1. Chemistry 100 Chapter 19 Spontaneity of Chemical and Physical Processes: Thermodynamics

2. What Is Thermodynamics? Study of the energy changes that accompany chemical and physical processes. Based on a set of laws. In chemistry, a primary application of thermodynamics is as a tool to predict the spontaneous directions of a chemical reaction.

3. What Is Spontaneity? Spontaneity refers to the ability of a process to occur on its own! Waterfalls “Though the course may change sometimes, rivers always reach the sea” Page/Plant ‘Ten Years Gone’. Ice melts at room temperature!

4. The First Law of Thermodynamics The First Law deals with the conservation of energy changes.  E = q + w The First Law tells us nothing about the spontaneous direction of a process.

5. Entropy and Spontaneity Need to examine ◦ the entropy change of the process as well as its enthalpy change (heat flow). Entropy – the degree of randomness of a system. ◦ Solids – highly ordered  low entropy. ◦ Gases – very disordered  high entropy. ◦ Liquids – entropy is variable between that of a solid and a gas.

6. Entropy Is a State Variable Changes in entropy are state functions  S = S f – S i S f = the entropy of the final state S i = the entropy of the initial state

7. Entropy Changes for Different Processes  S > 0 entropy increases (melting ice or making steam)  S < 0 entropy decreases (examples freezing water or condensing steam)

8. The Solution Process For the dissolution of NaCl (s) in water NaCl (s)  Na + (aq) + Cl - (aq) Highly ordered – low entropy Disordered or random state – high entropy The formation of a solution is always accompanied by an increase in the entropy of the system!

9. The Entropy Change in a Chemical Reaction Burning ethane! C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O (l) The entropy change ◦ r S    n p S  (products) -  n r S  (reactants) ◦ n p and n r represent the number of moles of products and reactants, respectively.

10. The Entropy Change (Cont’d) For the ethane combustion reaction 1 C 2 H 6 (g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l)  r S    n p S  (products) -  n r S  (reactants) = 3 S  [H 2 O (l)] + 2 S  [CO 2 (g)] - (7/2 S  [O 2 (g)] + 1 S  [C 2 H 6 (g)] )

11. Finding S  Values Appendix C in your textbook has entropy values for a wide variety of species. Units for entropy values  J / (K mole) Temperature and pressure for the tabulated values are 298.2 K and 1.00 atm.

12. Finding S  Values Note – entropy values are absolute! Note – the elements have NON-ZERO entropy values! e.g., for H 2 (g)  f H  = 0 kJ/mole (by def’n) S  = 130.58 J/(K mole)

13. Some Generalizations For any gaseous reaction (or a reaction involving gases).  n g > 0,  r S  > 0 J/(K mole).  n g < 0,  r S  < 0 J/(K mole).  n g = 0,  r S   0 J/(K mole). For reactions involving only solids and liquids – depends on the entropy values of the substances.

14. The Second Law of Thermodynamics The entropy of the universe (  univ S) increases in a spontaneous process. ◦ univ S unchanged in an equilibrium process

15. What is  univ S?  univ S =  sys S +  surr S  sys S = the entropy change of the system.  surr S = the entropy change of the surroundings.

16. How Do We Obtain  univ S? We need to obtain estimates for both the  sys S and the  surr S. Look at the following chemical reaction. C(s) + 2H 2 (g)  CH 4 (g) The entropy change for the systems is the reaction entropy change,  r S . How do we calculate  surr S?

17. Calculating  surr S Note that for an exothermic process, an amount of thermal energy is released to the surroundings!

18. A small part of the surroundings is warmed (kinetic energy increases). The entropy increases!

19. Calculating  surr S Note that for an endothermic process, thermal energy is absorbed from the surroundings!

20. A small part of the surroundings is cooled (kinetic energy decreases). The entropy decreases! For a constant pressure process q p =  H  surr S   surr H  surr S  -  sys H

21. The entropy of the surroundings is calculated as follows.  surr S = -  sys H / T For a chemical reaction  sys H =  r H   surr S = -  r H  / T

22. The Use of  univ S to Determine Spontaneity Calculation of T  univ S  two system parameters ◦ r S  ◦ r H  Define a system parameter that determines if a given process will be spontaneous?

23. The Definition of the Gibbs Energy The Gibbs energy of the system G = H – TS For a spontaneous process  sys G = G f – G i G f = the Gibbs energy of the final state G i = the Gibbs energy of the initial state

24. Gibbs Energy and Spontaneity  sys G < 0 - spontaneous process  sys G > 0 - non-spontaneous process (note that this process would be spontaneous in the reverse direction)  sys G = 0 - system is in equilibrium Note that these are the Gibbs energies of the system under non- standard conditions

25. Standard Gibbs Energy Changes The Gibbs energy change for a chemical reaction? Combustion of methane. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l) Define ◦ r G  =  n p  f G  (products) -  n r  f G  (reactants) ◦ f G  = the formation Gibbs energy of the substance

26. The Gibbs Energy Change (cont’d) For the methane combustion reaction 1 CH 4 (g) + 2 O 2 (g)  1 CO 2 (g) + 2 H 2 O(l)  r G  =  n p  f G  (products) -  n r  f G  (reactants) = 2  f G  [H 2 O(l)] + 1  f G  [CO 2 (g)] - (2  f G  [O 2 (g)] + 1  f G  [CH 4 (g)] )

27. Gibbs Energy Changes  f G  (elements) = 0 kJ / mole. Use tabulated values of the Gibbs formation energies to calculate the Gibbs energy changes for chemical reactions.

28. The Third Law of Thermodynamics Entropy is related to the degree of randomness of a substance. Entropy is directly proportional to the absolute temperature. Cooling the system decreases the disorder.

29. The Third Law of Thermodynamics The Third Law - the entropy of any perfect crystal is 0 J /(K mole) at 0 K (absolute 0!) Due to the Third Law, we are able to calculate absolute entropy values.

30. At a very low temperature, the disorder decreases to 0 (i.e., 0 J/(K mole) value for S). The most ordered arrangement of any substance is a perfect crystal!

31. Applications of the Gibbs Energy The Gibbs energy is used to determine the spontaneous direction of a process. Two contributions to the Gibbs energy change (  G) ◦ Entropy (  S) ◦ Enthalpy (  H)  G =  H - T  S

32. Spontaneity and Temperature HH SS GG ++ < 0 at high temperatures +- > 0 at all temperatures -+< 0 at all temperatures --< 0 at low temperatures

33. Gibbs Energies and Equilibrium Constants  r G  < 0 - spontaneous under standard conditions  r G  > 0 - non-spontaneous under standard conditions

34. The Reaction Quotient Relationship between Q J and K eq Q < K eq - reaction moves in the forward direction Q > K eq - reaction moves in the reverse direction Q = K eq - reaction is at equilibrium

35.  r G° refers to standard conditions only! For non-standard conditions -  r G  r G < 0 - reaction moves in the forward direction  r G > 0 - reaction moves in the reverse direction  r G = 0 - reaction is at equilibrium

36. Relating K to  r G   r G =  r G  +RT ln Q  r G = 0  system is at equilibrium  r G  = -RT ln Q eq  r G  = -RT ln K eq

37. Phase Equilibria At the transition (phase-change) temperature only -  tr G = 0 kJ tr = transition type (melting, vapourization, etc.)  tr S =  tr H / T tr