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Spontaneity, Entropy, and Free Energy Chapter 17

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Spontaneity, Entropy, and Free Energy Chapter 17 3/30/14 First Law of Thermodynamics First law of thermodynamics: energy cannot Chapter 17 be created or destroyed. The total energy of the universe cannot change Spontaneity, Entropy, and (though you can transfer it from one place to another). Free Energy ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings Chapter 17 Chapter 17 2 First Law of Thermodynamics First Law of Thermodynamics conservation of energy ΔE is a state function: internal energy change independent of how it was done exothermic rxn: releases heat to the surroundings (system è surroundings) two ways energy “lost” from a system: converted to heat, q and/or used to do work, w Energy conservation requires that the energy change in the system equal the heat released plus the work done. ΔE = q + w ΔE = ΔH + pΔV Chapter 17 3 Chapter 17 4 1 3/30/14 Thermodynamics and Spontaneity Comparing Potential Energy Thermodynamics predicts whether a process will proceed under the given conditions (spontaneous process). Nonspontaneous processes require energy input to go. Spontaneity is determined by comparing the chemical potential energy of the system before the reaction with the free energy of the system after the reaction If the system after the rxn has less potential energy than The direction of spontaneity can be determined by before the rxn, the rxn is thermodynamically favorable. comparing the potential energy of the system at the start and the end. spontaneity ≠ fast or slow Chapter 17 5 Chapter 17 6 Reversibility of Process Thermodynamics and Spontaneity Any spontaneous process is irreversible, i.e., it will proceed in only one direction. A reversible process will proceed back and forth between the two end conditions. - equilibrium rxns (they result in no change in free energy;) If a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction. Chapter 17 7 Chapter 17 8 2 3/30/14 Thermodynamics and Spontaneity Factors Affecting Whether a Rxn is Spontaneous diamond vs. graphite Reactions are spontaneous in the direction of lower chemical potential energy. There are two factors that determine the thermodynamic favorability: the enthalpy change and the entropy change. The enthalpy change, ΔH, is the difference in the sum of the internal energy and pV work energy of the reactants more stable compared to the products. conversion of diamond into graphite is spontaneous The entropy change, ΔS, is the difference in randomness of the reactants compared to the products. Chapter 17 9 Chapter 17 10 Enthalpy (review) Entropy Entropy, S, is a thermodynamic function that increases as ΔH is generally measured in kJ/mol. the number of energetically equivalent ways of arranging - stronger bonds = more stable molecules the components (degree of freedom) increases. ΔHrxn < 0: exothermic rxn; ΔHrxn > 0: endothermic rxn S is generally measured in J/mol. ΔΗ = Σ (ΔH° ) − Σ (ΔH° ) rxn f products f reactants Random systems require less energy than ordered systems. (read section 17.1) endothermic rxn: the bonds in the products are weaker than the bonds in the reactant; exothermic rxn: if the bonds in the products are stronger than the bonds in the reactants Chapter 17 11 Chapter 17 12 3 3/30/14 Changes in Entropy, ΔS Changes in Entropy, ΔS Entropy change is favorable when the result is a more Examples: random system (ΔS = Sfinal – Sinit. > 0 ) ΔS > 0 if: - products are in a more random state - going from solid to less ordered liquid to less ordered gas - rxns that have larger # of product molecules than reactant molecules - increase in temperature - solids dissociating into ions upon dissolving Chapter 17 13 Chapter 17 14 Changes in Entropy, ΔS Entropy Change and State Changes When a material changes physical state, the number of ΔSsystem > 0 for a process in which the final condition is macrostates it can have changes as well. more random than the initial condition (favorable entropy) - the more degrees of freedom the molecules have, the more macrostates are possible; ΔSsystem < 0 for a process in which the final condition is more orderly than the initial condition (unfavorable entropy) ΔSsystem = ΔSreaction = Σ (S°prod) − Σ (S°react) Chapter 17 15 Chapter 17 16 4 3/30/14 Problem A Predict whether S is + or − for Entropy Change and State Changes Δ system each of the following: - solids have fewer macrostates than liquids, which have • Heating air in a balloon fewer macrostates than gases; • Water vapor condensing • Separation of oil and vinegar salad dressing • Dissolving sugar in tea • 2 HgO(s) → 2 Hg(l) + O2 (g) • 2 NH3 (g) → N2 (g) + 3 H2 (g) + − • Ag (aq) + Cl (aq) → AgCl(s) Chapter 17 17 Chapter 17 18 nd the 2 Law of Thermodynamics Temperature Dependence of ΔSsurroundings • The total entropy change of the universe must be When a system process is exothermic, it adds heat to the positive for a process to be spontaneous. surroundings, increasing the entropy of the surroundings. - for reversible process, ΔSuniv = 0. - for irreversible (spontaneous) process, ΔSuniv >0 When a system process is endothermic, it takes heat from ΔSuniverse = ΔSsystem + ΔSsurroundings the surroundings, decreasing the entropy of the surroundings. • If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount. The amount the entropy of the surroundings changes - when ΔSsystem < 0 , ΔSsurroundings > 0 . depends on its initial temperature. Chapter 17 19 Chapter 17 20 5 3/30/14 Problem B Calculate the ΔS at 25ºC for the rxn: Temperature dependence of ΔSsurroundings surr. C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) The higher the original temperature, the less effect ΔHrxn = − 2044 kJ. addition or removal of heat has. Chapter 17 21 Chapter 17 22 Problem C The rxn below has H = +66.4 kJ at 25°C. Problem C Δ rxn The rxn below has ΔHrxn = +66.4 kJ at 25°C. 2 O2 (g) + N2 (g) → 2 NO2 (g) 2 O2 (g) + N2 (g) → 2 NO2 (g) i. Determine ΔSsurroundings. ii. Determine the sign of ΔSsystem. Chapter 17 23 Chapter 17 24 6 3/30/14 Problem C The rxn below has ΔH = + 66.4 kJ at 25°C. rxn Gibbs Free Energy and Spontaneity iii. Determine whether the process is ΔSuniv = ΔSsys + ΔSsurr spontaneous -Τ ΔSuniv = ΔGsys 2 O2 (g) + N2 (g) → 2 NO2 (g) ΔGsys = ΔHsys− TΔSsys ΔH ΔS = ΔS - sys univ sys T -T ΔH -Τ ΔS = -T ΔS - sys univ sys T -Τ ΔSuniv = -T ΔSsys + ΔHsys -Τ ΔSuniv = ΔHsys - Τ ΔSsys Chapter 17 25 Chapter 17 26 Gibbs Free Energy and Spontaneity Gibbs Free Energy, ΔG -Τ ΔSuniv = ΔGsys The Gibbs free energy is the maximum amount of work ΔSuniv is + when spontaneous, so ΔG is − energy that can be released to the surroundings by a system. And therefore, when ΔG < 0 process is spontaneous - valid at a constant temperature and pressure Possible ΔG values: ΔG = ΔH − T ΔS - often called the chemical potential energy because it is sys sys sys ΔG < 0: analogous to the storing of energy in a mechanical system ΔH < 0 and ΔS > 0 rxn exothermic & more random ΔG = ΔH − T ΔS sys sys sys ΔH < 0 and large and ΔS < 0 but small - since ΔSuniv determines whether a process is spontaneous, ΔG also determines spontaneity; ΔH > 0 but small and ΔS > 0 and large -Τ ΔSuniv = ΔGsys or high temperature Chapter 17 27 Chapter 17 28 7 3/30/14 G = H − T S Gibbs Free Energy, ΔG Δ sys Δ sys Δ sys Free Energy Change and Spontaneity Possible ΔG values: ΔG > 0 : ΔH > 0 and ΔS < 0 never spontaneous at any temperature ΔG = 0 : the reaction is at equilibrium Chapter 17 29 Chapter 17 30 Problem D Problem D continued The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. ii. Determine if the reaction is spontaneous. i. Calculate ΔG for the reaction. Chapter 17 31 Chapter 17 32 8 3/30/14 Problem E Problem E continued The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. ii. Determine if the reaction is spontaneous. i. Calculate ΔG for the reaction. Chapter 17 33 Chapter 17 34 Problem F Problem F continued The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C. Calculate the minimum temperature for it to be spontaneous. Chapter 17 35 Chapter 17 36 9 3/30/14 Problem G Problem G continued The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C. Calculate the maximum temperature for it to be spontaneous. Chapter 17 37 Chapter 17 38 Problem G continued The Third Law of Thermodynamics Absolute Entropy Read pages 656 - 659 Chapter 17 39 Chapter 17 40 10 3/30/14 Problem H Calculate ΔS° for the rxn Changes in Entropy 4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g) ΔSºreaction = ( ∑ np Sºproducts) − ( ∑ nr Sºreactants) Substance S°, J/mol⋅K NH 192.8 the standard entropy change is the difference in absolute 3 (g) O 205.2 entropy between the reactants and products under 2 (g) standard conditions; NO(g) 210.8 H2O(g) 188.8 Chapter 17 41 Chapter 17 42 Problem H continued Problem H continued Chapter 17 43 Chapter 17 44 11 3/30/14 Problem I Calculate ΔS° for the rxn Problem I continued 2 H2 (g) + O2 (g) → 2 H2O(g) Substance S°, J/mol⋅K H2 (g) 130.7 O2 (g) 205.2 H2O(g) 188.8 Chapter 17 45 Chapter 17 46 Calculating ΔGo Problem J Calculate ΔG° at 25 °C for the reaction CH4 (g) + 8 O2 (g) → CO2 (g) + 2 H2O(g) + 4 O3 (g) o o o 1.
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