1. Free Energy and Thermodynamics 17.4-17.5 By Emily Entner

2. Heat Transfer and Changes in the Entropy of the Surroundings We know an increase in the entropy of the universe causes spontaneity But there are many spontaneous processes in which entropy seems to decrease Example: When water freezes at temperatures below 0 C, the entropy of the water decreases but it’s still a spontaneous process Why are these spontaneous? We must go back to the second law: for any spontaneous process, the entropy of the universe increases (S univ > 0) So even though entropy of water decreases during freezing, the entropy of the universe must somehow increase in order for the process to be spontaneous

3. It’s useful to distinguish the system from its surroundings Example: The freezing water is our system and the surroundings are the rest of the universe. So, S sys is the entropy change for the water, S surr is the entropy change for the surroundings, and S univ is the entropy change for the universe. Entropy change for the universe is the sum of entropy changes for the system and the surroundings S univ = S sys + S surr Entropy of the system can decrease (S sys -S sys ): overall entropy of the universe undergoes net increase

4. For liquid water freezing we know change in entropy of the system (S sys ) is negative because the water becomes more orderly For the S univ to be positive, the S surr must be positive and greater in absolute value (or magnitude) than S sys as shown graphically here: Why does freezing increase the entropy? Its an exothermic process: gives off heat to the surroundings The release of heat energy by the system disperses that energy into the surroundings, increasing the entropy of the surroundings

5. The Temperature Dependence of S surr The greater the temperature, the smaller the increase in entropy for a given amount of energy dispersed into the surroundings. Entropy is a measure of energy dispersal (joules) per unit temperature (kelvins) The higher the temperature, the lower the amount of entropy for a given amount of energy dispersed

6. We can understand temperature dependence with a simple analogy: Say you have $1000 to give away. If you give the money to a rich man, it would have little impact on his networth (because he already has a ton of money). But, if you gave the $1000 to a poor man, his networth would change substantially (because he has so little money). The same is true if you disperse 1000 J of energy into surroundings that are hot, the entropy increase is negligible. If you disperse the same 1000 J of energy into surroundings that are cold, the entropy increase is large. S univ = S sys + S surr Negative Positive and large at low temp Positive and small at high temp

7. At low temps, the decrease in entropy is overcome by large increase in entropy of surroundings, resulting in pos S univ + spontaneous process At high temps, the decrease in entropy of the system is not overcome by the increase in entropy of the surroundings, resulting in neg S univ therefore it isn’t spontaneous at high temperatures

8. Quantifying Entropy Changes in the Surroundings We have seen that when a system exchanges heat with its surroundings, it changes the entropy of the surroundings. At constant pressure, we can use q sys to quantify the change in entropy of the surroundings (S surr ) Process that emits heat into surroundings (-q sys ) increases the entropy of the surroundings (+S surr ) Process that absorbs heat from the surroundings (+q sys ) decreases the entropy of the surroundings (-S surr ) Magnitude of the change in entropy of the surroundings is proportional to the magnitude of q sys

9. This equation allows us to see why exothermic processes have a tendency to be spontaneous at low temperatures (increase entropy of the surroundings) But as temperature increases, a given negative q produces smaller positive S surr So, exothermicity becomes less of a determining factor for spontaneity as temperature increases Under constant pressure q sys = H sys and we are given the formula: S surr = -H sys /T

10. Practice Problem- Calculating Entropy Changes in the Surroundings Consider the combustion of propane gas: C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O (g) H rxn = -2044 kJ (a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25 °C (Remember T is in kelvins!) (b) Determine the sign of the entropy change for the system (c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous?

11. Practice Problem- continued (a) T = 273 + 25 =298 K S surr = -H rxn /T = -(-2044kJ)/298 K = + 6.86 kJ/K = +6.86 x 10 3 J/K (a) C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O (g) 6 mol gas7 mol gas Increase in # of moles of gas implies positive S sys = positive (a) S univ = S sys + S surr Therefore, S univ is positive and the reaction is spontaneous.

12. Gibbs Free Energy Recall: for any process the entropy change of the universe is the sum of the entropy change of system and the entropy change of the surroundings: Looking at the thermodynamic function Gibbs free energy: H is change in enthalpy, T is the temperature (K), and S is entropy, and the change in Gibbs free energy symbolized by G is expressed as: G = H -TS This is on the reference table.

13. Summary of Gibbs Free Energy G is proportional to the negative S univ A decrease in Gibbs free energy (G>0) corresponds to a spontaneous process An increase in Gibbs free energy (G<0) corresponds to a nonspontaneous process Also called chemical potential because it determines direction of spontaneous change for chemical systems.

14. We can calculate changes in Gibbs free energy solely with reference to the system To determine whether a process is spontaneous, we only have to find the change in entropy for the system (S) and the change in enthalpy for the system (H) We can predict spontaneity of the process at any temperature In chapter 6, we learned how to calculate changes in enthalpy (H) and in section 17.6 we will learn to calculate changes in entropy (S) for chemical reactions We can use those quantities to calculate changes in free energy and predict their spontaneity Before that, we will examine some examples that demonstrate how H, S, and T affect spontaneity of chemical processes

15. Effect of H, S, and T on Spontaneity H Negative, S Positive If reaction is exothermic (H 0), then change in free energy is negative at all temperatures Example: dissociation of N 2 O: 2 N 2 0(g) → 2 N 2 (g) + O 2 (g) H° rxn = -163.2 kJ 2 mol gas 3 mol gas Change in enthalpy is negative, heat is emitted, increasing entropy of the surroundings. Change in entropy for reaction is positive (entropy increases). We can see change in entropy is positive because number of moles of gas increases. Since entropy of both system and surroundings increases, entropy of universe must also increase, making the reaction spontaneous at all temperatures.

16. H negative, S negative If reaction is exothermic (H<0) and if the change in entropy is negative (S<0) then the sign of the change in free energy depends on the temperature. Reaction is spontaneous at low temperatures and nonspontaneous at high temperatures. Example: H 2 0(l) → H 2 O(s) H° = -6.01 kJ Change in enthalpy is negative (heat is emitted), increasing entropy of surroundings. Change in entropy is negative (entropy decreases) we can see because (l) → (s). Unlike the ones before, the changes in entropy have the opposite sign. So, sign of change in free energy depends on relative magnitudes of the two changes. At a low temperature, heat emitted into surroundings causes large entropy change in surroundings (spontaneous). At a high temperature, same amount of heat is dispersed into warmer surroundings so positive entropy change in surroundings is smaller (nonspontaneous)

17. The Effect of H and S, and T on Spontaneity Note: When H and S are opposite signs, spontaneity doesn’t depend on temperature. When H and S have the same sign, spontaneity depends on temperature. Temperature at which reaction changes from being spontaneous to being nonspontaneous is the temperature at which G changes sign (found by setting G= 0 and solving for T) Page 831:

18. Practice Problem! Consider the reaction for decomposition of carbon tetrachloride gas: CCl 4 (g) → C(s, graphite) + 2 CL 2 (g) H = +95.7 kJ; S = +142.2 J/K (a) Calculate G at 25 ° C and determine whether the reaction is spontaneous. (Formula on reference table) (b)If the reaction is not spontaneous at 25 ° C, determine at what temperature (if any) the reaction becomes spontaneous.

19. T = 273 + 25 = 298 G = H - TS = 95.7 x 10 3 J - (298 K)142.2 J/K = 95.7 x 10 3 J - 42.4 x 10 3 J = +53.3 x 10 3 J not spontaneous G = H - T S 0 = 95.7 x 10 3 J - (T)142.2 J/K T = (95.7 x 10 3 J)/ (142.2 J/K) = 673 K

20. Course Guideline