1. Entropy Changes in Chemical Reactions.  Because entropy is a state function, the property is what it is regardless of pathway, the entropy change for a given reaction can be calculated by taking the difference between the standard entropy values of products and those of the reactants.   S o reaction =  n p  S o products -  n r  S o reactants

2. Entropy Changes in Chemical Reactions.  Calculating  S o.  Calculate  S o at 25 o C for the reaction 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s)

3. Entropy Changes II Calculate  S o for the reaction of aluminum oxide by hydrogen gas: Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g)

4. Free Energy and Chemical Reactions  Standard free energy (  G o ) is the change in the free energy that will occur if the reactants in their standard states are converted to the products in their standard states.  The value of  G o tells nothing about the rate of a reaction, only its eventual equilibrium position.

5. Free Energy and Chemical Reactions  Calculating  G o as a State Function.   G o =  H o - T  S o  Consider the reaction 2 SO 2 (g) + O 2 (g)  2SO 3 (g)  carried out at 25 o C and 1 atm. Calculate  H o and  S o, then calculate  G o.

6. Free Energy and Chemical Reactions  Calculating  G o as a State Function.  Solving  G o Using Hess’s Law.  Using the following data (at 25 o C)  C diamond (s) + O 2 (g)  CO 2 (g)  G o = -397 kJ  C graphite (s) + O 2 (g)  CO 2 (g)  G o = -394 kJ  Calculate  G o for the reaction C diamond (s)  C graphite (s)

7. Free Energy and Chemical Reactions  Calculating  G o as a State Function.  Standard Free Energy of Formation (  G f o ).   G o =  n p  G f o products -  n r  G f o reactants  Methanol is a high-octane fuel used in high- performance racing engines. Calculate  G o for the reaction 2CH 3 OH(g) + 3O 2 (g)  2CO 2 (g) + 4H 2 O(g)

8. Free Energy and Chemical Reactions  A chemical engineer wants to determine the feasibility of making ethanol (C 2 H 5 OH) by reacting water with the ethylene (C 2 H 4 ) according to the equation C 2 H 4 (g) + H 2 O(l)  C 2 H 5 OH(l) Is the reaction spontaneous under standard conditions?

9. The Dependence of Free Energy on Pressure  The equilibrium position represents the lowest free energy value available to a particular reaction.  Free energy changes throughout the course of a reaction because it is pressure and concentration dependent.  For any 1 mole of a gas at a given temperature  S large V > S small V or S low P > S high P

10. Free Energy and Equilibrium  For standard conditions at equilibrium   G o = -RT ln(K)  This is on the equations sheet   G o is the free energy at standard conditions, T is the Kelvin temperature, R is the ideal gas constant 8.31 J/mol K, and K is an equilibrium constant ( like we found in the previous unit  Note- R gives you a value in J, not kJ

11. Equilibrium and Free Energy  The reaction for the synthesis of ammonia is:  N 2 + 3 H 2  2 NH 3  Calculate the equilibrium constant for this reaction at 25 o C. H f o (kJ/mol)S o (J/mol) N2N2 0192 H2H2 0131 NH 3 -46193

12. Free Energy and Equilibrium  Equilibrium (LeChâtelier’s) is about reactions being spontaneous in one direction, then switching to the other direction in certain conditions (switching the sign of  G).  For standard conditions at equilibrium   G o = -RT ln(K)  For nonstandard conditions NOT at equilibrium,   G =  G o + RT ln(Q)  The difference between  G,  G o is the 2 nd one is at standard conditions. The first is not.

13. The Dependence of Free Energy on Pressure  One method for synthesizing methanol (CH 3 OH) involves reacting carbon monoxide and hydrogen gases:  CO(g) + 2H 2 (g)  CH 3 OH(l)  Calculate  G at 25 o C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.

14. The Meaning of  G for a Chemical Reaction  A system achieves the lowest magnitude of free energy possible by going to equilibrium, not by going to completion.  Spontaneous reactions do not go to completion, shift all the way over to products from reactants, but instead to equilibrium  At equilibrium  G= O

15. Free Energy and Equilibrium  The equilibrium point occurs at the lowest value of free energy available to the reaction.  Recall the following:  If K > Q, then the reaction as written proceeds to the right.  If K < Q, then the reaction as written proceeds to the left.  If K = Q, then the reaction as written is at equilibrium, and there is no net reaction in either direction.

16. Another derivation of the same equation   G o = -RT ln(K)  In the learning objectives for the AP test they mention an equation determined by solving the above for K.  In which case, you get:  K = e   G / (RT)  This derivation is not on the equations sheet, you are not likely to need it.

17. Free Energy and Work  For a spontaneous reaction,  G is the maximum work obtainable from the system. w max =  G  For a nonspontaneous process,  G is the minimum work that must be done to the system to make a change happen.  For  G sys =  H sys - T  S sys,  G sys is the portion of the total energy change that does the work. T  S sys is given off as heat and is not usable.

18. Work  In the real world, some free energy is always changed to heat and is unusable, or wasted. Therefore, no process is 100% efficient.  1. A spontaneous reaction will occur and can do work on the surroundings.  2. A nonspontaneous reaction will not occur unless the surroundings do work on it.  3. A reaction at equilibrium can no longer do work.

19. Review   H Enthalpy; negative is exothermic, positive is endothermic   S Entropy (Do NOT use spontaneity to determine); look for complexity of molecules or phases. Positive is less complex, negative is more complex   G Gibbs Free Energy; negative means the equation is spontaneous, positive is not.   G =  H - T  S

20. Questions from a previous test

21. Free Response