1. Chapter 16 Spontaneity, Entropy and Free energy

2. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect of temperature on spontaneity l Free energy l Entropy changes in chemical reactions: The Third Law of Thermodynamics l Free energy and chemical reactions l The dependence of free energy on pressure l Free energy and equilibrium l Free energy and work

3. 16.1 Spontaneous Process and Entropy l A process that will occur without outside intervention. l Thermodynamics can’t determine how fast the process is (may be fast or slow). l Kinetics tell us that the rate of reaction depends on: activation energy; temperature, concentration and catalysts; i.e, it depends on the pathway of process. l Thermodynamics compares initial and final states and does not require knowledge of the pathway. l Kinetics describes pathway between reactants and products. l We need both thermodynamics and kinetics to describe a reaction completely.

4. Spontaneity l process that will occur without outside intervention. l Rolling a ball down a hill (gravity); steel rusting (?); wood burning (exothermic process); transfer of heat from hot to cold (exothermic); freezing of water (exothermic); Melting of ice (endothermic!!); etc. l What common characteristic derives all those processes to be spontaneous? l It is an increase in a property called Entropy, S. l Thus all spontaneous processes occur as a result of an increase of the S of the universe.

5. What is Entropy? l Entropy, S, is measure of randomness, or disorder in the system. For example molecular randomness. l Naturally, things move from order to disorder! (from lower S to higher S.)

6. Entropy l S defined in terms of probability. l Substances take the most likely arrangement (that have more probabilities). random l The most likely arrangement is the most random. l Can we calculate the number of arrangements for a system?

7. Entropy and Gases l Gas placed in one bulb of a container will spontaneously expand to fill the entire vessel (vessel of two bulbs) evenly. l Probabilities of finding equal number of molecules in each bulb are huge. l Finding molecules in one bulb is highly improbable; thus the process does not occur spontaneously.

8. Entropy and state of matter l Gases completely fill their chamber because there are many more ways to do that than to leave half empty. l Ssolid <Sliquid <<Sgas l In solids, molecules are very close and thus they have relatively few positions available to them l Entropy also describes the number of possible positions of a molecule l There are many more ways for the molecules to be arranged as a liquid than a solid. l Gases have a huge number of positions possible.

9. Example l Which has higher positional S? – Solid CO 2 or gaseous CO 2 ? – N 2 gas at 1 atm or N 2 gas at 0.01 atm? Predict the sign of  S – Solid sugar is added to water + because larger volume – Iodine vapor condenses on cold surface to form crystals - because g  s, less volume

10. Positional Entropy l Entropy also describes the number of possible positions of a molecule l A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system; i.e., the highest positional entropy l Probability depends on the number of configurations in space ( positional microstates)

11. l Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate; there is an increase in entropy l Generally, in any spontaneous process, there is always an increase in the entropy of the universe second law of thermodynamics l This is the second law of thermodynamics ;  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.  S univ > 0 for any spontaneous process l First law: The energy of a universe is constant 16.2 Entropy and the second law of thermodynamics

12. For exothermic processes  S surr is positive For endothermic processes  S surr is negative l Consider this process (endothermic) H 2 O(l)  H 2 O(g)  S sys is positive (change from L to G)  S surr is negative (random motion of atoms in the surrounding decreases) Which one will control the process:  S sys or  S surr ?  S univ; l If the ∆S sys and ∆S surr have different signs, the temperature determines the ∆S univ 16.3 The effect of temperature on spontaneity

13. Determining ∆S surr l Sign of ∆S surr – depends on direction of heat flow – ∆S surr + for exothermic reactions – ∆S surr - for endothermic reactions l Magnitude of ∆S surr – Depends on temperature – Heat flow = ∆H at constant P – Very small at high T, increases as T decreases

14.  S sys  S surr  S univ Spontaneous? + ++ - -- +-? +-? No Yes Yes at Low temp. Yes at High temp. Entropy of  Sys and  S surr in determining the sign of  S univ  S surr = -  H/T

15. Example A process has a  H of +22 kJ and a  S of -13 J/K. At which temperatures is the process spontaneous? – if there is no subscript,  S =  S sys –  S univ > 0 to be spontaneous –  S sys +  S surr > 0

16. Example l For methanol, the enthalpy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 213 J/K. What is the normal boiling point of methanol? –  S sys = 213 J/K and  H = 71.8 kJ/molK = 71,800 J/molK – at the boiling point, the vaporization begins to be spontaneous –  S univ = 0 to be at bp   S sys +  S surr = 0

17. 16.4 Free Energy, G l G: Gibbs free energy l G: helps the determination of the T dependence on spontaneity l G ≡ H - TS definition of G l ∆G ≡ ∆ H - T ∆ S for constant T l All quantities refer to the system. l When no subscript the quantity refers to the system

18. If  G is negative at constant T and P, the process is spontaneous. The process is spontaneous in the direction  G decreases -  G means +  S univ l So what sign would the ∆G of reaction with a + ∆S univ have? – Negative at constant T and P

19. Example Is the following reaction spontaneous at -10°C in the forward direction? l H 2 O(s)  H 2 O(l) – where ∆H°=6.03x10 3 J/mol and ∆S°=22.1 J/mol. K ∆G o = ∆ H o - T ∆ S o

20. l Another way to check for spontaneity of a reaction? – Check to see if the ∆S univ is positive – How can we solve for ∆S surr ? – use ∆H° and T l At -10°C, is it spontaneous? NO!- the reverse is spontaneous

21. Example 2 l At what T is this reaction spontaneous at 1 atm Of pressure? – Br 2 (l)  Br 2 (g) – ∆H o =31.0 kJ/mol, – ∆S o =93.0 J/molK

22.  G=  H-T  S HH SS Spontaneous? + - At all Temperatures + + Spontaneous at high temperatures - - Spontaneous at low temperatures +- Not spontaneous at any temperature, Reverse is spontaneous

23. 16.5 Entropy changes in chemical reactions The Third Law of Thermodynamics l So far we dealt with physical changes l In a chemical reaction, when the number of gaseous molecules increases, the positional disorder will increase and consequently  S would be Positive and Visa Versa l Consider the following examples: N 2 (g) + 3H 2 (g) 2NH 3 (g) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) CaCO 3 (s) CaO (s) + CO 2 (g)

24. l The changes in enthalpy determines the exothermicity and endothermicity at constant P l The changes in entropy determines the spontaneity at constant P and T l Are there values given for S? l The entropy of a pure crystal at 0 K is 0. – Molecular motion is almost zero and only one arrangement is possible. l Thus, the entropy of a perfect crystal is zero; this statement is called the Third Law of Thermodynamics

25. l At T >0, some changes in order will occur, consequently, S >0 l This value gives us a starting point. l Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed.  Sº can then be determined for products and reactants  S  reaction =  n p  S  (products)   n r  S  (reactants) l Entropy is a state function of the system (It is not pathway dependent)

26. l Entropy is an extensive property. It depends on the amount of substance present l Number of moles of reactants and products must be taken into account l What is the expected  S (0, 0r +ve or –ve) for the following reaction? Al 2 O 3 (s) + 3H 2 (g) 2Al(s) + 3H 2 O(g) l More complex molecules possess higher Sº

27. Example l Find the ∆S° at 25°C for: – 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s) 53 205 248 38

28. 16.6 Free Energy and Chemical Reactions  Gº is needed when dealing with chemical reactions  Gº = standard free energy change. l Free energy change that will occur if reactants in their standard state turn to products in their standard state. N 2 (g) + 3H 2 (g) 2NH 3 (g)  Gº= -33.3 kJ  Gº can’t be measured directly, but can be calculated from other measurements.  Gº=  Hº-T  Sº

29. Free Energy in Reactions There are tables of  Gº f. l The standard free energy of formation for any element in its standard state is 0. Why  Gº is useful? To compare relative tendencies of reactions to occur. The more –ve the value of  Gº, the further a reaction will go to the right to reach equilibrium

30. How do we calculate  Gº? There are three Ways to Calculate ∆G° l Use equation: ∆G° ≡ ∆ H ° - T ∆ S ° l Use Hess’s Law – Rearrange equations to get the given equation – Add ∆G° values for the equations l Use ∆G f ° : standard free energy of formation

31. 1. Use the equation:   Gº=  Hº-T  Sº  This equation is applicable for reactions taking place at constant temperature

35. 2. Claculation of  Gº using Hess’s law l Free energy is a state function just like enethalpy, i.e., it depends upon the pathway of the reaction It can be found same as  H using Hess’s law

37. 3. Calculation of  Gº using the standard free energies of formation  Gº f  Gº f of a substance is – the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states  Gº for a specific reaction is obtained from the equation:  G  =  n p  G f  (products)  n r  G f  (reactants)

38. There are tables of  Gº f. l The standard free energy of formation for any element in its standard state is 0. Remember also that the number of moles of each reactant and product must be used in calculating  Gº for a reaction

40. Exercise l Is the following reaction spontaneous under the standard conditions? : C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) – Find  Gº for the reaction using  Gº f values from the table – If  Gº is –Ve, the process is spontaneous and the formation of ethanol is favorable – Of course to judge whether the reaction is feasible or not, we have to study its kinetics to know whether the reaction is fast or slow. – If it is slow adding a catalyst may be considered – Also, remember that  Gº depends on temperature:  G o =  H o -T  S o

41. 16.7 The Dependence of Free Energy on Pressure l At a large volume the gas has many more positions available for its molecules than at low volumes l S large V > S small V l S low P > S high P l Since S depends on P then G will depend on P. It can be shown that: l G = G o + RTln(P) l (G o = free energy at 1 atm)

42. By proper derivation we got:  G =  Gº +RTln(Q) – where Q is the reaction quotient (P of the products /P of the reactants).

45. 16.8 Free energy and equilibrium l According to thermodynamics the equilibrium occurs at the lowest value of free energy available At equilibrium  G = 0, Q = K  G =  Gº +RT ln(Q) Thus,  Gº = -RT lnK

46.  Gº K =0=1 0 >0<0  Gº = -RT lnK

47. Temperature dependence of K  Gº= -RT lnK =  Hº - T  Sº ln(K) = -  Hº/RT +  Sº/R l Plot ln(K) VS 1/T l A straight line gives a: Slope =  Hº/R Intercept =  Sº/R

50. Example

53. 16.9 Free energy and Work l Free energy is that energy free to do work. l The maximum amount of work possible at a given temperature and pressure. l 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.

54. Reversible v. Irreversible Processes l Reversible: The universe is exactly the same as it was before the cyclic process. l Irreversible: The universe is different after the cyclic process. l All real processes are irreversible -- (some work is changed to heat)