Chapter 18 Thermodynamics

1. Chapter 18Thermodynamics

2. Spontaneous vs Nonspontaneous Change • Spontaneousprocess – any process, once started, that proceeds without the external input of energy • Nonspontaneousprocess – any process which requires the continual, external input of energy to keep the process going Any process that is spontaneous in one direction will be nonspontaneous in the reverse direction !

3. Nonspontaneous Change • Occurs only with outside assistance • Never occurs by itself: • Room gets straightened up • Pile of bricks turns into a brick wall • Decomposition of H2O by electrolysis • Continues only as long as outside assistance occurs: • Person does work to clean up room • Bricklayer layers mortar and bricks • Electric current passed through H2O

4. Spontaneous Change • What factors influence spontaneity? • Spontaneous Change • Occurs by itself • Without outside assistance until finished • e.g. • Water flowing over waterfall • Melting of ice cubes in glass on warm day

6. Spontaneous Processes • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. • Above 0 C, it is spontaneous for ice to melt. • Below 0 C, the reverse process is spontaneous.

7. Spontaneous processes tend to favor: 1. Decrease in Energy 2. Increase in Disorder

9. Direction of Spontaneous Change • Many reactions which occur spontaneously are exothermic: • Iron rusting • Fuel burning • H and Esystem are negative • Heat given off • Energy leaving system • Thus, H is one factor that influences spontaneity

10. Direction of Spontaneous Change • Some endothermicreactions occur spontaneously: • Ice melting • Evaporation of water • Expansion of CO2 gas into vacuum • The operation of a chemical “cold pack” • H and E are positive • Heat absorbed • Energy entering system • Clearly other factors influence spontaneity

11. Entropy (Symbol S ) • Thermodynamic quantity • Describes number of equivalent ways that energy can be distributed • Can be thought of as a measure of the randomness or disorder of a system • The greater the statistical probability of a particular state the greater the entropy! • Larger S, means more possible ways to distribute energy and that it is a more probable result

12. Disorder Driven by Statistical Probability Statistical probability – refers to the number of possible arrangements of a system Adisordered state is more probable because it can be achieved in more statistical ways the larger the number of different possible combinations, the greater the probability of getting a disordered state

13. Examples of Spontaneity • Spontaneous reactions • Things get rusty spontaneously • Don't get shiny again • Sugar dissolves in coffee • Stir more—it doesn't undissolve • Ice  liquid water at RT • Opposite does NOT occur • Fire burns wood, smoke goes up chimney • Can't regenerate wood • Common factor in all of these: • Increase in randomness and disorder of system • Something that brings about randomness more likely to occur than something that brings order

14. There are 2,598,960 possible five-card poker hands

15. Entropy, S lessdisorder S = smaller # Ludwig Boltzmann1844 – 1906 more disorderS = larger #

17. order disorder is favored to occur spontaneously ΔS = + # order disorder disorder order ΔS = -# 2nd Law of Thermodynamics– the entropy of the universe is increasing over time

18. Entropy, S • State function • Independent of path • S = Change in entropy • For chemical reactions or physical processes • Factors that affect entropy – volume, temperature, physical state, number of particles

19. Effect of Volume on Entropy • For gases, entropy increases as volume increases • Gas separated from vacuum by partition • Partition removed, more ways to distribute energy • Gas expands to achieve more probable particle distribution • More random, higher probability, more positive S

20. Effect of Temperature on Entropy 3rd Law of Thermodynamics– entropy decreases as temperature decreases (and vice-versa)

21. Effect of Physical State on Entropy • Crystalline solid very low entropy • Liquid higher entropy,molecules can move freely • More ways to distribute KE among them • Gas highest entropy,particles randomly distributed throughout container • Many, many ways to distribute KE

22. Entropy Affected by Number of Particles • Adding particles to system • Increase number of ways energy can be distributed in system • So all other things being equal • Reaction that produces more particles will have positive S

23. Learning Check Which represents an increase in entropy? A. Water vapor condensing to liquid B. Carbon dioxide subliming C. Liquefying helium gas D. Proteins forming from amino acids

24. Entropy Changes in Chemical Reactions • Reactions without gases - Calculate number of mole molecules n = nproducts – nreactants If n is positive, entropy increases More molecules, means more disorder • Reactions involving gases • - Calculate change in # of moles of gas, ngas • If ngas is positive , S is positive • ngas is more important than nmolecules

25. Predict Sign of S for Following Reactions CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) + CO2(g) • ngas = 1 mol – 0 mol = 1 mol • sincengasis positive, Sis positive 2 N2O5(g) 4 NO2(g) + O2(g) • ngas = 4 mol + 1 mol – 2 mol = 3 mol • sincengasis positive, Sis positive OH–(aq) + H+(aq) H2O(l) • ngas = 0 mol • n = 1 mol – 2 mol = –1 mol • sincenis negative, Sis negative

26. Learning Check Which of the following has the most entropy at standard conditions? • H2O(l ) • NaCl(aq) • AlCl3(s) • Can’t tell from the information

27. Standard entropy of reaction – the (entropy, or disorder) that accompanies ANY reaction under standard conditions (units are J/K) = [(sum S products) – (sum S reactants)] 1. write and balance the reaction 2. use table of thermodynamic data of S to calculate Standard entropy, S– disorder of a substance at standard conditions(units are J/mole·K)

28. Determine the standard entropy of reaction, 2 3 4 = – 93.0 J/K N2O(g) + O2 (g) NO2(g) = – #, suggests the reaction is nonspontaneous

29. ΔHrxn only slightly varies with temperature ΔSrxn is highly dependent on temperature Recap: 2 forces in nature which drive processes to occur spontaneously 1. Decrease in energy ΔHrxn= – # (exo) 2. Increase in disorder ΔSrxn= + # (disorder)

30. Willard Gibbs 1839 – 1903 ΔGrxn = ΔHrxn – TΔSrxn ΔHrxn is the heat of reaction T is the temperature in Kelvin ΔSrxn is the entropy of reaction ΔGrxn is the Gibbs Free Energy

31. ΔGrxnor Gibbs Free Energy, is the ultimate, final deciding factor as to whether a reaction will occur spontaneously, anywhere in the universe Bottom Line: When ΔGrxn= – #, the reaction is spontaneous. When ΔGrxn= + #, the reaction is nonspontaneous. When ΔGrxn= 0, the reaction is at equilibrium (has no tendency to go one way or the other)

32. Is the following reaction spontaneous at 175 °C ? 3 2 4 favors spontaneous favors nonspontaneous = – 93.0 J/K Who wins ?? N2O(g) + O2 (g) NO2 (g) = – 28.0 kJ ΔGrxn = ΔHrxn – TΔSrxn ΔGrxn= +13.7 kJ ΔGrxn= + #, the reaction is nonspontaneous

35. Learning Check For the reaction 2NO(g) + O2(g)  2NO2(g), H° = -113.1 kJ/mol and S° = -145.3 J/K mol. Which of these statements is true?  A. The reaction is spontaneous at all temperatures. B. The reaction is only spontaneous at low temperatures. C. The reaction is only spontaneous at high temperatures. D. The reaction is at equilibrium at 25°C under standard conditions.

36. Learning Check At what temperature (K) will a reaction become nonspontaneous when H = –50.20 kJ mol–1 and S = +20.50 J K–1mol–1? A. 298 K B. 1200 K C. 2448 K D. The reaction cannot become non-spontaneous at any temperature We are looking for when ΔG > 0 So ΔH – TΔS > 0 or ΔH > TΔS ΔH/ΔS > T (-50.20 kJ mol-1)/(0.02050 kJ K-1 mol-1) > T -2448 K > T so not possible

37. Calculate S° for reduction of aluminum oxide by hydrogen gas Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)

38. Calculate ΔS° Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g) S ° = 56.5 J/K + 566.1 J/K – (51.00 J/K + 391.8 J/K ) S ° = 179.9 J/K = + #, suggests the reaction is spontaneous

39. Standard Free Energy Changes • Standard Free Energy Change, G ° • G measured at 25 °C (298 K) and 1 atm • Two ways to calculate, depending on what data is available • Method 1: • G ° = H °– TS ° • Method 2:

40. Calculate G ° Method 1 Calculate G ° for reduction of aluminum oxide by hydrogen gas Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g) Step 1: Calculate H° for reaction using heats of formation below

41. Calculate G ° Method 1 H ° = 0.0 kJ – 725.4 kJ – 0.00 kJ – (– 1669.8 kJ) H ° = 944.4 kJ

42. Calculate G ° Method 1 Step 2:Calculate S °  see Example 3 S ° = 179.9 J/K Step 3:Calculate G° = H°– (298.15 K)S ° G° = 944.4 kJ – (298 K)(179.9 J/K)(1 kJ/1000 J) G° = 944.4 kJ – 53.6 kJ = 890.8 kJ • G° is positive • Indicates that the reaction isnot spontaneous

43. Calculate G ° Method 2 • Use Standard Free Energies of Formation • Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 °C

44. Calculate G ° Method 2 Calculate G ° for reduction of aluminum oxide by hydrogen gas. Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)

45. Calculate G ° Method 2 G° = 0.0 kJ – 685.8 kJ – 0.00 kJ – (– 1576.4 kJ) G° = 890.6 kJ Both methods same within experimental error

46. G ° = Maximum Possible Work • G °is maximum amount of energy produced during a reaction that can theoretically be harnessed as work • Energy that need not be lost to surroundings as heat • Energy that is “free” or available to do work

47. How K is related to G° G °=–RT lnK K = e–G°/RT

48. G° and Position of Equilibrium • When G ° > 0 (positive) • Position of equilibrium lies close to reactants • Little reaction occurs by the time equilibrium is reached • Reaction appears nonspontaneous • When G ° < 0 (negative) • Position of equilibrium lies close to products • Mainly products exist by the time equilibrium is reached • Reaction appears spontaneous

49. G ° and Position of Equilibrium • When G ° = 0 • Position of equilibrium lies ~ halfway between products and reactants • Significant amount of both reactants and products present at time equilibrium is reached • Reaction appears spontaneous, whether start with reactants or products • Can Use G ° to Determine Reaction Outcome • G ° large and positive • No observable reaction occurs • G ° large and negative • Reaction goes to completion

50. Calculating G ° from K • Ksp for AgCl(s) at 25 °C is 1.8  10–10 Determine G ° for the process • Ag+(aq) + Cl–(aq)  AgCl(s) • Reverse of Ksp equation, so G ° =–RT lnK = –(8.3145 J/K mol)(298 K) × ln(5.6  109) × (1 kJ/1000 J) • G °= –56 kJ/mol • Negative G °indicates precipitation will occur