Chemistry 100 Chapter 19

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 = Sf – Si Sf = the entropy of the final state Si = 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! C2H6 (g) + 7/2O2 (g)  2CO2 (g) + 3H2O (l) • The entropy change • rS   np S (products) -  nr S (reactants) • np and nr represent the number of moles of products and reactants, respectively.

10. The Entropy Change (Cont’d) • For the ethane combustion reaction 1C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (l) rS   np S(products) -  nr S(reactants) = 3 S [H2O (l)] + 2 S [CO2 (g)] - (7/2 S [O2(g)] + 1 S [C2H6 (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 H2 (g) fH = 0 kJ/mole (by def’n) S = 130.58 J/(K mole)

13. Some Generalizations • For any gaseous reaction (or a reaction involving gases). ng > 0, rS > 0 J/(K mole). ng < 0, rS < 0 J/(K mole). ng = 0, rS  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 (univS) increases in a spontaneous process. • univS unchanged in an equilibrium process

15. What is univS? univS = sysS + surrS sysS = the entropy change of the system. surrS = the entropy change of the surroundings.

16. How Do We Obtain univS? • We need to obtain estimates for both the sysS and the surrS. • Look at the following chemical reaction. C(s) + 2H2 (g)  CH4(g) • The entropy change for the systems is the reaction entropy change, rS. • How do we calculate surrS?

17. Calculating surrS • 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 surrS • 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 qp = H surrS surrH surrS -sysH

21. The entropy of the surroundings is calculated as follows. surrS = -sysH / T • For a chemical reaction sysH = rH surrS = -rH/ T

22. The Use of univS to Determine Spontaneity • Calculation of TunivS  two system parameters • rS • rH • Define a systemparameter 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 sysG = Gf – G i Gf = the Gibbs energy of the final state Gi = the Gibbs energy of the initial state

24. Gibbs Energy and Spontaneity sysG < 0 - spontaneous process sysG > 0 - non-spontaneous process (note that this process would be spontaneous in the reverse direction) sysG = 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. CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) • Define • rG =  np fG (products) -  nr fG (reactants) • fG = the formation Gibbs energy of the substance

26. The Gibbs Energy Change (cont’d) • For the methane combustion reaction 1CH4(g) + 2 O2(g) 1 CO2(g) + 2 H2O(l) rG =  np fG (products) -  nr fG (reactants) = 2 fG [H2O(l)] + 1 fG [CO2(g)] - (2 fG [O2(g)] + 1 fG [CH4(g)] )

27. Gibbs Energy Changes • fG (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

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

34. The Reaction Quotient • Relationship between QJ and Keq Q < Keq - reaction moves in the forward direction Q > Keq - reaction moves in the reverse direction Q = Keq - reaction is at equilibrium

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

36. Relating K to rG rG = rG +RT ln Q rG = 0  system is at equilibrium rG = -RT ln Qeq rG = -RT ln Keq

37. Phase Equilibria • At the transition (phase-change) temperature only - trG = 0 kJ tr = transition type (melting, vapourization, etc.) trS = trH / Ttr