP 4. 1 mol H2(g) or 2 mol H2(g) at STP 100°C and 0.5 atm Room temperature SO2(g) 2 mol H2(g) Suggested Homework pp. 834-836 (3, 4, 29, 33, 39, 41) Entropy in Reactions ΔS° = Σ n S° (products) - Σ m S° (reactants) O Measure heat capacity to determine ΔS O Standard molar entropy, S°, p.817/Appendix C Notable Trends: O Increase with molar mass O Increase with number of atoms in substance Entropy in Reactions O If entropy of reaction system decreases, assume entropy gained by surroundings is larger O Entropy of universe must increase if spontaneous O If exothermic, entropy of surroundings always increases (unless isolated system) Example Calculate ΔS° for the synthesis of ammonia. The values of S° are 192.5 kJ/mol for ammonia, 191.5 kJ/mol for nitrogen, and 130.6 kJ/mol for hydrogen. -198.3 kJ/mol (decreases) Suggested Homework pp. 836-837 (49) Gibbs Free Energy O Spontaneous processes that decrease the system’s entropy are always exothermic O Entropy and enthalpy must be related O New state function Gibbs Free Energy G=H–TS ΔG° = ΔH° – TΔS° Gas @ 1 atm Solution @ 1 M 25°C Gibbs Free Energy ΔSuniv = ΔSsys + ΔSsurr ΔSuniv = ΔSsys + -ΔHsys/T If spontaneous ΔSuniv + positive ΔGsys - negative -TΔSuniv = ΔHsys - TΔSsys ΔGsys = ΔHsys - TΔSsys Gibbs Free Energy ΔG sign Negative Reaction is spontaneous in forward direction (supply work to make it go in reverse) Zero Reaction is at equilibrium Positive Reaction is spontaneous in reverse direction (supply work to make it go forward) Gibbs Free Energy O Free energy changes until it reaches a minimum value O Free energy always decreases for a spontaneous process Free Energy in Reactions ΔG° = Σ n Gf° (products) - Σ m Gf° (reactants) O Standard free energies of formation, Gf° are in Appendix C O Can use free energies of formation to calculate free energy of reaction O Can also use Hess’s law Example A reaction occurs at 298K and has ΔH° = 24.6 kJ and ΔS° = 132 J/K. Calculate ΔG° and determine if the reaction is spontaneous under these conditions. -14.7 kJ Example P4(g) + 6 Cl2(g) 4 PCl3(g) The free energy of formation for P4 is 24.4 kJ/mol and for PCl3 is -269.6 kJ/mol. What is the free energy change for this reaction? -1102.8 kJ Example Would you expect the standard free energy change for the combustion of propane gas, C3H8(g), to be more or less negative than the standard enthalpy change? More negative Free Energy & Temperature O Enthalpy and entropy only change small amounts with temperature O T in free energy equation dominates Free Energy & Temperature ΔH ΔS -TΔS ΔG - + - - Always spontaneous + + + Never spontaneous - - + + or - Spontaneous at low T + + - + or - Spontaneous at high T Depends on bond energies and IMF Depends on microstates Remember: Can be very, very slow Suggested Homework pp. 834, 837-838 (5, 6, 8, 53, 57, 59, 61) Free Energy & Equilibrium For nonstandard conditions: ΔG = ΔG° + RT ln Q O If at equilibrium, Q = 1 and ln Q = 0 0 = ΔG° + RT ln K K = e -ΔG°/RT If free energy is negative, K is greater than 1 Important Nonspontaneous Reactions Must find a way to do work on the reaction O In body, glucose combustion is very exergonic (negative free energy) O Plants use light to complete photosynthesis O ATP ADP is exergonic O Need to convert ADP back to ATP O Couple the reaction with ones that are spontaneous Bozeman – life requires free energy https://www.youtube.com/watch?v=JBmykor-2kU Example Indicate if ΔG increases, decreases, or stays the same when the partial pressure of hydrogen is increased in the following reactions: N2(g) + H2(g) 2 NH3(g) 2 HBr(g) H2(g) + Br2(g) 2 H2(g) + C2H2(g) C2H6(g) Example Hydrogen gas and bromine liquid reaction to form gaseous hydrogen bromide. The free energy of formation for hydrogen bromide is -53.22 kJ/mol. 1. Calculate the standard free energy change for the reaction 2. Calculate the equilibrium constant for the reaction at 298 K -106.4 kJ/mol 4 x 1018 Example Hydrogen gas and bromine liquid reaction to form gaseous hydrogen bromide. The equilibrium constant for the reaction is 4 x 1018. 1. Calculate the free energy change at 200°C 2. Calculate the free energy change if the partial pressure of HBr is 1.50 atm and hydrogen is 0.500 atm at 298 K. The standard free energy change is 106.4 kJ/mol -2 x 105 J/mol -104 kJ/mol Suggested Homework pp. 834, 837-838 (7, 73, 75, 79) 20.5 Free Energy and Redox O Voltaic cells are spontaneous E° = E°red (red) - E°red (ox) O Positive value of E° is spontaneous O Activity series 20.5 Free Energy and EMF ΔG° = -n F E° O n is the moles of electrons O Faraday’s constant is charge on 1 mole of eO F = 96,485 C/mol or J/V-mol O Positive E negative ΔG ΔG° = -RT ln K Example Use standard reduction potentials in Table 20.1 (p. 857) to calculate ΔGº and the equilibrium constant at 298 K for the following reaction. 4 Ag(s) + O2(g) + 4 H+(aq) 4 Ag+(aq) + 2 H2O(l) Example Use standard reduction potentials in Table 20.1 (p. 857) to calculate ΔGº and the equilibrium constant at 298 K for the following reaction. 1 2 2 Ag(s) + O2(g) + 2 H+(aq) 2 Ag+(aq) + H2O(l) Suggested Homework p. 887 (51, 53) Integrative Exercise p.831