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Chapter 17 Thermodynamics: Spontaneity, Entropy & Free Energy Hill, Petrucci, McCreary & Perry 4th Ed. Thermodynamics & Equilibrium First Law of Thermodynamics ∆u = q + w Change in the Internal Energy of a System = Heat absorbed or lost to a system + Work done on or by a system on its surroundings. The internal energy,"u", is a "state function". This means that it is a property that depends only on the present state (T, P, phase) of the system. Another State Function: Enthalpy The “Heat Content” of a System ∆H = Hfinal - Hinitial In practice only ∆H's are known! o ∆H = Σn∆Hf (products) o Σm∆Hf (reactants) o - The small zeros "o" indicate the system is at "Standard State". "Standard State" The system state (phase) at 25 C; 1 atm Pressure. Note: System phase may not be the normal phase! o o ∆Hf H2O(l) = ∆Hf H2O(g) Spontaneous Processes and Entropy •A spontaneous process is a chemical or physical change that occurs without outside agency. •Entropy – A measure of a system’s disorder or degree of randomness. ∆S = S final - S initial Second Law of Thermodynamics - For any spontaneous process the total Entropy of the system and its surroundings always increases. Entropy is created when heat flows into a System ∆S = Entropy Created Entropy created cannot usually be measured A mathematical way to define the Second Law of Thermodynamics ∆S > q T + q T For a change at equilibrium: ∆S = Kelvin Temp. At constant T & P ∆H > T q T heat ∆H = T Gibb’s Free Energy - ∆G ∆S > rearranging: ∆H T For a spontaneous process at constant temperature and pressure. ∆H ∆S < 0 T A negative quantity! Rearranging again: ∆H - T∆S < 0 Defining a quantity! ∆H - T∆S This quantity is the energy available to a system to do useful work. ∆G The "Gibb's Free Energy" or just the "Free Energy". Quantifying Spontaneity ∆H – T∆S / ∆G ∆G is negative Y Process is spontaneous ∆G is zero Y Process is at equilibrium ∆G is positive Y Process is non-spontaneous In Practice: ∆G# -10 kJ/mole Y complete reaction ∆G $ +10 kJ/mole Y no reaction ∆G = K10 kJ/mole Y equilibrium mixture Thermodynamic Quantities o ∆Hf Standard Enthalpy of Formation The Heat liberated (or consumed) when forming one mole of a substance from the elements. o ∆Gf Standard Free Energy of Formation Defined analogously for formation of one mole of a substance from the elements. See Appendix C1 p A13. o ∆G = Σn∆Gf (products) o - Σm∆Gf (reactants) o The Third Law of Thermodynamics A perfectly crystalline substance at Absolute Zero (0 oK) has an entropy of zero. So Standard Entropy A measure of the disorder of one mole of a substance at 25 C and 1 atm pressure. See Appendix C1 p A13. For a chemical reaction: o ∆S = ΣnS(products) o Note: these are absolute entropies. - ΣmS(reactants) o A Rule of Thumb in Predicting Entropies S0gases >> S0liquids > S0solids Processes producing gases tend to have large, positive entropy changes. Commonly, not always, ∆S is positive when gases are formed. ∆S is positive when solids go to liquids or when solids go into solution Predict the Sign of Then Calculate ∆S Predicted Sign of ∆S Reaction: CaCO3(s) Remarks Calc. ∆S CaO(s) + CO2(g) CS2(g) CS2(l) 2 Hg(l) + O2(g) 2 HgO(s) 2 Na2O2(s) + 2 H2O(l) o ∆S = ΣnS(products) o 4 NaOH(aq) + O2(g) - ΣmS(reactants) o Calculating an Entropy Change 2 Hg(l) + O2(g) o ∆S = ΣnS(products) o o o 2 HgO(s) - ΣmS(reactants) o o o ∆S = [2 SHgO(s)] - [2 SHg(l) + SO2(g) ] o ∆S = [ 2mol(70.29 J/molK)] [2mol(76.02 J/molK) +1mol(205.0 J/molK)] o ∆S = [ 140.58 J/K] - [152.04 J/K + 205.0 J/K] o ∆S = [ 140.58 J/K] - [ 357.0 J/K] o ∆S = -216.5 J/K Can a Process with Negative Entropy be Spontaneous? 2 Hg(l) + O2(g) o ∆Hr = Σn∆Hf(products) o 2 HgO(s) - Σm∆Hf(reactants) o ∆Gr = ∆Hr - T∆Sr o ∆Sr = -216.5 J/K o Use tabulated ∆Hf values to calculate the Enthalpy and use the Gibbs Eqn. to calculate the Free Energy. Calculate the ∆G for the Reaction Directly 2 Hg(l) + O2(g) o ∆Gr = Σn∆Gf(products) o o 2 HgO(s) - Σm∆Gf(reactants) o Use tabulated values of ∆G f to calculate the Free Energy directly for the reaction. Thermodynamic Quantities o ∆Hf Zero for an element in its standard state*. *The normal phase of the element at T = 25 C and 1 atm. P. ∆Gfo Zero for an element in its standard state*. *These data will NOT be found in Appendix C1 p A13 because they are defined values. Like Enthalpies, Entropy and Free Energy values for chemical compounds must ultimately be determined by experiment! o ∆Hrxn - o T∆Srxn o ∆Grxn The Chemical Coupling of Reactions The rusting of iron is a very spontaneous reaction! 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) o The reverse reaction, the refining ∆Grxn = -1487 kJ of iron is very non-spontaneous! 2 Fe2O3(s) 4 Fe(s) + 3 O2(g) How does one enable a nonspontaneous reaction? o = +1487 kJ ∆Grxn Couple it with a spontaneous reaction! 2 CO(g) + O2(g) Spontaneous Reaction + 2 CO2(g) non-Spontaneous Reaction ∆Grxn = -514.4 kJ o Spontaneous Reaction Couple Couple the two reactions so the couple is spontaneous! Free Energy & the Equilibrium Constant For Non-Standard States: ∆G = ∆Go + RT ln Q But at Equilibrium ∆G = zero & Q = K 0 = ∆Go + RT ln K Rearranging: ∆Go = - RT ln K ∆Go = -2.303 RT log K or or log K = -∆Go 2.303 RT What are the Values of the Equilibrium Constant for various ∆Go? log K = -∆Go 2.303 RT ~ K = [products] [reactants] If ∆Go => Spontaneous: If ∆Go => Non-Spontaneous: Variation of ∆G with Temperature From the Gibbs Equation : o ∆G = o ∆H o - T∆S o o But ∆H and ∆S don't vary much with a change in Temperature. o Since ∆H and ∆S o are nearly constants o ∆G varys mostly with T . ∆GT o o ~ = ∆H - T∆S The above equation is a good approximation of ∆GT at other temperatures. Predicting Spontaneity w.r.t Temperature ∆H o negative negative positive positive Color code: ∆S o positive negative negative positive ∆G always negative negative low Temp positive high Temp always positive positive low Temp negative high Temp A Non-spontaneous effect A Spontaneous effect Predict the Reaction Spontaneity N2(g) + 3 H2(g) ∆H is negative CaCO3(s) ∆H is positive 2 NH3(g) ∆S is negative ∆G = ? High Temp Low Temp CaO(s) + CO2(g) ∆S is positive ∆G = ? High Temp Low Temp