1. Advanced Topics in Chemistry Thermochemistry

2. Objectives 1· Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion 2· Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion 1/2013Dr. Dura 2

3. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Standard Enthalpy Changes in Chemical Reactions Temperature is a measure of the thermal energy. 90 0 C 40 0 C greater thermal energy Temperature = Thermal Energy 1/2013Dr. Dura 3

4. Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing 1/2013Dr. Dura 4

5. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l) 1/2013Dr. Dura 5

6. Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial 1/2013Dr. Dura 6

7. First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.  E system +  E surroundings = 0 or  E system = -  E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings 1/2013Dr. Dura 7

8. Another form of the first law for  E system  E = q + w  E is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -P  V when a gas expands against a constant external pressure 1/2013Dr. Dura 8

9. Work Done On the System w = F x d w = -P  V P x V = x d 3 = F x d = w F d2d2  V > 0 -P  V < 0 w sys < 0 Work is not a state function!  w = w final - w initial initialfinal 1/2013Dr. Dura 9

10. A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P  V (a)  V = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 Latm = 0 joules (b)  V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 Latm w = -14.1 Latm x 101.3 J 1Latm = -1430 J 1/2013Dr. Dura 10

11. Enthalpy and the First Law of Thermodynamics  E = q + w  E =  H - P  V  H =  E + P  V q =  H and w = -P  V At constant pressure: 1/2013Dr. Dura 11

12. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 1/2013Dr. Dura 12

13. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 1/2013Dr. Dura 13

14. Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 1/2013Dr. Dura 14

15. H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ 1/2013Dr. Dura 15

16. H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations H 2 O (l) H 2 O (g)  H = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ 1/2013Dr. Dura 16

17. A Comparison of  H and  E 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g)  H = -367.5 kJ/mol  E =  H - P  V At 25 0 C, 1 mole H 2 = 24.5 L at 1 atm P  V = 1 atm x 24.5 L = 2.5 kJ  E = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol 1/2013Dr. Dura 17

18. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x  t q = C x  t  t = t final - t initial 1/2013Dr. Dura 18

19. Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l)  H = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J Substance  H combustion (kJ/g) Apple-2 Beef-8 Beer-1.5 Gasoline-34 1/2013Dr. Dura 19

20. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (  H 0 ) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 1/2013Dr. Dura 20

21. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 1/2013Dr. Dura 21

22. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 1/2013 Dr. Dura 22

23. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C 6 H 6 1/2013Dr. Dura 23

24. Sample Problem Given the following thermochemical equations: B 2 O 3 (s) + 3 H 2 O (g) → 3 O 2 (g) + B 2 H 6 (g) (ΔH = 2035 kJ/mol) H 2 O (l) → H 2 O (g) (ΔH = 44 kJ/mol) H 2 (g) + (1/2) O 2 (g) → H 2 O (l) (ΔH = -286 kJ/mol) 2 B (s) + 3 H 2 (g) → B 2 H 6 (g) (ΔH = 36 kJ/mol) Determine the ΔH f of: 2 B (s) + (3/2) O 2 (g) → B 2 O 3 (s) 1/2013Dr. Dura 24

25. 1/2013Dr. Dura 25 After the multiplication and reversing of the equations (and their enthalpy changes), the result is: B 2 H 6 (g) + 3 O 2 (g) → B 2 O 3 (s) + 3 H 2 O (g) (ΔH = -2035 kJ/mol) 3 H 2 O (g) → 3 H 2 O (l) (ΔH = -132 kJ/mol) 3 H 2 O (l) → 3 H 2 (g) + (3/2) O 2 (g) (ΔH = 858 kJ/mol) 2 B (s) + 3 H 2 (g) → B 2 H 6 (g) (ΔH = 36 kJ/mol) Adding these equations and canceling out the common terms on both sides, we get 2 B (s) + (3/2) O 2 (g) → B 2 O 3 (s) (ΔH = -1273 kJ/mol)

26. Objectives 3· Define and apply the terms lattice enthalpy and electron affinity 4. Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds 5. Construct a Born-Haber cycle for group 1 and group 2 oxides and chlorides, and use it to calculate an enthalpy change (9.1.12.A.1, 9.1.12.B.1) 6· Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character 1/2013Dr. Dura 26

27. Experimental lattice energies cannot be determined directly. An energy cycle based on Hess’s Law, known as the Born-Haber cycle is used. The Key Concepts in Born-Haber Cycle are: Heat of formation Δ H f Ionization Energy I.E. Electron Affinity E.A. Atomization (Sublimation) s  g Bond enthalpy B.E. Lattice Energy – is the enthalpy change that occurs when one mole of a solid ionic compound is separated into ions under standard conditions. Example: NaCl(s)  Na+ (g) + Cl -(g) (endothermic). Note that in forming ionic compounds, the oppositely charged gaseous ions come together to form an ionic lattice – this is a very exothermic process as there is a strong attraction between the ionsi.e. Na+ (g) + Cl - (g)  NaCl(s) 1/2013Dr. Dura 27

28. Na(s) + ½ Cl 2 (g) NaCl(s) Na(g)Cl(g)  H sub ½ BDE Na + (g)Cl - (g) IEEA HfHf Lattice E. = 786.5 kJ/mol + Born-Haber Cycle for NaCl

29. 1/2013Dr. Dura 29 Construct the Born-Haber cycle for KF and using appropriate data, calculate the electron affinity of fluorine. Thermodynamic data K(s) → K(g) ΔH°a = 90 kJ mol-1 K(g) → K+(g) + e- IE = 419 kJ mol-1 ½ F2(g) → F(g) ΔH°a = 79.5 kJ mol-1 K(s) + F(g) → KF(s) ΔH°f = -569 kJ mol-1 K + (g) + F - (g) → KF(s) ΔH°lattice = -821 kJ mol-1 Choose the correct answer: a) EA = 336.5 kJ mol-1 b) EA = -336.5 kJ mol-1 c) EA = -840.5 kJ mol-1 Answer: b Do NOW

30. 1/2013Dr. Dura 30 Draw the Born-Haber cycle for potassium sulphide K 2 S. Thermodynamic data Potassium K Sulphur S Potassium Sulphide K2S IE I = 419 kJ mol-1 ΔH°a= 279 kJ mol-1 ΔH°f = -257 kJ mol-1 ΔH°a = 78 kJ mol-1 EA I = -199.5 kJ mol-1 ΔH°lattice = -1979 kJ mol EA II - 648.5 kJ mol-1 Practice problem

31. 1/2013 Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous

32. 1/2013 Does a decrease in enthalpy mean a reaction proceeds spontaneously? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H 0 = -890.4 kJ H + (aq) + OH - (aq) H 2 O (l)  H 0 = -56.2 kJ H 2 O (s) H 2 O (l)  H 0 = 6.01 kJ NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H 0 = 25 kJ H2OH2O Spontaneous reactions

33. Objectives 7. State and explain the factors that increase entropy in a system 8· Predict whether the entropy change for a given reaction or process is positive or negative· 9. Calculate the standard entropy change for a reaction using standard entropy values (5.2.12.D.2, 9.1.12.A.1) 9. Calculate the standard entropy change for a reaction using standard entropy values (5.2.12.D.2, 9.1.12.A.1) 1/2013Dr. Dura 33

34. 1/2013 Entropy (S) is a measure of the randomness or disorder of a system. orderS disorder S  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S > 0

35. 1/2013 Processes that lead to an increase in entropy (  S > 0)

36. 1/2013 How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (  S < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (  S < 0) (c) Heating hydrogen gas from 60 0 C to 80 0 C Randomness increases Entropy increases (  S > 0) (d) Subliming dry ice Randomness increases Entropy increases (  S > 0)

37. 1/2013 Entropy State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, enthalpy, pressure, volume, temperature, entropy

38. 1/2013 First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process:

39. 1/2013 Entropy Changes in the System (  S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - The standard entropy of reaction (  S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol

40. 1/2013 Entropy Changes in the System (  S sys ) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes,  S 0 > 0. If the total number of gas molecules diminishes,  S 0 < 0. If there is no net change in the total number of gas molecules, then  S 0 may be positive or negative BUT  S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down,  S is negative.

41. 1/2013 Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. S = k ln W W = 1 where W is the number of microstates. There is only one way to arrange the atoms or molecules to form a perfect crystal. S = 0

42. 1/2013Dr. Dura 42 Objectives 10.Predict whether a reaction or process will be spontaneous using the sign of ΔG 11· Calculate ΔG for a reaction using the equation with enthalpy, temperature, and entropy and by using values of free energy change of formation, ΔG f (9.1.12.A.1) 12. Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and enthalpy changes and the equation

43. 1/2013  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: Gibbs Free Energy For a constant-temperature process:  G =  H sys -T  S sys Gibbs free energy (G)  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium.

44. 1/2013 aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - The standard free-energy of reaction (  G 0 ) is the free- energy change for a reaction when it occurs under standard-state conditions. rxn Standard free energy of formation (  G 0 ) is the free- energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f.  G 0 of any element in its stable form is zero f

45. 1/2013 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous

46. 1/2013  G =  H - T  S