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Chapter 16 Spontaneity, Entropy and Free Energy 16.1 Spontaneity and Entropy A reaction that will occur without outside intervention. We can’t determine how fast. We need both thermodynamics and kinetics to describe a reaction completely. Thermodynamics compares initial and final states but not how long it will take. Kinetics (YDVD) describes the pathway between. Thermodynamics 1st Law - the energy of the universe is constant. (Cannot create or destroy) Keeping track of thermodynamics doesn’t correctly predict spontaneity. Entropy (S) is disorder or randomness. 2nd Law - the entropy of the universe increases. Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. Defined in terms of probability. Substances take the arrangement that is most likely. The most likely is the most random. Think about this. It would seem like reactions with a negative DH value would proceed spontaneously. But NH3 = O2 NO + H2O will not occur spontaneously even though DH = -909kJ/mol It would also seem like reactions with a positive DH value would not occur spontaneously. But KCl(s) K+(aq) + Cl-(aq) will occur spontaneously even though DH =+19.2 kJ/mol Activation energy is involved in both reactions but there must be something else involved. Entropy is that driving force! Positional Entropy A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system. Ssolid < Sliquid << Sgas There are many more ways for the molecules to be arranged as a liquid than a solid. Gases have a huge number of positions possible. Practice For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature. • solid CO2 and gaseous CO2 • N2 gas at 1 atm and N2 gas at 1.0 x 10-2 atm Predict the sign of the entropy change for each of the following processes. • Solid sugar is added to water to form a solution. • Iodine vapor condenses on a cold surface to form crystals. Proof? The change in entropy (DS) in a reaction can be found by subtracting the absolute values of the entropy of the reactants from the absolute entropies of the products DS reaction = DS products - DS reactants For the reaction below, the absolute entropies of the chemicals involved are 193 J/K mol forNH3 (g), 192 J/K mol-1 for N2 (g) and 131 J/K mol-1 for H2 (g). N2(g) + 3H2(g) 2NH3(g) DS reaction = (2 x 193) – (192 + (3 x 131)) = -199 J/K This makes sense since the negative value implies that the system has become more ordered (four gas molecules are converted to two gas molecules). 16.2 Entropy Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate. 2nd Law - in any spontaneous process there is always an increase in the entropy of the universe. DSuniv = DSsys + DSsurr If DSuniv is positive the process is spontaneous. If DSuniv is negative the process is spontaneous in the opposite direction. If DSuniv is 0 the system is at equilibrium. For exothermic processes DSsurr is positive. For endothermic processes DSsurr is negative. Consider this process H2O(l) H2O(g) DSsys is positive DSsurr is negative DSuniv depends on temperature. 16.3 Temperature and Spontaneity Entropy changes in the surroundings are determined by the heat flow. An exothermic process is favored because by giving up heat the entropy of the surroundings increases. The size of DSsurr depends on temperature. DSsurr = -DH/T Practice In the metallurgy of antimony, the pure metal is recovered by two different reactions. Ex16.4 Sb2S3(s) + 3Fe(s) 2Sb(s) +3FeS(s) DH = -125 kJ Sb4O6(s) + 6C(s) 4Sb(s) + 6CO(g) DH = 778 kJ Calculate the DSsurr for each of these reactions at 25°C and 1 atm. 16.4 Gibb's Free Energy DG=DH-TDS at constant temperature Dividing by -T and substituting DH/T for DSsurr we’ve shown that: -DG/T = DSuniv (at constant T and P) This is very important! It means if the process is carried out at constant T and P, the process will be spontaneous only if DG is negative. -DG means +DSuniv Let’s Check For the reaction H2O(s) H2O(l) DSº = 22.1 J/K mol DHº =6030 J/mol Calculate DG at 10ºC and -10ºC Look at the equation DG=DH-TDS Spontaneity can be predicted from the sign of DH and DS. DG=DH-TDS DS DH + - At all Temperatures + At high temperatures, “entropy driven” - At low temperatures, “enthalpy driven” + Not at any temperature, Reverse is spontaneous + - Spontaneous? DG and Spontaneity At what temperature is the following process spontaneous at 1 atm? Br2(l) Br2(g) DkJ/mol DS° = 93.0 J/K•mol Hint: Use DG=DH-TDS and set DG to 0 16.5 Entropy Changes in Chemical Reactions Reactions involving gaseous molecules: The change in positional entropy is dominated by the relative number of molecules of gaseous reactants and products. 2C2H6 + 7O2 4CO2 + 6H2O DS increases because there are more molecules on the product side than the reactant side. Practice Predict the sign of DS° for each. Ex16.6 The thermal decomposition of solid calcium carbonate: CaCO3(s) CaO(s) + CO2(g) The oxidation of SO2 in air: 2SO2(g) + O2(g) 2SO3(g) Predict the sign of ΔS。 for each of the following changes. (a)Na(s) + 1/2 Cl2(g) NaCl(s) (b) N2(g) + 3 H2(g) 2 NH3(g) (c) NaCl(s) Na+(aq) + Cl -(aq) (d) NaCl(s) NaCl(l) Third Law of Thermodynamics The entropy of a pure crystal at 0 K is 0. No disorder. This gives us a starting point. All others must be > 0. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. Products - reactants to find DSº (a state function). More complex molecules higher Sº. Practice 1. Predict then calculate DS° at 25°C for the reaction. (use the DS° values in appendix A21) Ex 16.7 2NiS(s) + 3O2(g) 2SO2(g) + 2NiO(s) 2. Predict then calculate DS° for the reduction of aluminum oxide by hydrogen gas. Ex 16.8 Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g) Practice 1. For the reaction #35 C2H2(s) + 4F2(g) 2CF4(g) + H2(g) DS° is equal to - 358J/K. Use this value and data from the appendix to calculate the S° for CF4(g) Consider the reaction #39 1. 2O(g) O2(g) aPredict the signs of DH and DS. b. Would the reaction be more spontaneous at high or low temperatures? From data in Appendix 4, calculate ΔH°, ΔS°, and ΔG° for the following reaction at 25。C. #41 CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) 16.6 Free Energy in Reactions DGº = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Can’t be measured directly, can be calculated from other measurements. Method #1 DGº=DHº-TDSº Method #2 Use Hess’s Law with known reactions. Free Energy in Reactions There are tables of DGºf .(A21) Method #3 DG = npDGf(products) nrDGf(reactants) Products - reactants because it is a state function. The more negative G, the further to the right the reaction will proceed to reach equilibrium. The standard free energy of formation for any element in its standard state is 0. Remember - Spontaneity tells us nothing about rate. Practice 1.Consider the reaction: 2SO2(g) + O2(g) 2SO3(g) Carried out at 25°C and 1 atm. Calculate DH° & DS° using the data in A21. Find DG° using DG°=DH°-T DS° 2.Using the following data (at 25°C) Cdiamond(s) + O2(g) CO2(g) DG° = -397 kJ Cgraphite(s) + O2(g) CO2(g) DG° = -394 kJ calculate DG° for the reaction: Cdiamond(s) Cgraphite(s) 3. Calculate DG° for the reaction using data on A21 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g) 16.7 Free energy and Pressure DG = DGº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants). CO(g) + 2H2(g) CH3OH(l) Would the reaction be spontaneous at 25ºC with the H2 pressure of 5.0 atm and the CO pressure of 3.0 atm? DGºf CH3OH(l) = -166 kJ DGºf CO(g) = -137 kJ DGºf H2(g) = 0 kJ 16.8 Free Energy and Equilibrium DG tells us spontaneity at current conditions. When will it stop? It will go to the lowest possible free energy which may be an equilibrium. At equilibrium DG = 0, Q = K DGº = -RTlnK from [DG = DGº + RTlnK] DGº =0 <0 >0 K =1 >1 <1 Temperature Dependence of K DGº= -RTlnK = DHº - TDSº ln(K) = DHº/R(1/T)+ DSº/R A straight line of lnK vs 1/T 16.9 Free Energy and Work “Free energy” is energy “free” to do work. The maximum amount of work possible at a given temperature and pressure. Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.