1. 17.4-17.5 Heat Transfer and Change in Entropy – Gibbs Free Energy

2. Heat Transfer and Changes in Entropy of the Surroundings In the last PPT it was mentioned that in order to determine if a process is thermodynamically favored depends on the change in entropy of the universe. So what does this mean? The entropy change of the universe is dependent on two things: 1.) the change in entropy in the system, compared to 2.) the change in entropy in the surroundings. ΔS univ = ΔS sys + ΔS surr This is the reason that the freezing of water below zero degrees Celsius is spontaneous. Even though the entropy change in the water (the system) is negative because order is increasing, the release of heat energy by the system disperses that energy into the surroundings, increasing the entropy of the surroundings. The change in entropy of the surroundings is however, temperature dependent. The freezing of water is only spontaneous at low temperatures, because the heat give off drastically increases the entropy of the surroundings. However at high temps., the increase in entropy of t he surroundings is minimal, and the negative entropy change of the system far outweighs the positive entropy change of the surroundings.

3. Quantifying Entropy Changes in the Surroundings When pressure is constant, we can use q sys to quantify the change in entropy of the surroundings (ΔS surr ). A process that emits heat into the surroundings (q sys negative) increases the entropy of the surroundings (positive ΔS surr ). A process that absorbs heat from the surroundings (qsys positive) decreases the entropy of the surroundings (negative ΔS surr ). The magnitude of the change in entropy of the surroundings is proportional to the magnitude of q sys. Under conditions of constant pressure q sys = ΔH sys The following formula can be used to quantify the enthalpy change of the surroundings at constant temp. and pressure : - ΔH sys ΔS surr = ----------- T

4. Let’s Try a Practice Problem! Consider the reaction between nitrogen and oxygen gas to form dinitrogen monoxide: 2N 2 (g) + O 2 (g)  2N 2 O(g) ΔH rxn = +163.2 kJ a.) Calculate the entropy change in the surroundings associated with this reaction occurring at 25 o C. (The units for entropy are J/K) - ΔH sys -(163,200 J) ΔS surr = ----------- = ----------------- = -548 J/K T (25 + 273 K) b.) Determine the sign of entropy change of the system. (Hint: In order to do this, determine the number of moles of gas on both sides of the reaction. If the moles of gas increases from reactants to products, the entropy of the system is positive) ΔS surr is negative (3 mol gas on reactant side, and 2 mol gas on product side) c.) Determine the sign of the entropy change of the universe. Will the reaction be thermodynamically favored spontaneous)? ΔS univ = ΔS sys + ΔS surr = since the entropy change of both the system and surroundings are negative, so will the entropy change of the universe. The reaction will not occur spontaneously.

5. Gibb’s Free Energy If we combine the previous two equations, the following equation (which is on your reference table) can be derived: ΔG = ΔH-TΔS Here (G) represents Gibb’s Free Energy: (SI units J/mol) is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system (one that can exchange heat and work with its surroundings, but not matter) H is enthalpy of the system (kJ/mol)  (most of the time mol is dropped as it is assumed), T is temperature in K, and S is entropy of the system (J/K).

6. Summarizing Gibb’s Free Energy ΔG is proportional to the negative ΔS univ. A decrease in Gibbs free energy (ΔG < 0) corresponds to a thermodynamically favored process. An increase in Gibb’s free energy (ΔG > 0) corresponds to a nonspontaneous process.

7. The Effect of ΔH, ΔS and ΔT on spontaneity Case 1: If ΔH 0 for a reaction, then ΔG is negative at all temperatures and the reaction is thermodynamically favored at all temperatures. Case 2: If ΔH > 0, and ΔS < 0 for a reaction, then ΔG is positive at all temperatures and the reaction is nonspontaneous at all temperatures. Case 3: If both enthalpy and entropy are negative, then the sign of the change in free energy depends on temperature. At low temperatures, the reaction would be thermodynamically favored, but nonspontaneous at high temperatures. Case 4: If both the enthalpy and entropy are positive, then here again, the sign of the change in free energy depends on temperature…except here, the reactions will be thermodynamically favored at high temperatures, and nonspontaneous at low temperatures.

8. Let’s Try a Practice Problem! Consider the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) ΔH = -137.5 kJ (  remember, this value is a molar quantity) ΔS = -120.5 J/K Calculate the ΔG at 25 o C and determine whether the reaction is spontaneous. Does ΔG become more negative or more positive as the temperature increases? ΔG = ΔH-TΔS ΔG = (-137,500 J/mol) – ((298K)(-120.5 J/K) = -1.02x10 5 J/mol Since ΔG is negative, the reaction is spontaneous. Due to the fact that both enthalpy and entropy are negative, ΔG becomes more positive at high temperatures.

9. Let’s Try Another!!! Which Statement is true regarding the sublimation of dry ice (Solid CO 2 )? a.) ΔH is positive, ΔS is positive, and ΔG is positive at low temperature and negative at high temperature. b.) ΔH is negative, ΔS is negative, and ΔG is negative at low temperature and positive at high temperature. c.) ΔH is negative, ΔS is positive, and ΔG is negative at all temperatures. d.) ΔH is positive, ΔS is negative, and ΔG is positive at all temperatures temperature. a.) ΔH is positive, ΔS is positive, and ΔG is positive at low temperature and negative at high temperature.

10. 17.4-17.5 pgs. 853-855 #’s 42, 44, 80 Read 17.6-17.7 pgs. 832-842