# Three Laws of Thermodynamics 1. ΔE univ = 0 For the system: ΔE = q + w 2. ΔS univ ≥ 0 For the system: ΔS ≥ q/T, where the “=“ applies to reversible processes.

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Three Laws of Thermodynamics 1. ΔE univ = 0 For the system: ΔE = q + w 2. ΔS univ ≥ 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.

Gibbs Energy Change at Constant T,P 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 (at constant T and P) must be supplied to the system for the reaction to take place (but the reverse reaction is spontaneous).

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. Δ rxn G° = ΣmΔ f G°(products) – ΣnΔ f G°( reactants) (m and n represent the stoichiometric coefficients) 2. ΔG° = ΔH° – TΔS°

Conventions Used in Establishing Standard Gibbs Energy G° State of MatterStandard State ElementΔ f G° is 0 if it is in its standard state at the specified T. Solution1 M Gas1 bar Liquidpure liquid Solidpure solid

Finding ΔG° at a T other than 25°C ΔG°(T) = ΔH°(25°C) – TΔS°(25°C) You may not use the ΔG° values in the Appendix. You 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.)

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? H 2 O(l, 75°C)  H 2 O(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

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 ? H 2 O(g, 1 bar, 65°C)  H 2 O(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.

How do ΔH° and ΔS° Affect ΔG°? What causes the condensation of 1 mol of water vapor at 65°C and 1 bar pressure to be spontaneous? H 2 O(g, 1bar, 65°C)  H 2 O(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.

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

Finding ΔG Under Nonstandard Conditions

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

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 N 2, 0.75 bar H 2, and 2.00 bar NH 3. 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.

Finding ΔG Under Nonstandard Conditions

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

Calculating K from ΔG° 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: ½N 2 (g) + H 2 (g) NH 3 (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)

Calculating K from ΔG°

Use Appendix C to calculate the equilibrium constant for the formation of 1 mole of ammonia at 450°C. ½N 2 (g) + H 2 (g) NH 3 (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.

Calculating K from ΔG° The CRC Handbook of Chemistry and Physics gives Δ f G° = -623.42 kJ for carbonic acid and Δ f G° = -587.06 kJ for the bicarbonate ion, both at 25°C. Find K a1 using this data. First, we write the equation for the reaction: H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) ΔG°(1M, 25°C) = -587.06 - (-623.42) = 36.36 kJ Appendix D, Table 1 has K a1 = 4.3 x 10 -7

Calculating ΔG° from K

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