1. A spontaneous reaction (or favourable change) is a change that has a natural tendency to happen under certain conditions. Eg. The oxidation of iron (rust) First law of Thermodynamics: The amount of energy in the universe is constant. Energy can be neither created nor destroyed, but can be transferred from one object or place to another, or transformed from one form to another. In a chemical reaction, changes in the potential energy of reactants and products result in the transfer of energy, as heat, from the system to the surroundings (exothermic change) or the surroundings to the chemical system (endothermic change). The total energy of the system remains constant. THERMODYNAMICS AND EQUILIBRIUM

2. Enthalpy Changes and Spontaneity When products have less enthalpy than reactants, the reaction releases energy, therefore Reaction is exothermic. Enthalpy H reactant product ΔH (-) ΔH (+) exothermicendothermic (fwd direction usually favoured)(rev direction usually favoured)

3. In general, exothermic reactions tend to proceed spontaneously and endothermic reactions tend to be non-spontaneous, occurring only when energy is supplied. However, some endothermic reactions ARE spontaneous even though the products are less energetically stable than the reactants. Why?? The answer is because of Entropy, S. Entropy is the measure of the randomness or disorder in a system.

4. Second law of Thermodynamics: Total entropy of the universe is increasing. Favourable changes involve an increase in the total entropy In chemical reactions, entropy increases when: 1. The physical state of a system changes (s  l  aq  g) 2. Fewer moles of reactant molecules form a greater number of moles of product molecules. (Eg2 NH 3(g)  N 2(g) + 3H 2(g) ) (less mol  more mol)  volume of container increases (gases only), temp increases, complex substances are broken down into simpler substances.

5. Ex)Determine whether the entropy, S, is increasing (+) or decreasing (-) a)H 2 O (s) → H 2 O (l) change of state (s)  (l), increase in entropy S = + b)2 NH 3(g) → N 2(g) + 3 H 2(g) 2 mol  4 mol, increase in moles of gas, increase in entropy S = + c)C 2 H 4(g) + H 2(g) → C 2 H 6(g) 2 mol  1 mol, decrease in moles of gas, decrease in entropy S = - d) H 2 O 2(g) →H 2 O 2(l) change of state (g)  (l), decrease in entropy S = -

6. How do we know if a Reaction is spontaneous or non- spontaneous? GIBBS FREE ENERGY, G The spontaneity of any reaction can be predicted by calculating Gibbs Free Energy - the energy available to do work  G  =  H  - T  S  where  G  = standard Gibbs free energy  H  = standard enthalpy change T = temperature (Kelvins)  S  = standard entropy change  G  < 0 change is spontaneous  G  > 0change is non-spontaneous

7. -  H and +  S (exothermic,  entropy or disorder) Reaction spontaneous in the forward direction +  H and -  S (endothermic,  entropy or more ordered) Reaction spontaneous in the reverse direction -  H and -  S Fwd rxn favourable at low temperatures +  H and +  S Fwd rxn favourable at high temperatures

8. Note:  if  G <0, the reaction is spontaneous in the forward direction  if  G >0, the reaction is non-spontaneous in the forward direction but IS spontaneous in the reverse direction  if  G = 0, there is no “preferred” direction. The reaction is occurring in the forward and reverse direction at the same rate, therefore the system is at EQUILIBRIUM!

9. Ex)For the following reaction, CaCO 3(s)  CaO (s) + CO 2(g) a) Determine whether the reaction is spontaneous if it occurs at 25  C, and the value of  H is determined to be 10.5 kJ and the  S is 30 J/K. b) Determine the temperature at which a reaction, whose  H = 126 kJ, and  S = 84 J/K, will be spontaneous. a)T = 25°C = 298 K ΔH = 10.5 kJ = 10500 J ΔS = 30 J/K ΔG = ΔH - TΔS = 10500 J – (298)(30 J/K) = 1560 J not spontaneous b) ΔG= ΔH - TΔS 0= 126000 J – T(84 J/K) T(84 J/K)= 126000 J T= 1500 Kspontaneous when T > 1500 K

10. HOMEWORK WS “Equilibrium Worksheet” WS “Thermodynamics and Equilibrium” P 327 #2-5 p333 #3-5