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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.

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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).

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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°

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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

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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.)

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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

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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.

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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.

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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)

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Finding ΔG Under Nonstandard Conditions

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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.

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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.

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Finding ΔG Under Nonstandard Conditions

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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

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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)

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Calculating K from ΔG°

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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.

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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

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Calculating ΔG° from K

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