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Chapter 19 Spontaneity, entropy and free energy (rev. 11/09/08)
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Spontaneous l This is a reaction that will occur without outside intervention. l We can’t determine how fast. l We need both thermodynamics and kinetics to describe a reaction completely. l Thermodynamics compares initial and final states. l Kinetics describes pathway between.
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Thermodynamics l 1st Law- the energy of the universe is constant. l Entropy (S) is disorder or randomness l 2nd Law - the entropy of the universe increases.
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Entropy l Defined in terms of probability. l Substances take the arrangement that is most likely. l The most likely is the most random.
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Randomness l You need to calculate the number of arrangements for a system. l See the next few slides for clarification.
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l 2 possible arrangements l 50 % chance of finding the left empty
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l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed
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l 4 possible arrangements l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed
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Gases l Gases completely fill their chamber because there are many more ways to do that than to leave half empty. l Gases have a huge number of positions possible. l S solid < S liquid << S gas l There are many more ways for the molecules to be arranged as a liquid than a solid.
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Entropy of Solutions l Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.
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2nd Law S univ = S sys + S surr If S univ is positive the process is spontaneous. If S univ is negative the process is spontaneous in the opposite direction.
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For exothermic processes S surr is positive. For endothermic processes S surr is negative. Consider this process H 2 O(l) H 2 O(g) S sys is positive S surr is negative S univ depends on temperature.
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Temperature and Spontaneity l Entropy changes in the surroundings are determined by the heat flow. l An exothermic process is favored because by giving up heat the entropy of the surroundings increases. The magnitude of S surr depends on temperature S surr = - H/T
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S sys S surr S univ Spontaneous? +++ --- +-? +-? Yes No, Reverse At Low temp. At High temp.
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Gibb's Free Energy l G=H-TS l The equation is never used this way. G= H-T S at constant temperature l Divide by -T - G/T = - H/T- S - G/T = S surr + S - G/T = S univ If G is negative at constant T and P, the process is spontaneous.
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Let’s Check For the reaction H 2 O(s) H 2 O(l) Sº = 22.1 J/K mol Hº =6030 J/mol Look at the equation G= H-T S Spontaneity can be predicted from the sign of H and S. Predict the sign of G.
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Try the Calculation For the reaction H 2 O(s) H 2 O(l) Sº = 22.1 J/K mol Hº =6030 J/mol Calculate G at 10ºC and -10ºC
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G= H-T S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous
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3rd Law l The entropy of a pure crystal at 0 K is 0. l The pure crystal entropy gives us a starting point. l All other entropy values must be>0. l (liquids and gases are more random, therefore their values should be more positive)
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l Standard Entropies Sº (at 298 K and 1 atm) of substances are listed in the textbook Appendix. Sº = ∑ n S º prod - ∑ m S º reac l Sº is a state function l More complex molecules higher Sº.
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Free Energy in Reactions Gº = standard free energy change. l Free energy change that will occur if reactants (in their standard state) turn into products (in their standard state). l Can’t be measured directly, can be calculated from other measurements. Gº= Hº-T Sº l Use Hess’s Law with known reactions.
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Free Energy in Reactions There are tables of Gº f. Use Gº = ∑ n G º prod - ∑ m G º reac l Products-reactants because it is a state function. l The standard free energy of formation for any element in its standard state is 0. l Remember- Spontaneity tells us nothing about rate.
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G= H-T S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous
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Free energy and Pressure G = Gº + RT ln(Q) where Q is the reaction quotient CO(g) + 2H 2 (g) CH 3 OH(l) l Would the reaction be spontaneous at 25ºC with the H 2 pressure of 5.0 atm and the CO pressure of 3.0 atm? Gº f CH 3 OH(l) = -166 kJ Gº f CO(g) = -137 kJ Gº f H 2 (g) = 0 kJ
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How far with the free energy go? G tells us spontaneity at current conditions. l When will it stop? l It will go to the lowest possible free energy which may be an equilibrium. At equilibrium G = 0, Q = K Gº = -RT lnK
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Gº K =0=1 0 >0<0 The connection between ∆Gº and K.
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Temperature dependence of K Gº= -RT lnK = Hº - T Sº ln(K) = Hº/R(1/T)+ Sº/R l A straight line of ln K vs 1/T
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Free energy And Work l Free energy is: energy free to do work. l It is the maximum amount of work possible at a given temperature and pressure. l It is never really achieved because some of the free energy is changed to heat during a reaction, so it can’t be used to do work.
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