Introduction to Thermodynamics: Conservation & Calculations

1. Chapter 18 – Thermodynamics and Equilibrium Note: You need to briefly review chapter 6 (Thermochemistry) from Chem 122 - Especially the sections on calculating ΔH values. Online HW 18: - Due on Friday, May 17, 2013 Quiz #2: - Over 17&18 on Friday, May 17 Exam #2: - This exam is broken into two parts – Ch 17 on M 5/20 & Ch 18 on W 5/22.

2. Problems pg 768-777 Chapter 18 (next pg & those below) 1 2 6 7 8 9 14 15 16 17 23 24 25 26 31 35 39 41 43 45 47 53 57 61 65 (T at which ∆G = 0) Do the two on the next page.

3. Typical Bond Energy, ∆H Problem • Use the bond energy table in chapter 9 (pg 367) to solve for the ∆Hr in kJ for the following two reactions. 1C2H6 + 6Br2 -----) 1C2Br6 + 6HBr Answer ≈ - 216 kJ 2H2 + O2 -----) 2H2O Answer ≈ - 482 kJ

4. I. Introduction • Thermodynamics deals with the relationship between heat and other forms of energy in a chemical reaction. • Thermodynamics allows a chemist to 1) predict the spontaneity of a reaction (This Chapter, 18) 2) predict the energy lost or gained in a reaction (Chapters 6 & 18) 3) calculate equilibrium constants (This Chapter, 18) 4) predict voltages of redox reactions (Next Chapter, 19) 5) calculate other thermodynamic values (Chapters 6 & 18) • Terms to know: Internal Energy – sum of kinetic & potential energies of the particles making up the system = U; ∆U = sum of heat & work done during change State Function – a property whose value depends only on the present state of a system and not how that state was reached. - ΔU, ΔH or Δq, ΔG, and ΔS all are state functions.

5. II. First Law of Thermodynamics – Conservation of Energy First Law of Thermodynamics: total energy is constant. - In a chemical reaction, energy is neither created nor destroyed (but it may be transformed). - One mathematical representation of the first law is in terms of the Internal Energy (U) of the system. U, is sum of kinetic & potential energies of the particles making up the system. ΔU = q + w q = heat w = work q = + if heat is absorbed or added to a system KNOW q = - if heat is evolved or released by a system KNOW - In a chemical reaction, the work is frequently due to expansion & is small compared to the value of q.

6. II. First Law of Thermodynamics A. Introduction • Δq = heat lost/gained in a chemical rxn & is usually = ΔH or enthalpy. • Enthalpy - Defined as “the heat lost or gained in a chemical reaction taking place at constant pressure.” Since most reactions take place at constant pressure (atmospheric pressure), Δq = ΔH. • ΔH is a state function; so, pathway to ΔH does not matter. • 4 Ways to calculate ΔH’s (will get 2 more): (from chem 122, chapters 6&9) 1. Experimentally from Calorimetry 2. Theoretically from Bond Energy tables (pg 367) 3. Theoretically using Hess’s Law 4. Theoretically from Heats of Formation tables (ΔHof pg 250 or A-8 through A-11) • We will go back over these & add a couple more calculation methods.

7. II. First Law B. Calculation of ΔH – 1.Calorimeters - ΔH’s can be determined experimentally using calorimeters: Δq = s x m x ΔT (Know this) Where s = Specific Heat in J/(g oC) = Characteristic Constant m = mass in g (usually grams of solution; use density if given) ΔT = Change in temperature in oC Example 1: 0.10 mols of a compound reacts in 500. g of aqueous solution in a calorimeter. The solns specific heat is 4.2 J/(g oC). The temperature goes from 25 to 35 oC. Calculate the ΔH inkJ/mole for this reaction. ΔH = - [4.2 J/(g oC)] x [500. g] x [10. oC] = - 21,000 J =- 21 kJ (A negative value since T goes up; exothermic; heat is released) ΔH(per mole)= - 21 kJ / 0.10 mole = - 210 kJ / mole

8. II. First Law B. Calculation of ΔH – 1. Calorimeters

9. II. First Law B. Calculation of ΔH – 1. Calorimeters • Example 2: 50.0 mL each of 0.250 M HCl & 0.250 M NaOH react at 19.50 oC in calorimeter & Tf=21.21oC. Calculate ΔHr in kJ/mol HCl. S = 4.18 J/goC & d =1.00 g/mL for soln. • 1HCl + 1NaOH -----) 1NaCl + 1H2O • Δq = m x S x ΔT (Know This) Δqliq = 100.0g x 4.18 J/goCx 1.71oC = 715 J = 0.715 kJ Δqrxn or Δqr = - 0.715 kJ (q = negative since exothermic) 0.250 mol/L x 0.0500L = 0.0125 mol of HCl or NaOH ΔH(mol) = - 0.715 kJ / 0.0125 mol = - 57.2 kJ/mol

10. II. First Law B. Calculation of ΔH – 2.Bond Energies - Calculate heat of reaction in kJ from bond energies (pg 367) for: 1CH4 + 4Cl2 ---) 1CCl4 + 4HCl - Bond energies in kJ/m = 413 (C-H) 242 (Cl-Cl) 328 (C-Cl) 431 (H-Cl) - 1. use Lewis Structures to see how many & types of bonds involved; 2. take both # of bonds/molecule & stoichiometry into account; 3. use signs (+) & values in the table; 4. plug into the equation below. ∆Hr = ∑ ∆Hbonds broken - ∑ ∆Hbonds formed(know this) (4xC-H) (4xCl-Cl) (4xC-Cl) (4xH-Cl) ΔHr = [4x413 + 4x242] -[4x328 + 4x431] ΔHr = [2620] - [3036] = - 416 kJ

11. II. First Law B. Calculation of ΔH – 3.Hess’s Law • Hess’s Law: Since ΔH is a state function, canadd & subtract reactions & heats to get the reaction you want. Note: use the given rxn as a template for deciding how to manipulate rxns a & b. • Calculate ΔHr for: 2S + 3O2 -----) 2SO3 From: #a) 1S + 1O2 -----) 1SO2ΔH = - 297 kJ #b) 2SO3 -----) 2SO2 + 1O2ΔH = + 198 kJ Solution: Multiply#a by 2 & reverse #b;ADD together; Cancel. 2S + 2O2 -----) 2SO2ΔH = 2x-297 = - 594 kJ 2SO2 + 1O2 -----) 2SO3 -(ΔH = +198 kJ) = - 198 kJ 2SO2 + 2S + 3O2 ----) 2SO2 + 2SO3ΔH = -594 + -198 = -792 kJ Final Result: 2S + 3O2 -----) 2SO3ΔHr = - 792 kJ

12. II. First Law B. Calculation of ΔH – 4.Heat of Formation Example #1: Calculate ΔHor for the rxn at 25oC & 1.00 atm (Std State) 3 CH4(g) + 2 H2O(l)+ 1 CO2(g) -----) 4 CO(g) + 8 H2(g) ΔHor = ∑ Δ Hof (products)– ∑ Δ Hof(reactants)(Know This) - Note: 1. Values for Hof are in kJ per 1 mol (table 6.2, pg 250) 2. Hof for an element in standard state = 0.00 kJ (Know This) 3. Must take states (g,l,s) into account 3 CH4(g) + 2 H2O(l) + 1 CO2(g) -----) 4 CO(g) + 8 H2(g) (Hof) 3(-74.9) 2(-286) 1(-394) 4(-111) 8(0) -225 -572 -394 -444 0.00 ΔHor = [ -444 + 0 ] - [ -225 + -572 + -394 ] = + 747 kJ

13. II. First Law B. Calculation of ΔH – 4.Heat of Formation Example #2: Calculate heat of reaction from ΔHof for: 2NH3 + 1CO2 ---) 1N2H4CO + 1H2O(l) ΔHof: 2 x -46 -393 -319 -286 (kJ) ΔHr = ∑ Δ Hof products – ∑ Δ Hofreactants ΔHor = [-319 + -286] - [- 2x46 + -393] = - 120. kJ

14. More ΔH Examples 1. Calculate the ΔHr for 1H2C=CH2 + 1HCl -----) 1H3C-CH2-Cl from following bond energies in kJ/m: H-C 413, C-C 348, C=C 614 C≡C 839, H-Cl 431, C-Cl 328 ΔHr = ∑ΔHbreak - ∑ΔHformed ΔHr = [4C-H + 1C=C + 1H-Cl] - [5C-H + 1C-C +1C-Cl] 4x413 614 431 5x413 348 328 ΔHr = [4x413 + 614 + 431] - [5x413 + 348 + 328] = 2697 - 2741 = -44 kJ 2. Calculate ΔHr for the above reaction using the following ΔHof values in kJ/m: C2H4(g) = 52 HCl(g) = -92 C2H5Cl(g) = -84 1C2H4(g) + 1HCl(g) -----) 1C2H5Cl(g)ΔHr = ∑ΔHoprod - ∑ΔHoreact 52 -92 -84 ΔHr = [-84] - [52 + -92] = -84 + 40 = -44 kJ

15. More ΔH Examples 3. Calculate the ΔHr for 3H2 + 1O3 ----) 3H2O from the following two rxns. Note: all are in the gaseous state. 1H2 + 1/2O2 ----) 1H2O ΔH = -242 kJ 3/2O2 ----) 1O3ΔH = -143 kJ Soln: 3 x [1H2 + 1/2O2 ----) 1H2O ΔH = -242 kJ] -1 x [3/2O2 ----) 1O3ΔH = -143 kJ] 3H2 + 3/2O2 + 1O3 -----) 3H2O + 3/2O2ΔHr = 3x-242 + 143 3H2 + 1O3 -----) 3H2O ΔHr = -583 kJ 4. What is ΔHr (kJ/mol) if 1.0 mol of acid reacts with excess NaOH in 2.0 L soln. The solution s = 4.0 J/goC & the temperature goes from 20.0oC to 25.0oC. ΔHr = g x s x ΔT = 2000g x (4.0x10-3 kJ/goC) x 5.0oC = - 40 kJ per 1.0 mole

16. III. Second Law of Thermodynamics A. Introduction to Entropy (S) - S is a measure of the disorder of a system. - Units of S are Joules/K (J/K). - ΔS x T = a valid energy term. - The greater the disorder of a system, the greater is its entropy (ΔS = +) - More randomness/disorderΔS = + ; More orderΔS = - (Know These) B. Second Law Def: The total entropy of a system & its surroundings always increases (more disorder) for a spontaneous process. Examples (Systems on their own go to a state of maximum disorder): - Tires fall apart; Gases diffuse; Messy room; Mixing paint - It requires much more energy to organize than to disorganize. - Note: We usually concentrate on energy; however, in order to order need: 1) Energy; 2) Mechanism/Plan; and 3) A Director

17. III. Second Law of Thermodynamics - Most spontaneous reactions liberate heat; however, some reactions proceed spontaneously and absorb heat! Why? NH4NO3 (solid) ----) NH+4 (aq) + NO3-(aq)ΔHr = Endothermic = + - Why does this go spontaneously when it absorbs heat & becomes cooler! - The problem exists because we are only considering heat energy; we need to also consider the energy involved with the randomness of a changing system. - Energy in a Reaction = ΔH - (T x ΔS). IF TxΔS is large, then total ΔE may be negative (favorable) even if ΔH is positive (endothermic). ΔG = ΔH - (TxΔS) ΔG = Gibbs Free Energy; T = temp in K (Know) ΔG is a measure of the maximum energy available for useful work.

18. C. Calculation of Entropy 1)ΔS = ΔH/T for a system at or near equilibrium (Know Equation) Example: Calc. ΔS (J/K) for 1.0 mol melting ice (0oC). ΔHfus= 6.0kJ/mol ΔS = ΔHfus/T = + 6.0x103 J / 273 K = + 22 J/K (more random) 2)ΔS is a State function so the following applies. ΔSor = ∑Sofproducts - ∑Sofreactants(Know Equation) Get Sof from table of standard entropies such as 18.1, pg 751. Note: Sof is NOT 0.00 (like ΔHof & ΔGof) for element in standard state. Example: Calculate ΔSo for H2O(l) ---) H2O (g)(1 mol of water vaporizing) ΔSo = Sg - Sl = 189 – 70. = +119 J/K (more random) 3) Predict sign of Entropy: 2Na(s) + 2H2O(l) -----) 2NaOH(s) + 1H2(g)Going from 0 mols gas to 1 mol gas; ΔS = + (more random)

19. IV. Third Law of Thermodynamics - 3rd Law: A pure crystalline substance has an entropy of 0.0 at 0 K. - This seems reasonable since a pure crystal at 0 K will be as ordered as possible (all motion stops at 0 K). V. Additional Applications of Thermodynamics A. Predicting Spontaneity of a Reaction ΔG = ΔH - TxΔS (Know This & Signs) If ΔG = - , then reaction is spontaneous (< ≈ -10 kJ) If ΔG = 0, then system is at equilibrium (≈ 0 ± few kJ) If ΔG = +, then reaction in that direction is not spontaneous (> ≈ +10 kJ) Note: Standard State includes: 25oC (298K), 1 atm, 1 M

20. B. Calculate ΔGr by TWO ways for: 1N2(g) + 3H2(g) -----) 2NH3(g) 1)First calculate ΔGor(25 oC) using: ΔGo = ΔHo - TΔSo - Get both ΔHor & ΔSor from table, pg A-8, using ΔHor & ΔSor = ∑products – ∑reactants (put Δsor in kJ) ΔHor = - 92 kJ & ΔSor = - 0.199 kJ/K ΔG = ΔH - T x ΔS ΔGor = -92 kJ - (298 K) (-0.199 kJ/K) = - 33 kJ 2)Nowcalculate ΔGorusing:ΔGor = ∑ΔGofProducts - ∑ΔGofReactants - Use tables of ΔG values: ΔGof for NH3 = -16.4 kJ/mole & ΔGof = 0.00 for element in std state. 1N2(g) + 3H2(g) -----) 2NH3(g) ΔGor= ∑ΔGofProd - ∑ΔGofReact ΔGor= [2x(-16.4)] – [(1x0.00 + 3x0.00)] = - 32.8 kJ

21. C. ΔG & Keq C. Free Energy & Equilibrium Constants - In order to calculate the free energy at conditions other than standard conditions, then use the following: ΔG = ΔGo + RT ln Q (do not need to know this) Where: R = gas constant in J or kJ / (K x mole) R = 8.31 J/(Kxm) = 8.31x10-3kJ/(Kxm) T = temperature in K Q = “equilibrium expression” using initial M

22. C. ΔG & Keq ΔG = ΔGo + RT ln Q & at equil.: ΔG = 0 and Q = Keq 0 = ΔGo + RT ln Keq ΔGo = - RT ln Keq a) ΔGo = - RT ln Keq(Know; 3rd way to calculate ΔGo) b) ln Keq = - ΔGo / RT Keq = inverse ln of [- ΔGo / RT]; Keq = e[-ΔG/RT] R = gas constant = 8.31 J/(mol K) or 8.31x10-3kJ/(mol K) T = temperature in K Examples using a) & b) above on next two slides.

23. C. ΔG & KeqΔGo = - RT ln Keq Calculate ΔGo (kJ/m) at 25oC for: AgCls (-----) Ag+aq + Cl-aq 1. From Ksp (1.8x10-10): ΔGo = - RT x lnKeq ΔGo = - (8.31x10-3kJ/mK)(298 K) x ln (1.8x10-10) ΔGo = - (2.476kJ/m) x -22.44 ΔGo = + 55.6 kJ/m Compare to: ∆GofTable, pg A-9 AgCls (-----) Ag+aq + Cl-aq ∆Gofin kJ/m : -109.8 77.12 -131.3 ΔGor= ∑ΔGofProd- ∑ΔGofReact = (-131.3 + 77.12) – (-109.8) = +55.6 kJ/m

24. C. ΔG & Keq ln Keq = - ΔGo / RT 2) Calculate Keq from ∆Go. - Determine Keq at 25 oC (298 K) for: 2 NO2(g) (----) 1N2O4(g) From Table 18.2:ΔGof (kJ/m) : 2x51.30 97.82 ΔGor = ∑ΔGoproducts - ∑ΔGoreactants ΔGor = (97.82 kJ/m) - (2 x 51.30 kJ/m) = - 4.78 kJ/m Keq=inverse ln [-ΔGo/RT] = e [-ΔG/RT] [-ΔGo/RT]= [ -(- 4.78) / ( 8.31x10-3 x 298) ]kJ/m / kJ/m = 1.93 Keq = e1.93 = 6.89 = 6.9

25. Summary of Ways to Get ΔH, ΔS, ΔG • ΔH 1. ΔH = g x s x ΔT (calorimeter) 2. ΔH = ∑ ∆Hbonds broken - ∑ ∆Hbonds formed 3. ΔH = ∑ ∆Hof products - ∑ ∆Hof reactants 4. Hess’s Law • ΔS 1. ΔS = ΔH / T (only good at equilibrium) 2. ΔSor = ∑Sofproducts - ∑Sofreactants 3. ΔG = ΔH - TxΔS • ΔG 1. ΔG = ΔH - TxΔS 2. ΔG = ∑ΔGofProducts - ∑ΔGofReactants 3. ΔG= - RT ln Keq or Keq = e[-ΔG/RT]