3. A science that includes the study of energy transformations and the relationships among the physical properties of substances which are affected by these transformations.

4. Zeroth Law - If two thermodynamic systems are in thermal equilibrium with a third, they are also in thermal equilibrium with each other. A ⇔ B and B ⇔ C implies A ⇔ C. - The increase in the internal energy of a system can be achieved in many ways, among which is by heating the system. “Energy can neither be created nor destroyed" or The energy of the universe is constant First Law Second Law - the fact that over time, ignoring the effects of self-gravity, differences in temperature, pressure, and density tend to even out in a physical system that is isolated from the outside world. Entropy is a measure of how far along this evening- out process has progressed. - as a system approaches absolute zero, all processes cease and the entropy of the system approaches a minimum value. This is based on the assumption that the entropy of a perfect crystal is zero at 0 K. Third Law

5. 1. Spontaneous - Processes that occur without outside intervention Spontaneous processes may be fast or slow Many forms of combustion are fast Conversion of diamond to graphite is slow 2. Reversible a process or cycle such that the net change at each stage in the combined entropy of the system and its surroundings is zero. 3. Irrevesible there will be a certain amount of heat energy loss or dissipation due to intermolecular friction and collisions; energy that will not be recoverable if the process is reversed.

6. Entropy (S)  A measure of the randomness or disorder  The driving force for a spontaneous process is an increase in the entropy of the universe  Entropy is a thermodynamic function describing the number of arrangements that are available to a system  Nature proceeds toward the states that have the highest probabilities of existing

7. Positional Entropy  The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved S solid < S liquid << S gas

8. Second Law of Thermodynamics  "In any spontaneous process there is always an increase in the entropy of the universe"  "The entropy of the universe is increasing"  For a given change to be spontaneous,  S universe must be positive  S univ =  S sys +  S surr

9. Things in the world tend toward lowest energy (-ΔH) and also tend toward greatest disorder (+ ΔS). More disorder can be recognized as: · greater # of moles of gas formed · gas > liquid > solid and (aq) > (s) · greater # of particles · greater volume formed examples: N 2 (g) + 3H 2 (g) → 2NH 3 (g) NaCl (s) → Na + + Cl - NaCl (s) → NaCl (l) Note: ΔS° can be calculated using Hess’s Law, but S° of elements is NOT 0. Also, S° is often reported in J/mol×K and cal/mol×K, not kJ and kcal… watch your units!

10. Do problem 22!!!

11. Calculating Entropy Change in a Reaction  Entropy is an extensive property (a function of the number of moles)  Generally, the more complex the molecule, the higher the standard entropy value

12. Do problem 38!!!

13. Standard Free Energy Change   G 0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states   G 0 cannot be measured directly  The more negative the value for  G 0, the farther to the right the reaction will proceed in order to achieve equilibrium  Equilibrium is the lowest possible free energy position for a reaction

14. G=H-TS Never used this way.  ΔG= ΔH-TΔS at constant temperature Divide by -T - ΔG/T = - ΔH/T- ΔS - ΔG/T = ΔS surr + ΔS - ΔG/T = ΔS univ If ΔG is negative at constant T and P, the Process is spontaneous.

15. For reactions at constant temperature:  G 0 =  H 0 - T  S 0 Calculating Free Energy Method #1 Gibb’s-Helmholtz equation:

16. Let’s Check For the reaction H 2 O(s) → H 2 O(l)  Sº = 22.1 J/K mol  Hº =6030 J/mol Calculate  G at 10ºC and -10ºC Look at the equation  G=  H-T  S Spontaneity can be predicted from the sign of  H and  S.

17.  H,  S,  G and Spontaneity Value of  H Value of T  S Value of  G Spontaneity Negative Positive Negative Positive  G =  H - T  S H is enthalpy, T is Kelvin temperature NegativeSpontaneous Positive Nonspontaneous ??? Spontaneous if the absolute value of  H is greater than the absolute value of T  S (low temperature) Spontaneous if the absolute value of T  S is greater than the absolute value of  H (high temperature)

18. Calculating Free Energy: Method #2 An adaptation of Hess's Law: C diamond (s) + O 2 (g)  CO 2 (g)  G 0 = -397 kJ C graphite (s) + O 2 (g)  CO 2 (g)  G 0 = -394 kJ CO 2 (g)  C graphite (s) + O 2 (g)  G 0 = +394 kJ C diamond (s)  C graphite (s)  G 0 = C diamond (s) + O 2 (g)  CO 2 (g)  G 0 = -397 kJ -3 kJ

19. Calculating Free Energy Method #3 Using standard free energy of formation (  G f 0 ):  G f 0 of an element in its standard state is zero

20. Do problem 45a!!!

21. The Dependence of Free Energy on Pressure  Enthalpy, H, is not pressure dependent  Entropy, S  entropy depends on volume, so it also depends on pressure S large volume > S small volume S low pressure > S high pressure

22. Free Energy and Equilibrium  Equilibrium point occurs at the lowest value of free energy available to the reaction system  At equilibrium,  G = 0 and Q = K G0G0 K  G 0 = 0K = 1  G 0 < 0K > 1  G 0 > 0K < 1

23. Temperature Dependence of K So, ln(K)  1/T

24. Do problem 63!!!

25. Nonstandard Free Energy

26. Do problem 75!!!