2. * Studying energy flow in chemical changes allows us to predict what is possible and what is not. * 1 st Law of Thermodynamics PE tends only to decrease spontaneously. Since ΔH is primarily ΔPE, then any chemical system that is exothermic (ΔH = -ve) should be spontaneous These reactions are exothermic and spontaneous. Spontaneous – given the necessary activation energy to begin, a reaction occurs without continuous outside assistance, proceeding to completion in open systems.

4. * Entropy (S) is… * Randomness/disorder * All substances exist in a various degrees of organization. * Eg. Solids, liquids & gases * As temperature INCREASES… entropy INCREASES as well Substances will become as random or disorganized as possible.

5. SUBSTANCES° (J/mol. K) Water (gas)188.84 Water (liquid)69.95 Table 1: Entropy Values for Water in Different States ***Water is more RANDOM in the gaseous state, therefore S°value is HIGHER for steam*** The probability that a given substance will exist in a specified state. The HIGHER the S°value, the GREATER the probability of existence.

6. * The state of order within a chemical system. A crystal at absolute zero (O K) would have S=0 J/mol K. Everything above 0 K has a certain amount of chaos within it hence has S>0 J/mol K **Very dependent on the temperature of the system**

7. Physical Changes: In general, a system will increase Δ S if: * The volume of the gaseous system increases * The temperature of a system increases * The physical state of a system changes from solid to liquid or gas. (S gas >S liquid >S solid ) Chemical Changes: In general, a system will increase Δ S if: * Fewer moles of reactant molecules form a greater number of moles of product molecules eg. 2H 2 O 2 (l)  2H 2 O (l) + O 2 (g) * Complex molecules are broken down into simpler subunits eg. 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O (l) * Solid reactants become liquid or gaseous products

8. * Predict whether the change in entropy is positive or negative for each of the following changes and predict their spontaneity. Explain your answers. a. Solid carbon dioxide sublimes into gaseous carbon dioxide. b. N 2 O 4 (g)  2 NO 2 (g) c. The synthesis reaction between oxygen and hydrogen forms liquid water.

9. * Done the same way ΔH was calculated so… * Special Notes: 1. Elements are NOT assigned an S°= 0 2. All matter above 0 Kelvin has some entropy

10. * Determine the change in entropy for:

11. * Discuss with a partner why entropy went down in this system and relate to our notes or other examples. 2 Ag (s) + 0.5 O 2 (g)  Ag 2 O (s) ΔS = -66.4 J/mol K

12. * 2 nd Law of Thermodynamics: the entropy in the universe tends only to increase spontaneously. So if ΔS = +ve the driving force is to the product side. The 2 nd law of thermodynamics governs chemical behaviour Possibilities of both FORCES driving chemical rxns: ΔH=-ve/ ΔS=+ve(rxn spontaneous at all temperatures) ΔH=+ve/ ΔS=-ve(rxn cannot be spontaneous at any temp) ΔH=-ve/ ΔS=-ve(rxn tends to be spontaneous at low temp) ΔH=+ve/ ΔS=+ve(rxn tends to be spontaneous at high temp Must use Gibbs

13. Gibbs Free Energy ( ΔG) * The maximum total energy in a chemical change that is ‘free’ to do useful work. * The sum of the energies from a change of enthalpy and the change in entropy. - ΔG is dependant on the 2 driving forces ΔH & ΔS ΔG = ΔH - T ΔS at any temp T(in K)

14. Remember… FAVOURABLE: ΔH = -ve (exothermic) ΔS = +ve (more disorder) ThereforeΔG = -ve UNFAVOURABLE: ΔH = +ve(endothermic) ΔS = -ve (more order) Therefore ΔG = +ve Spontaneous Non- Spontaneous

15. * For the reaction: NH 4 Cl (s)  NH 3 (g) + HCl (g) a. Determine ΔH°and indicate if H favours spontaneity. -find the standard enthalpy values from the textbook: So ΔH r ° Therefore ΔH r °= +ve (reaction needs heat) and enthalpy increases, which does NOT favour reaction.

16. b. Determine ΔS°and indicate if S favours spontaneity -find the ΔS°values from the textbook:

17. * c. Determine ΔG°and indicate if G favours spontaneity -since °, SATP, which means 25°C So… ΔG = ΔH – T ΔS = 176.2 kJ/mol-298.15K(0.28508 kJ/mol K) = 176.2 kJ/mol-85.00 kJ/mol = + 91.2 kJ/mol

18. * At what temperature would it become spontaneous? Calculate the Temp when this happens: > 618.1 K = 345 degrees Celcius