Chapter 19 Spontaneity, entropy and free energy (rev. 11/09/08)

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.

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.

Entropy l Defined in terms of probability. l Substances take the arrangement that is most likely. l The most likely is the most random.

Randomness l You need to calculate the number of arrangements for a system. l See the next few slides for clarification.

l 2 possible arrangements l 50 % chance of finding the left empty

l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed

l 4 possible arrangements l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed

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.

Entropy of Solutions l Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.

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.

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.

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

 S sys  S surr  S univ Spontaneous? ? +-? Yes No, Reverse At Low temp. At High temp.

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.

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.

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

 G=  H-T  S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

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)

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.

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.

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.

 G=  H-T  S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

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

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

 Gº K =0=1 0 >0<0 The connection between ∆Gº and K.

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

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.