Three Laws of Thermodynamics

1. Three Laws of Thermodynamics 1. ΔEuniv = 0 For the system: ΔE = q + w 2. ΔSuniv ≥ 0 For the system: ΔS ≥ q/T, where the “=“ applies to reversible processes and the “>” applies to irreversible processes. 3. The entropy of a pure substance in perfect crystalline form at absolute zero is 0 J/K.

2. Gibbs Energy ΔG < 0 Process is spontaneous.ΔG is the maximum P-V work that can be obtained from the reaction at constant T or is all of the non-PV work that can be obtained at constant T and P. ΔG = 0 Process is at equilibrium.(at constant T and P.) ΔG > 0 Process is not spontaneous.A minimum energy ΔG must be supplied to the system for the reaction to take place (but the reverse reaction is spontaneous).

3. Gibbs Energy The change in Gibbs Energy ΔG is the maximum non-PV work that can be obtained from a chemical reaction at constant T and P (or the minimum energy that must be supplied to make the reaction happen). There two ways to obtain ΔG°: 1. ΔrxnG° = ΣmΔfG°(products) – ΣnΔfG°( reactants) (m and n represent the stoichiometric coefficients) 2. ΔG° = ΔH° – TΔS°

4. Conventions Used in Establishing Standard Gibbs Energy G°

5. Finding ΔG° at a T other than 25°C ΔG°(T) = ΔH°(25°C) – TΔS°(25°C) You cannot use the ΔG° values in the Appendix, but must calculate ΔG° from ΔH°(25°C) and ΔS°(25°C). (We make the assumption that ΔH°(25°C) and ΔS°(25°C) don’t change with T.)

6. Finding ΔG° at a T other than 25°C How much energy must be added to 1 mole of water at 75°C to form 1 mole of water vapor at 1 bar and 75°C? H2O(l, 75°C) H2O(g, 1 bar, 75°C) ΔG°(75°C) = ΔH°(25°C) – TΔS°(25°C) ΔG°(75°C) = 44.01 kJ – 348 K (0.11892 kJ/K) ΔG°(75°C) = 2.62 kJ The physical change is not spontaneous. 2.62 kJ of energy must be added to the water.

7. Finding ΔG° at a T other than 25°C What is the maximum P-V work that can be obtained from 1 mole of water vapor condensing at 1 bar and 65°C ? H2O(g, 1 bar, 65°C) H2O(l, 65°C) ΔG°(65°C) = ΔH°(25°C) - TΔS°(25°C) ΔG°(65°C) = - 44.01 kJ - 338 K (- 0.11892 kJ/K) ΔG°(65°C) = - 3.82 kJ The physical change is spontaneous and produces 3.82 kJ of energy per mole of water vapor.

8. How do ΔH° and ΔS° Affect ΔG°? What causes the condensation of 1mol of water vapor at 65°C and 1 bar pressure to be spontaneous? H2O(g, 1bar, 65°C) H2O(l, 65°C) ΔG°(65°C) = - 44.01 kJ - 338 K (- 0.11892 kJ/K) = - 44.01+ 40.194 = - 3.82 kJ Answer: The exothermicity of the process is enough to overcome the decrease in entropy at 65°C.

9. How do ΔH and ΔS Affect ΔG? ΔG = ΔH - TΔS ΔHΔSΔG -- - at low T, + at higher T -+ - at all T (always spontaneous) ++ + at low T, - at higher T +- + at all T (never spontaneous)

10. Finding ΔG Under Nonstandard Conditions ΔG(T)= ΔG°(T) + RT ln Q Q is the reaction quotient R is the gas constant R = 8.3145 J/mol-K

11. The Reaction Quotient Q Q is found by putting values into the expression for the equilibrium constant K. N2(g) + 3H2(g) 2NH3(g) The values here MUST BE equilibrium values. The values here do NOT have to be equilibrium values.

12. Finding ΔG Under Nonstandard Conditions ΔG(T)= ΔG°(T) + RT ln Q Find the change in Gibbs Energy for the formation of ammonia at 298 K when the reaction mixture consists of 0.50 bar N2, 0.75 bar H2, and 2.00 bar NH3. You must always write the equation for the reaction first. ΔG is an extensive property, like ΔH and ΔS. Its value depends on the stoichiometry.

13. Finding ΔG Under Nonstandard Conditions ΔG(T)= ΔG°(T) + RT ln Q Find the Gibbs Energy for the formation of ammonia at 298 K when the reaction mixture consists of 0.50 bar N2, 0.75 bar H2, and 2.00 bar NH3. N2(g) + 3H2(g) 2NH3 (g)

14. Finding ΔG Under Nonstandard Conditions ΔG(T)= ΔG°(T) + RT ln Q Find the Gibbs Energy for the formation of ammonia at 298 K when the reaction mixture consists of 0.50 bar N2, 0.75 bar H2, and 2.00 bar NH3. N2(g, 1bar, 25°C) + 3H2(g, 1 bar, 25°C) 2NH3 (g, 1 bar, 25°C) ΔG°(1 bar, 25°C) = - 33.32 kJ ΔG = - 33.32 kJ + (8.3145 J/mol-K)(298K) (ln 18.963) = - 33.32 kJ + 7291 J = - 33.32 kJ + 7.291 kJ = - 26.03 kJ The fact that ΔG < 0 tells us that the reaction is spontaneous…and that the system is NOT at equilibrium.

15. Gibbs Energy and Equilibrium ΔG(T)= ΔG°(T) + RT ln Q At equilibrium, Q = K and ΔG = 0. 0= ΔG°(T) + RT ln K(T) ΔG°(T) = - RT ln K(T) K(T) = e -ΔG°/RT

16. CalculatingKfromΔG° Use Appendix C to calculate the equilibrium constant for the formation of 1 mole of ammonia at 25°C and at 450°C. First, we write the equation for the reaction: ½N2(g) + H2(g) NH3(g) For 450°C, we will need ΔH° and ΔS°. We will use these to calculate ΔG° for both temperatures. ΔH° = - 46.19 kJ ΔS° = -99.12 J/K = - 0.09912 kJ/K ΔG°(25°C) = - 16.66 kJ (from Appendix C)

17. CalculatingKfromΔG° Use Appendix C to calculate the equilibrium constant for the formation of 1 mole of ammonia at 25°C. ½N2(g) + H2(g) NH3(g) ΔG°(1 bar, 25°C) = - 16.66 kJ

18. CalculatingKfromΔG° Use Appendix C to calculate the equilibrium constant for the formation of 1 mole of ammonia at 450°C. ½N2(g) + H2(g) NH3(g) ΔG°(450°C) = - 46.19 - (723 K)(- 0.09912 kJ/K) = 25.47 kJ The reaction is not spontaneous at 450°C under standard conditions. It takes 25.47 kJ of energy to make the reaction happen at this temperature.

19. Calculating K fromΔG° The CRC Handbook of Chemistry and Physics givesΔfG° = -623.42 kJ for carbonic acid and ΔfG° = -587.06 kJ for the bicarbonate ion, both at 25°C. Find Ka1 using this data. First, we write the equation for the reaction: H2CO3(aq) H+(aq) + HCO3-(aq) ΔG°(1M, 25°C) = -587.06 - (-623.42) = 36.36 kJ Appendix D, Table 1 has Ka1 = 4.3 x 10-7

20. Calculating ΔG°fromK Appendix D Table 3 gives the solubility product Ksp for copper(II) carbonate. Use this Ksp to find ΔG° for copper(II) carbonate dissolving in water at 25°C. First, we write the equation for the reaction: CuCO3(s) Cu2+(aq) + CO32-(aq) ΔG° = - RT ln Ksp = - (8.3145 J/mol-K)(298 K) ln (2.3x10-10) ΔG°(1M, 25°C) = 55.0 kJ From the CRC Handbook of Chemistry and Physics: ΔG° = 54.9 kJ...If this value is not zero, does it mean the system cannot be at equilibrium at 25°C?