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Thermodynamics of Reactions

Thermodynamics of Reactions. Spontaneity, Entropy, and Free Energy Chapter 16. 1 st Law of Thermodynamics. The energy of the universe is constant (conservation of energy). Spontaneous Processes. Can be fast OR slow. Occurs without outside intervention. True, or Not True?. Why??.

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Thermodynamics of Reactions

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  1. Thermodynamics of Reactions Spontaneity, Entropy, and Free Energy Chapter 16

  2. 1st Law of Thermodynamics • The energy of the universe is constant • (conservation of energy)

  3. Spontaneous Processes • Can be fast OR slow. • Occurs without outside intervention

  4. True, or Not True? Why?? • Ball rolls downhill spontaneously. • Ball rolls uphill spontaneously. • Wood burns spontaneously in O2. • CO2 and H2O spontaneously form wood. • Steel rusts spontaneously. • Rust spontaneously turns into iron and water.

  5. Its All About Entropy • Entropy is… • All about the ability of energy to spread out • Related to probability. • Over-simplified to be about disorder.

  6. Positional Entropy • When a substance has more opportunities to exist, (ie, particles can have more than one arrangement) it is also said to have more positional entropy. • Increasing moles of gas • Increasing volume • Decreasing pressure • Changing state of matter ( s  l  g ) • Exothermic rxns **

  7. Try Me Conceptual • Determine which of the following pairs has the most positional entropy: • 1 mol H2 at STP or 1 Mol H2 at 100oC, 0.5 atm • 1 mol N2 at STP or 1 mol N2 at 100 K 2atm • 1 mol H2O(s) at 0oC or 1 mole H2O(l) at 20oC

  8. 2nd Law of Thermodynamics • Spontaneous processes increase the entropy of the universe. DSuniverse = DSsystem + DSsurroundings

  9. The Value of S • Sign is determined by enthalpy: • exothermic • endothermic • Magnitude is determined by temperature

  10. Try Me Calculation • The melting point of tungsten (W) is the second highest among the elements, at 3680 K. The enthalpy of fusion for this metal is 35.2 kJ/mol. What is the entropy of fusion?

  11. 3rd Law of Thermodynamics The entropy of a perfect crystal at 0 Kelvin is zero. In Phase Out of Phase Angle Bending Rotation Translation vibrations

  12. Entropy for a chemical reaction • DSorxn = SnSop - SnSor

  13. Try Me • Calculate the change in entropy at 25oC for the reaction: 2 NiS + 3 O2 2 SO2 + 2 NiO Given Entropy Values: SO2 = 248 J/Kmol NiO = 38 J/Kmol O2 = 205 J/Kmol NiS = 53 J/Kmol

  14. Free Energy • Gibbs Free Energy- another method for determining spontaneity. • Also indicates the amount of available energy that is capable of doing work. • As an energy source is used, the energy is not destroyed, only converted to a non-usable form.

  15. DG = DH - TDS • Defines Gibbs energy in terms of enthalpy and entropy. • All three factors will contribute to reaction spontaneity. • When DG is negative, the reaction is spontaneous.

  16. Rearrange that formula • Lets start with the following 3 formulae: • DG = DH –TDS • DSsurr = -DH/T • DSuniv = DSsurr + DSsyst DSuniverse = - DG /T

  17. Fancy Pants Charts DO NOT MEMORIZE

  18. Free Energy in Chemical Reactions • Standard Free energy is used so that we can compare the relative tendency to occur. • DGo = DHo – TDSo

  19. 2SO2(g) + O2(g) 2SO3(g) • The above rxn. Occurs at 25oC and 1 atm. Calculate DHo, DSo, and DGo using the following data: substance DHof So kJ/mol J/Kmol SO2(g) -297 248 SO3(g) -396 257 O2(g) 0 205

  20. It also works like Hess’s Law problems: What is the DGo for the reaction if the mechanism is: Cdi(s) + O2(g) CO2(g)DGo = -397kJ Cgr(s) + O2(g)  CO2(g)DGo = -394 kJ

  21. Dependence on Pressure • Enthalpy does not depend on pressure. • Entropy does depend on temperature. S low pressure > S high pressure

  22. DG = DGo + RT ln(P) • This can be adapted to reflect partial pressures for the reaction it describes. When that occurs, the formula can be re-written: DGo = -RT ln(Q)

  23. @ Equilibrium • K = Q • The free energy is the lowest possible it will ever be for the system.

  24. Try Me Out • The overall rxn for rusting iron by oxygen is 4Fe(s) + 3O2(g) 2 Fe2O3(s) @25oC, find the equilibrium constant given: Substance DHo So Fe2O3(s) -826 90 Fe(s) 0 27 O2(g) 0 205

  25. Free Energy and Work • Achieving the maximum amount of work from a process is highly unlikely because of transfers of energy. • DG represents the maximum possible quantity of work a system is capable of doing.

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