1. Thermodynamics SPONTANEITY, ENTROPY, AND FREE ENERGY

2.  ENTHALPY: ΔH is heat exchange—exothermic reactions are generally favored.  ENTROPY: ΔS is the dispersal (disorder) of the matter and energy of a system.  Thermodynamically favored processes are those that involve both a decrease in the internal energy of the components (ΔH o 0). These processes are thermodynamically favored if ΔG o <0 or negative.  Thermodynamically favored reactions occur without outside intervention once the activation energy has been reached. Some are slow like diamond turning to graphite and some are fast like the combustion of methane.

3. First Law of Thermodynamics  Answers questions like:  How much energy is involved in the change?  Does energy flow into or out of the system?  What form does the energy finally assume?  It does not explain why a process occurs in a given direction.

4.  Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process.  For example, according to the law of thermodynamics, a diamond should change spontaneously into graphite. The fact that we don’t see it happen doesn’t mean it won’t, its just very slow.  Consider the following:  A ball rolls down a hill but never spontaneously rolls back up the hill.  Wood burns spontaneously in an exothermic reaction to form CO 2 and H 2 O but wood is not formed from those two molecules when they are heated together.  At temperatures below 0 o C, water spontaneously freezes, and at temperatures above 0 o C ice spontaneously melts.  What thermodynamic principle will provide an explanation of why, under a given set of conditions, each of these diverse processes occurs in one direction but never the reverse?

5. Spontaneous Processes and Entropy  A process is said to be spontaneous if it occurs without outside intervention. Spontaneous processes may be fast or slow, they are not instantaneous.  Thermodynamics can tell us the direction in which a process will occur, but it can say nothing about the speed at which the process occurs—that’s the job of kinetics.  Entropy can be viewed as a measure of molecular randomness or disorder.  Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a given state.

6. EntropyEntropy (13:41) and the Second Law of Thermodynamics  The second law of thermodynamics is that in any spontaneous process there is always an increase in the entropy of the universe.  OR: The entropy of the universe is increasing.  The change in the entropy of the universe can be expressed as  ΔS universe = ΔS system + ΔS surroundings  If the sign of ΔS universe is positive, the entropy of the universe increases and the process is spontaneous in the direction written.  If ΔS universe is negative, the process is spontaneous in the opposite direction.  If ΔS universe is 0, then the process has no tendency to occur and the system is at equilibrium.

7.  Consider the set of flasks above. In the first flask there are two gas molecules confined to the right side. If the stopcock is opened, there are then 4 possible places for the molecules to be, each equally probable. The probability that one molecule being in the right side is ½ while the probability that two molecules would be in the right half would be ½ 2 = ¼ Three gas molecules would be ½ 3 = 1/8 etc.  If you were to consider a mole of gas molecules being in the right- hand flask at the same time it would be1/2 n where n = 6.022 x 10 23.  This is such a small number that essentially there is zero likelihood that all the gas molecules will be found in the right-hand flask at the same time. Not entirely impossible, but not probable. This is called positional probability.

8. Positional entropy  For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature.  A. Solid CO 2 and gaseous CO 2  B. N 2 gas at 1 atm and N 2 gas at 1.0 x 10 -2 atm  A. Since a mole of gaseous CO 2 has the greater volume by far, the molecules have many more available positions than in a mole of solid CO 2. Thus gaseous CO 2 has a higher positional entropy.  B. A mole of N 2 gas at 1 x 10 -2 atm has a volume 100 times that of a mole of N 2 gas at 1 atm. Thus N 2 gas at 1x 10 -2 atm has the higher positional entropy. (Greater volume means more available positions)

9. Predicting Entropy Changes  Predict the sign of the entropy change for each of the following processes:  A. Solid sugar is added to water to form a solution.  B. Iodine vapor condenses on a cold surface to form crystals.  A. The sugar molecules become randomly dispersed in the water when the solution forms and thus have access to a larger volume and a larger number of possible positions. The positional disorder is increased, and there will be an increase in entropy. ΔS is positive, since the final state has a larger entropy than the initial state. ΔS = S final – S initial  B. Gaseous iodine is forming a solid. This process involves a change from a relatively large volume to a much smaller volume, which results in lower positional disorder. For this process, ΔS is negative (the entropy decreases).  In general, the greater the number of arrangements, the higher the entropy of the system.

10. Temperature and Spontaneity  ΔS universe = ΔS system + ΔS surroundings  To determine the sign of ΔS universe consider this process:  H 2 O (l)  H 2 O (g)  A mole of liquid water has a volume of about 18 mL while a mole of water vapor at 1 atm and 100 o C has a volume of about 31L. Therefore in this process the entropy of the system increases and ΔS system increases.  The entropy of the surroundings are determined by the flow of energy in or out of the system. An exothermic reaction releases energy into the surroundings, increasing the K.E. of the surrounding atoms and increasing entropy. Thus ΔS surr is positive.  Endothermic reactions decrease entropy of the surroundings and ΔS surr is negative.  The sign of ΔS universe determines if the vaporization of water is spontaneous.  In the water to gas scenario, the ΔS surr is negative because the water absorbs heat to turn to vapor but the ΔS system is positive.  What determines the ΔS universe is temperature. The impact of the transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures.

11.  Driving force provided by energy flow (heat) = magnitude of the entropy change of the surroundings = quantity of heat (J) temperature (K) Exothermic process: + quantity of heat (J) ÷ temperature (K) Endothermic process: - quantity of heat (J) ÷ temperature (K) ΔS surr = -ΔH ÷ T

12.  In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore.  Iron is used to reduce antimony in sulfide ores:  Sb 2 S 3(s) + 3Fe (s)  2Sb (s) + 3FeS (s) ΔH = -125 kJ  Carbon is used as the reducing agent for oxide ores:  Sb 4 O 6(s) + 6C (s)  4Sb (s) + 6CO (g) ΔH = 778 kJ  Calculate ΔS for each of these reactions at 25 o C and 1 atm. ΔS surr = -ΔH ÷ T  ΔS surr = - (-125kJ ÷ 298K )= 0.419 kJ/K = 419 J/K (+ because its exothermic)  ΔS surr = - (778kJ ÷ 298K ) = - 2.61 kJ/K = -2610 J/K ( - because its endothermic)  ΔS surr is negative because heat flow occurs from the surroundings to the system.

13.  If both ΔS surr and ΔS sys are positive, then ΔS univ is positive and the process is spontaneous.  If both ΔS surr and ΔS sys are negative, then ΔS univ is negative and the process does not occur in the direction indicated but is spontaneous in the opposite direction.  If ΔS surr and ΔS sys have opposite signs, the spontaneity depends on the sizes of the opposing terms.  Exothermicity is most important as a driving force in low temperatures. ΔS sys ΔS surr ΔS univ Process spontaneous? +++Yes ---No (reaction occurs in opposite direction) +??Yes, if ΔS sys has a larger magnitude than ΔS surr -??Yes, if ΔS surr has a larger magnitude than ΔS sys

14. Free Energy (G)  Free energy is another thermodynamic function related to spontaneity and useful in dealing with temperature dependence of spontaneity.  G = H – TSwhere H = enthalpy, T = K temp, S = entropy  ΔG = ΔH – T ΔS At low temperatures, ΔH dominates At high temperatures, ΔS dominates. The calculation of ΔG ultimately decides if a reaction is thermodynamically favored. CaseResult ΔS positive, ΔH negativeSpontaneous at all temperatures ΔS positive, ΔH positiveSpontaneous at high temperatures (where exothermicity is relatively unimportant) ΔS negative, ΔH negativeSpontaneous at low temperatures (where exothermicity is dominant) ΔS negative, ΔH positiveProcess not spontaneous at any temperatures (reverse process is spontaneous at all temperatures)

15.  Big Mamma equation:  ΔG rxn = ∑ ΔG 0 (products) - ∑ ΔG 0 (reactants)  Grandaddy equation:  ΔG 0 = ΔH 0 - TΔS 0 derives to T = ΔH 0 ÷ ΔS 0  Hess’s Law of Summation: same rules as you learned before for ΔH 0 is used for ΔG 0.  Rat Link equation for calculating ΔG 0 at standard conditions using the given temperature and equilibrium constant, K: ΔG 0 = -RT ln K where R = 8.1345 J/mol· K  K p = P reactants ÷ P products still raised to power of coefficients.  Solving for ΔG 0 using “minus nunfe” equation gien the standard cell potential, Faraday’s constant and number of moles of electrons involved: ΔG 0 = - nFE 0  Faraday’s constant: 96,485 Coulombs/mole e - and E 0 is the standard cell potential for the electrochemical process. Good to know that 1volt = 1 joule/coulomb

16. 2H 2 O (l) + O 2(g)  H 2 O 2(l)  Calculate the free energy of formation for the oxidation of water to produce hydrogen peroxide given the following information ΔG 0 f values:  H 2 O (l) - 56.7 kcal/mol rxn  O 2(g) 0 kcal/mol rxn  H 2 O 2(l) - 27.2 kcal/mol rxn  2(-27.2) – (2 x -56.7) = 59 kcal/mol

17. 2SO 2(g) + O 2(g)  2SO 2(g)  Calculate the ΔH 0, ΔS 0, and ΔG 0 using the following data at 25 0 C and 1 atm. ΔH 0 = products – reactants ΔS 0 = products - reactants ΔG 0 = ΔH 0 - T ΔS 0 Both ΔG 0 and ΔH 0 = 0 for elements in their standard state. Notice that S is not ΔS 0 that’s because only a perfect diamond at absolute zero will have an S value of 0. Now you know…  ΔH 0 = -792 – (-594) = -198 kJ/mol  ΔS 0 = 2(257) – [ 2(248) + 205] = -187 J/K· mol  ΔG 0 = -198kJ/mol – (298K)(-187J/K·mol x 1kg/1000 J) = -142 kJ/mol SubstanceΔH 0 (kJ/mol)S (J/K· mol) SO 2(g) -297248 SO 3(g) -396257 O 2(g) 0205

18. Free energy and spontaneity  At what temperatures is the following process spontaneous at 1 atm?  Br 2(l)  Br 2(g) ΔS o = 93.0 J/K· mol, ΔH o = 31.0 kJ/mol  The vaporization process will be spontaneous where ΔG o is negative. ΔS o favors the vaporization process because of the increase in positional entropy, ΔH favors the opposite process, which is exothermic. These opposite tendencies will exactly balance at the boiling point of Br 2 since at this temperature liquid and gaseous bromine are in equilibrium and ΔG o = 0 Thus: 0 = ΔH o – T ΔS o  ΔH o = T ΔS o Solve for T  3.0 x 10 4 J/mol ÷ 93.0 J/K· mol = 333K Above 333K, the term ΔS controls. The increase in entropy when liquid Br 2 is vaporized is dominant. Below 333K, the term ΔH controls because the process is spontaneous in the direction in which it is exothermic. At 333K, and ΔG=0, the opposing driving forces are balanced (liquid and gaseous bromine coexist) and this is the normal b.p.

19. Predicting Entropy Changes  Predict the sign of the entropy change (ΔS) ro each of the following processes. Justify your answers.  Solid sugar is added to water to form a solution.  Positive  Iodine vapor condenses on a cold surface to form crystals.  Negative

20. Prediction  Predict the sign of ΔS o for each of the following reactions:  CaCO 3(S)  CaO (S) + CO 2(g)  2SO 2(g) + O 2(g)  2SO 3(g)  Since in this reaction a gas is produced from a solid reactant, the positional entropy increases, and ΔS o positive.  Here three molecules of gaseous reactants become two molecules of gaseous products. Since the number of gas molecules decreases, positional entropy decreases, and ΔS o is negative.

21. Third law of thermodynamics  The entropy of a perfect crystal at 0 K is zero. This is an unattainable ideal that is taken as a standard but never actually observed.  With increased temperature the random vibrations increase and the entropy also increases.  Because entropy is a state function of the system (not pathway dependent) the entropy change for a given chemical reaction can be calculated with:  Δs o reaction = ∑n p ΔS 0 products - ∑n r ΔS 0 reactants  Note that entropy is an extensive property and depends on the amount of substance present. (number of moles)  ΔS is + when dispersal/disorder increases (favored)  ΔS is – when dispersal/disorder decreases

22. 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s)  Calculate ΔS 0 at 25 o C for the above reaction given the following entropy values:  Δs o reaction = ∑n p ΔS 0 products - ∑n r ΔS 0 reactants  2S 0 SO2(g) + 2S 0 NiO(s) - 2S 0 NiS(s) - 3S 0 O2(g)  2 mol (248 J/K· mol) + 2 mol (38 J/K· mol)  -2mol (53 J/K· mol) - 3 mol (205 J/K· mol)  = 496 J/K + 76 J/K – 106 J/K – 615 J/K  = - 149 J/K  We would expect ΔS o to be negative because the number (mols) of gaseous molecules decreases in this reaction. SubstanceS o (J/K · mol) SO 2(g) 248 NiO (s) 38 O 2(g) 205 NiS (s) 53

23. Al 2 O 3(s) + 3H 2(g)  2Al (s) + 3H 2 O (g)  Calculate ΔS 0 at 25 o C for the above reaction given the following entropy values:  179 J/K SubstanceS 0 (J/K · mol) Al 2 O 3(s) 51 H 2(g) 131 Al (s) 28 H 2 O (g) 189

24. 2SO (g) + O 2(g) --> 2SO 3(g)  Calculate the entropy change at 25 0 C, in J/mol rxn · K (not kJ!)  Given the following data:  SO 2(g) 248.1 J/mol rxn · K  O 2(g) 205.3 J/mol rxn · K  SO 3(g) 256.6 J/mol rxn · K  2(256.6) – [2(248.4) + 205.3] =  - 188.3 J/mol rxn · K

25. Entropy changes as they relate to reversible phase changes  ΔS surroundings = heat transferred ÷ temperature at which change occurs. q/T = -ΔH/T expressed in J/mol rxn · K Where the heat supplied (endothermic) q > 0 Where heat is evolved (exothermic) q < 0 It is important to note if the reaction is endothermic or exothermic. The temperature at which the process occurs is significant. Example: water (liquid at 100 0 C) ↔ water (gas at 100 0 C) The entropy will increase for the forward reaction (vaporizing) since the reaction produces water in a less condensed state, thus the molecules are more dispersed.

26. Calculating ΔH 0, ΔG 0, ΔS 0 and K eq ΔG = ΔH – T ΔS  Calculate ΔG for the following reaction: C (S) + O 2(g)  CO 2(g)  ΔH = - 393.5kJ /mol ΔS= 3.05 J/K· mol Temperature is held constant at 298K ΔG 0 = ΔH 0 - T ΔS 0 ΔG 0 = -393.5 kJ/mol – (298K)(3.05 x 10 -3 kJ/K· mol - 394.4 kJ/mol of CO 2  Calculate ΔG, ΔH and ΔS for:  2SO 2(g) + O 2(g))  2SO 3(g) at 1 atm and constant temperature of 298K SubstanceΔH f 0 (kJ/mol)S 0 (J/K· mol) SO 2(g) --297248.1 SO 3(g) --196256.6 O 2(g) 0205.3

27. Find ΔH  ΔH o reaction = ∑n p ΔH 0 products - ∑n r ΔH 0 reactants  ΔH f o reaction = 2ΔH f 0 SO3 - 2ΔH f 0 SO2 + ΔH f 0 O2  = 2 mol (-396 kJ/mol) – 2 mol (-297 kJ/mol) - 0  = -792 kJ + 594 kJ  = -198 kJ

28. Find ΔS 0 and ΔG 0  ΔS o reaction = ∑n p ΔS 0 products - ∑n r ΔS 0 reactants  = 2 mol ( 256.6 J/K· mol) – 2 mol (248.1 J/K· mol) – 1 mol(205.3 J/K · mol)  = 513 J/K mol – (496.2 J/K mol+205.3 J/Kmol)  = -188 J/K/mol  You would expect ΔS 0 to be negative because 3 molecules of gaseous reactants give 2 molecules of gaseous products.  ΔG 0 = ΔH 0 - T ΔS 0  = - 198kJ/mol – (298K) ( -187 J/K mol )(1kJ/1000J)  = - 198 kJ/mol + 55.7 kJ/mol = - 142 kJ/mol

29. Find K eq  ΔG 0 = -RT ln K  ln K = ΔG 0 ÷ - RT  lnK = -142 kJ/mol x 1000 J/kJ ÷ (-8.3145 J/mol· K)(298)  lnK = 57.3107303  7.76 x 10 24 – huge number!

30. Calculate ΔG for the reaction C diamond(s)  C graphite(s)  C diamond(s) + O 2(g)  CO 2(g) ΔG = -397 kJ  C graphite(s) + O 2(g)  CO 2(g) ΔG = -394 kJ  Reverse the last equation and change the sign.  CO 2(g)  C graphite(s) + O 2(g) ΔG = +394 kJ  Add them together (summative ΔG) = 394 kJ + (-397) = 3 kJ

31. 4Fe (s) + 3O 2(g)  2Fe 2 O 3(s) Use the data to the right to calculate the equilibrium constant for the above reaction at 25 o C. ΔG = ΔH - T ΔS ΔH = 2(-826) = -1652 kJ/mol ΔS = 180 – [2(2.7) + 3(205)] = - 5.43 x 10 - 1 kJ/mol ΔG = -1652kJ/mol – (298K)(-5.43 x 10 -1 kJ/mol · K)= -1.49 x 10 3 kJ/mol lnK eq = - (- 1.47 x 10 -3 kJ/mol ÷ (8.3145 x 10 -3 kJ/mol · K) = 601 K = e 601 = 10 261 SubstanceΔH 0 f (kJ/mol)S o (J/K · mol) Fe 2 O 3(s) -82690 Fe (s) 027 O 2(g) 0205

32. H 2 O (l)  H 2 O (g)  Calculate the thermodynamic boiling point of water forming a gas from a liquid.  ΔH = + 44 kJ/mol ΔS = 118.8 kJ/K· mol  ΔG = 0 at equilibrium so ΔH = T ΔS  ΔH = 44kJ/mol ÷.1188 kJ/mol · K = 370 K

33. 2CO (g) + O 2(g)  2CO 2(g)  ΔG is – 2572 kJ/mol. Calculate the equilibrium constant at 25 o C.  ΔG = - RT lnK eq  lnK eq = ΔG ÷ -RT  - 2.572 x 10 -5 ÷ (-8.3145 J/mol · K)(298K)  1.27 x 10 45

34. CalorimetryCalorimetry 11:47