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Chemical Thermodynamics

Chemical Thermodynamics. Spontaneous Processes. First Law of Thermodynamics Energy is Conserved – ΔE = q + w Need value other than ΔE to determine if a process if favored (spontaneous) Spontaneous processes have a direction Spontaneity can depend on temperature. Reversible Processes.

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Chemical Thermodynamics

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  1. Chemical Thermodynamics

  2. Spontaneous Processes • First Law of Thermodynamics • Energy is Conserved – ΔE = q + w • Need value other than ΔE to determine if a process if favored (spontaneous) • Spontaneous processes have a direction • Spontaneity can depend on temperature

  3. Reversible Processes • Reversible Process • When a change in a system is made in such a way that the system can be restored to its original state by exactly reversing the change.

  4. Irreversible Processes • Irreversible Process • A process that cannot simply be reversed to restore the system and surroundings. • Must take alternative pathway • Only system is returned to its original state

  5. Entropy • Processes where the disorder of the system increases tend to occur spontaneously • Ice melting • Salts dissolving • Change in disorder and change in energy determine spontaneity • Entropy (S) is a state function (ΔS) that measures of disorder • Units – J/K

  6. Determining Entropy Change • Predict whether ΔS is positive or negative for each of the following processes. • H2O(l) → H2O(g) • Ag+ (aq) + Cl-(aq) → AgCl(s) • 4Fe(s) + 3O2(g) → 2Fe2O3(s) • CO2(s) → CO2(g) • CaO(s) + CO2(g) → CaCO3(s)

  7. Calculating Entropy • In a reversible process there is only one heat for both processes (qr e v) • When a process occurs at constant temperature entropy is related to both heat and absolute temperature – ΔS = qr e v / T • When 1mol of water is converted to 1mol of steam at 1 atm, ΔHv a p = 40.67kJ/mol, what is the change in entropy?

  8. Calculating Entropy • The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9°C, and its molar enthalpy of fusion is ΔHf u s = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? • Answer: -2.44 J/K

  9. Calculating Entropy • The normal boiling point of ethanol is 78.3°C and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 68.3g of ethanol at 1 atm condenses to liquid at the normal boiling point? • Answer: -163 J/K

  10. Second Law of Thermodynamics • Second law = Entropy of the universe increases in any spontaneous process. • ΔSuniv = ΔSsys+ ΔSsurr = 0 → reversible process • ΔSuniv = ΔSsys+ ΔSsurr > 0 → irreversible process • Unlike energy, entropy is not conserved • Examples • Straightening up your room • 4Fe(s) + 3O2(g) → 2Fe2O3(s) • Exceptions – Isolated Systems • ΔSsys = 0 → reversible process • ΔSsys > 0 → irreversible process

  11. Calculating Entropy • Consider the reversible melting of 1 mol of ice in a large isothermal water bath at 0°C and 1 atm pressure. The enthalpy of fusion of ice is 6.01 kJ/mol. Calculate the entropy change in the system and in the surroundings, and the overall change in entropy of the universe for this process. • Answer = 22.0 J/K, 0 J/K

  12. Molecular Interpretation of Entropy • Why does the entropy increase when a gas expands? • Why does entropy decrease in the following reaction - 2NO(g) + O2(g) → 2NO2(g)? • Degrees of Freedom • Translational Motion • Vibrational Motion • Rotational Motion

  13. Molecular Interpretation of Entropy • Lowering temperature decreases energy which lowers the degrees of freedom. • Third law of thermodynamics = entropy of a pure crystalline substance at absolute zero is zero. • Entropy generally increases with temperature.

  14. Entropy Changes in Reactions • Entropy increases for processes in which: • Liquids or solutions are formed from solids. • Gases are formed from either solids of liquids. • The number of molecules of gas increases during a chemical reaction. • Choose the sample of matter that has greater entropy and explain your choice. • 1 mol of NaCl(s) or 1 mol HCl(g) at 25°C • 1 mol of HCl(g) or 1 mol Ar (g) at 25°C • 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C

  15. Predicting Entropy Changes • Predict whether the entropy change of the system in each of the following isothermal reactions if positive or negative. • CaCO3(s) → CaO(s) + CO2(g) • N2(g) + 3H2(g) → 2NH3(g) • N2(g) + O2(g) → 2NO(g) • HCl(g) + NH3(g) → NH4Cl(s) • 2SO2(g) + O2(g) → 2SO3(g)

  16. Entropy Changes in Chemical Reactions • Using variation of heat capacity with temperature absolute entropy are measured. • Molar entropy at standard states (S°) • Molar entropies of elements are not zero • Molar entropies of gases are greater than liquids and solids • Molar entropies generally increase with increasing molar mass. • Molar entropies generally increase with increasing number of atoms

  17. Entropy Changes in Chemical Reactions • Entropy Change in a reaction can be calculated using a table of values • ΔS°= ΣnS°(prod)-ΣmS°(react) • Calculate ΔS° for the synthesis of ammonia from nitrogen and hydrogen at 298K. • N2(g) + 3H2(g) → 2NH3(g) • Answer: -198.3 J/K

  18. Entropy Changes in the Surroundings • Surroundings are essentially a large constant temperature heat source. • The change in entropy will then depend on how much heat is absorbed or given off by the system. • ΔSs u r r = -qs y s /T • If the reaction happens at constant P, what is q? • Calculate the ΔSu n i v given the heat of formation of ammonia (-46.19kJ/mol)

  19. Gibbs Free Energy • Spontaneity depends on enthalpy and entropy. • Gibbs Free Energy – G = H – TS • In a chemical reaction at constant T • ΔG = ΔH – TΔS • Algebra fun • When both T and P are constant • If ΔG is negative, the reaction is spontaneous in the forward direction • If ΔG is zero, the reaction is at equilibrium • If ΔG is positive, the forward reaction is nonspontaneous, work must be done to make it occur. The reverse reaction is spontaneous.

  20. Standard Free Energy Changes • Gibbs free energy is a state function. • For Standard free energies • Free energies for standard states are set to zero. • Gases should be at 1 atm • Solutions should be 1M in concetration • Solids and Liquids should be in their pure forms • ΔG°= ΣnG°(prod)-ΣmG°(react)

  21. Calculating Standard Free Energies • Using the data from Appendix C, calculate the standard free-energy change for the following reaction at 298K: • P4(g) + 6Cl2(g) → 4PCl3(g) • What about the reverse reaction? • Answer: -1054.0kJ

  22. Calculating Standard Free Energies • Without using data from appendix C, predict whether ΔG° for this reaction is more or less negative than ΔH° • C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH°=-2220kJ • Using the data from appendix C, calculate the standard free energy change for this reaction. What your prediction correct? • Answer: -2108kJ

  23. Calculating Standard Free Energies • Consider the combustion of propane to form CO2(g) and H2O(g) at 298K. Would you expect ΔG° to be more negative or less negative than ΔH°?

  24. Free Energy and Temperature • ΔG = ΔH – TΔS • Ice melting • ΔG° = ΔH° – TΔS° • Use to estimate at other temperatures

  25. Homework • Part 1 - 2, 5, 7, 17, 20, 23, 25, 29, 31, 34, 37 • Part 2 - 41, 43, 47, 49, 53, 55, 56

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