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

2. Definition l Entropy (S) is disorder, or randomness, of a system

3. Entropy l Gases with particles free to move randomly, generally have greater entropy (disorder) than liquids. l Crystalline solids, where particles are constrained to vibrate about a fixed position, generally have lower entropy than liquids. S solid < S liquid << S gas

4. Entropy and Gases l Within a sample of gas, the entropy increases with increased volume (and consequently decreases with increased pressure), as a greater volume allows the particles greater freedom of position and increased disorder. l The entropy of a gas sample also increases with increased temperature

5. Larger or More Complex Molecules and Entropy As a general rule: l Larger molecules have higher entropy than smaller molecules. l Molecules with more complex structures have larger entropies than simpler molecules. l This is due to the disorder associated with the degrees of freedom of the molecules

6. Thermodynamics l 1st Law- the energy of the universe is constant. l 2nd Law - the entropy of the universe increases.

7. 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)

8. l The entropy of an element or compound under any set of conditions is the entropy gained by converting the substance from 0 K to the defined conditions.

9. Connecting Entropy and Enthalpy of a System l ∆S = q rev T l q rev is the heat absorbed or emitted by the system l T is the Kelvin temperature at which the change occurs.

10. l Because it is necessary to add heat to raise the temperature, all substances have positive entropy values at temperatures above 0 K. l Based on the 3 rd law of thermodynamics, negative values of entropy cannot occur.

11. l Standard Entropy Sº (at 298 K and 1 atm) of substances are listed in the textbook Appendix. The standard entropy, Sº, of a substance, is the entropy gained by converting it from a perfect crystal at 0 K to standard state conditions. Units for entropy are: J/mole K

12.  Sº = ∑ n S º prod - ∑ m S º reac l Sº is a state function l “n” and “m” are the number of moles from the balanced equation.

13. 2H 2 ( g) + O 2 ( g) → 2H 2 O(l) + 572 kJ / mol SubstanceS[Subst], J / mol  K H2( g)131 O2( g)205 H2O(l)70  Sº = [ 2 (70 J /mol  K] - [ 2 (131 J / mol  K) + (205 J/mol  K) ] = - 327 J/mol  K

14. Intuitively Speaking l Intuitively, the balanced equation reflects the conversion of three moles of highly disordered gases to form two moles of less disordered liquid, it is expected that the change in entropy of the reaction would be negative.

15. l The second law of thermodynamics states that “in any spontaneously occurring process, the total entropy of the universe will increase.” Thus, a chemical reaction or physical change will proceed spontaneously in the direction that increases the total entropy of the universe.

16. Spontaneous Reactions 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.

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

18. 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.

19. 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.

20. 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 l In a previous slide you saw  S = q rev /T the  S was for the system

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

22. 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.

23. Gibbs Free Energy l Gibbs free energy is a thermodynamic property that indicates the amount of energy available for the system to do useful work at constant T and P and GFE is used to predict whether a process will occur spontaneously at constant temperature and pressure. Gibbs free energy “G” is defined as G = H - TS where H, T and S are the enthalpy, temperature, and entropy.

24. 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 sys -  G/T =  S univ If  G is negative at constant T and P, the process is spontaneous.

25. 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.

26. 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

27. Objectives l SWBAT l Practice solving for ΔG using various methods. l Practice solving for ΔG when the conditions are not standard state.

28.  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

29. 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.

30. 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.

31. Not Standard Conditions? l Thus far, only reactions occurring under standard concentration or pressure conditions, as indicated by the superscript “°” have been considered. l However, the concentration of each species in the equation does have an effect on the values of the thermodynamic properties of the reaction.

32. l An increase in the concentration of the reactant species, or a decrease in the concentration of the product species, will provide a greater driving force to the reaction and will decrease the value of ΔG Rxn from its value at standard concentrations.

33. l Similarly, an increase in product concentrations or a decrease in reactant concentrations will have the effect of increasing the value of ΔG Rxn from its value at standard concentrations.

34. Gibbs-Helmoltz Eqn  G =  Gº + RT ln(Q) where Q is the reaction quotient R is the gas constant R is 8.31 x 10 -3 kJ /mole  K

35. l Reactions that may occur spontaneously under standard concentrations (ΔG rxn 0

36. 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

37. How far will 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 position. At equilibrium  G = 0, Q = K l ΔG = ΔG° + RT(ln Q)  Gº = - RT lnK

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

39. Temperature dependence of K - RT ln K =  Gº =  Hº - T  Sº ln( K ) = - (  Hº/R)(1/T) +  Sº/R y = m x + b The graph is a straight line of ln K vs 1/T

40. ΔG = ΔG° + R T ln Q 0 = ΔG° + R T ln K eq ΔG° = - R T ln K eq or rearrange to: K eq = e –ΔG° ⁄R T