AP Chemistry CHAPTER 17 Thermodynamics

Spontaneous process (“Thermodynamically favored”) -occurs without outside intervention -may be fast or slow

Entropy, S - a measure of randomness or disorder -an increase in the number of “microstates” -associated with probability (There are more ways for something to be disorganized than organized.)

*Entropy increases going from a solid to a liquid to a gas. *Entropy usually increases when solutions are formed. *Entropy usually increases in a reactions when more atoms or molecules are formed. *Entropy increases with increasing temperature

2nd law of thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. The energy of the universe is constant but the entropy of the universe is increasing.

Free energy, G  G =  H -T  S Free energy change is a measure of the spontaneity (thermodynamic favorability) of a reaction. It is the maximum work available from the system.

A spontaneous (thermodynamically favored) reaction carried out at constant temperature and pressure has a negative  G. For example, when ice melts,  H is positive (endothermic),  S is positive and  G = 0 at 0 o C.

If:  S positive and  H negative = spontaneous at all temps  S positive and  H positive = spontaneous at high temps  S negative and  H negative = spontaneous at low temps  S negative and  H positive = nonspontaneous

Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero.  S o reaction =  S o prod -  S o react

Ex. Given the following standard molar entropies, calculate  S o for the reaction: 2Al(s) + 3MgO(s)  3Mg(s) + Al 2 O 3 (s) Al(s) = 28.0 J/K MgO(s) = 27.0 J/K Mg(s) = 33.0 J/K Al 2 O 3 (s) = 51.0 J/K  S o = [3(33.0)+51.0] –[2(28.0) + 3(27.0)] = 13.0 J/K. mol rxn

 G o = standard free energy change *change in free energy that occurs if the reactants in their standard states are converted to products in their standard states  G o =  G f o prod    G f o react  G f o for a free element in its standard state is zero.

Ex. Given the equation N 2 O 4 (g)  2NO 2 (g) and the following data, calculate  G o. G o f for N 2 O 4 (g) = kJ/mol G o f for NO 2 (g) = kJ/mol  G o = [2(51.30)] – [ 97.82] = 4.78 kJ/mol rxn

 G o =  H o  T  S o (***When working this, change S from J to kJ***) This is called the Gibbs-Helmholtz equation.

 H o = [2(  100.0) + 3(0)]  [2(  218.0) + 2(  297.0)] = kJ/mol rxn  S o = [2(91.0) + 3(205.0)] – [2(70.0)+2(248.0)] = J/Kmol rxn G= H  T SG= H  T S  G o =  298(0.1610) = 782 kJ/mol rxn Ex. For the given reaction and the following information, calculate  G o at 25 o C. 2PbO(s) + 2SO 2 (g)  2PbS(s) + 3O 2 (g)  H o (kJ/mol)  S o (J/molK) PbO(s)  SO 2 (g)  PbS(s)  O 2 (g)

Remember:  H o =   H f o prod    H f o react You can "add" equations similar to Hess' law method of  H.

The equilibrium point occurs at the lowest value of free energy available to the reaction system.  G = G prod -G react = 0

Ex. Given for the reaction Hg(l)  Hg(g) that  H o = 61.3 kJ/mol and  S o = J/molK, calculate the normal boiling point of Hg.  G o =  H o  T  S o  G = 0 at phase change 0 = 61.3 –T(0.1000) T = 613K or 340 o C

 G o =  RT ln K

When  G o = 0, free energy of reactants and products are equal when all components are in their standard states. During a phase change,  G = 0.

Ex. Calculate the approximate standard free energy for the ionization of hydrofluoric acid, HF (K a = 1.0 × 10  3 ), at 25 o C.  G o =  RT ln K  G o =  8.314(298)ln 1.0 × 10  3 = 1.7 × 10 4 J/mol or 17 kJ/mol

External sources of energy can be used to drive reactions with positive ∆G values. 1. Electricity may be used to cause a reaction to occur through electrolysis. We will look at this in our electrochemistry unit (Ch 18). 2. Light may be used (photoionization or absorption of photons (as in photosynthesis)) 3. The reaction may be couple with another that has a very negative ∆G value, as in the conversion of ADP to ATP.