1. Enthalpy and Hess’ Law Yep, it’s a law not a rule... meaning it always works! WOO HOO!

2. Enthalpy (  H), Let’s Review -The thermodynamic variable used to describe the heat of a reaction at constant pressure, q P - The potential thermodynamic energy of a reacting system - The potential energy stored (as heat) in chemical bonds - Exothermic reactions have negative  H rxn values -Typically (but not always) spontaneous reactions have negative values of  H rxn (the heat term is added to the products side) - We express enthalpy for a chemical reaction (  H rxn ) as a stoichiometric component in a thermochemical equation 2H 2 O (l)  2H 2(g) + O 2(g) + 572 kJ - Enthalpy is a state function which means it is independent of path  If we do not know the enthalpy for a reaction we can calculate it using the known enthalpies for smaller reactions

3. Enthalpy (  H), Let’s Review a Little More - When we discuss enthalpy we like to express it in terms of molar enthalpy  The molar enthalpy for the transformation of water into its constituent elements is 286 kJ/mol (not 572 kJ as written in the thermochemical equation... why?) - Enthalpy data is often expressed in tables in a standard state,  H o, this is 101.3 kPa, 298 K (and 1 mol/L for solutions) - We can determine  H rxn by measuring the temperature change of a pure substance to which the heat (energy) of the reaction has been transferred  Process known as calorimetry In a calorimeter: Heat absorbed (heat given off by rxn) = (mc  T) liquid + (mc  T) calorimeter The calorimeter can absorb some heat so choose your material wisely!!!

4. Enthalpy and Heat... Some Problems 1)How much heat energy is required to increase the temperature of 10 g of nickel (specific heat capacity 440 J kg -1 K -1 ) from 323 K to 343 K? 2)The enthalpy of combustion (  H comb ) of ethanol (C 2 H 5 OH) is 1370 kJ/mol. How much heat is released when 0.20 moles of ethanol undergo complete combustion? 3)Consider the following reaction: H 2(g) + ½ O 2(g)  H 2 O (l) + 286 kJ  H rxn = -286 kJ/mol, what mass of O 2(g) must be consumed to produce 1144 kJ of energy?

5. Enthalpy Diagrams

6. Hess’ Law Because enthalpy (  H) and energy (  E) are state functions, we do not concern ourselves with the reaction pathway when determining these values. They are said to be independent of path. Therefore when we cannot measure a reaction to determine its enthalpy change (  H) we can use literature data to calculate these energetic quantities. For a given reaction with an unknown energy term we can calculate  H by assembling the desired reaction from several smaller well-defined ones.

7. Hess’ Law and the Formation of Ethylene (C 2 H 4 ), A Polymer Precursor Ethylene or ethene is the monomer unit of polyethylene commonly known to us as plastic. The overall formation reaction is as follows: 2C (graphite) + 2H 2(g)  C 2 H 4(g)  H rxn = ? *Remember a formation reaction is the formation of a molecule from its elements in their most stable form (H 2 not H). By definition the heat of formation of an element  H form = 0. How do we find  H rxn for this reaction?

8. Hess’ Law and the Formation of Ethylene (C 2 H 4 ), A Polymer Precursor The overall formation reaction is as follows: (a)2C (graphite) + 2H 2(g)  C 2 H 4(g)  H rxn = ? We are given the following known data: (b) C (graphite) +O 2(g)  CO 2(g)  H = -393.5 kJ (c) C 2 H 4(g) + 3O 2(g)  2CO 2(g) + 2H 2 O (g)  H = -1410.9 kJ (d) H 2(g) + ½ O 2(g)  H 2 O (l)  H = -285.8 kJ Let’s begin by finding an equation that will place 2 moles of C(graphite) on the left hand side. We need to multiply (b) by 2 to accomplish this: 2(b) 2C (graphite) +2O 2(g)  2CO 2(g)  H = -787 kJ

9. Hess’ Law and the Formation of Ethylene (a) 2C (graphite) + 2H 2(g)  C 2 H 4(g)  H rxn = ? (b) C (graphite) +O 2(g)  CO 2(g)  H = -393.5 kJ (c) C 2 H 4(g) + 3O 2(g)  2CO 2(g) + 2H 2 O (g)  H = -1410.9 kJ (d) H 2(g) + ½ O 2(g)  H 2 O (l)  H = -285.8 kJ Next, let’s identify an equation that will place the C 2 H 4(g) term on the right hand side. We can achieve this by reversing (c) and the changing the sign of  H. -(c) 2CO 2(g) + 2H 2 O (g)  2C 2 H 4(g) + 3O 2(g)  H = 1410.9 kJ Finally, equation (a) has a 2H 2 term so we need to double (d) to obtain this. 2(d) 2H 2(g) + O 2(g)  2H 2 O (l)  H = -571.6 kJ

10. Hess’ Law and the Formation of Ethylene (C 2 H 4 ), A Polymer Precursor We can now sum (perform a summation) of the 3 known equations and obtain (a), which will yield an new energy term. 2(b) 2C (graphite) +2O 2(g)  2CO 2(g)  H = -787 kJ -(c) 2CO 2(g) + 2H 2 O (g)  2C 2 H 4(g) + 3O 2(g)  H = 1410.9 kJ 2(d) 2H 2(g) + O 2(g)  2H 2 O (l)  H = -571.6 kJ (a)2C (graphite) + 2H 2(g)  C 2 H 4(g)  H = (-787 + 1410.0 – 571.6) kJ  H = 52.3 kJ Yielding the thermochemical equation: 2C (graphite) + 2H 2(g) + 52.3 kJ  C 2 H 4(g) The formation of ethylene is an endothermic reaction.

11. Enthalpy and Hess’ Law This completes our work in chapter 6 of our textbook and roughly our work with enthalpy as the sole energy term for a reacting system. Please review your notes from this section and read chapter 6 if you have not done so.

12. Why do molecules form the way they do? Bond Enthalpies, Hess’ Law, The Born-Haber Cycle, and Heats of Reaction Textbook Reference: Chapter 6 with parts from Chapter 9

13. Molecular Compounds Why does oxygen form O 2 rather than O 8 (more accurately 4O 2 rather than O 8 )?  We know that oxygen is a diatomic, but this is not a reason this is merely an observation of trend.  We need to consider  H BDE (Bond Dissociation Energy) which is the energy required to cleave a covalent bond. BDE O 2 = 498 kJ/mol BDE 4O 2 = 4 x 498 kJ = 1992 kJ Meaning 1992 kJ is required to break 4 moles of O 2 OR 1992 kJ of energy is given off when we form 4 moles of O 2 from O atoms. BDE O—O = 146 kJ/mol BDE O 8 = 8 x 146 kJ = 1168 kJ Meaning only 1168 kJ is given off when we form 8 O—O single bonds in O 8. O 2 is energetically favored.

14. Some other elements to consider Why is phosphorus P 4 rather than 2P 2 ?  P 4 (white phosphorus) is tetrahedral  P—P BDE = 209 kJ  P≡P BDE = 490 kJ Why is sulfur S 8 rather than S 2 ?  This is the converse of oxygen which prefers O 2.  S—S BDE = 266 kJ  S=S BDE = 427 kJ

15. The Ionic Lattice... One More Time The find the lattice energy (  H latt )of an ionic compound we can use the following formula, known as the Born-Lande Equation  H latt = (-LA)(z + )(z - )(e 2 )(1 – 1/n) 4  r Where: L = 6.022 x 10 23 A = Madelung Constant z = summation of charges on the ions e = electron charge = 1.6 x 10 -19 C  = permittivity in a vacuum = 8 x 10 -12 F/m r = distance between the ions n = Born constant

16. Lattice Energy there has to be an easier way... (this would be a pretty lousy slide if there wasn’t) We use what’s called the Born-Haber cycle, which makes use of some specific heats of reaction (  H rxn ).  H f ° ≡ the standard heat of formation of a compound from its elements  H sub ≡ the heat of sublimation (solid  gas)  H BDE ≡ the Bond Dissociation Energy for a covalent bond  H I1 ≡ first ionization energy (neutral atom losing an e -, always positive)  H I2 ≡ second ionization energy (+1 to +2, large and positive)  H EA ≡ electron affinity (always a negative term except Be and N)  H latt ≡ lattice energy (always negative, usually quite large)

17. Formation of NaCl (s) Na (s) + ½ Cl 2(g)  H sub = 107 kJ Na (g) + ½ Cl 2(g)  H BDE = 122 kJ* * BDE Cl 2(g) = 244 kJ, so ½ (244 kJ) = 122 kJ Na (g) + Cl (g)  H I1 = 496 kJ Na + (g) + Cl (g) + e -  H EA = -349 kJ Na + (g) + Cl - (g)  H latt = -787 kJ NaCl (s)  H f ° = ???

18. How do we calculate  H f ° from the Born-Haber Cycle? From our work with Hess’ Law we know that energies are additive.  Therefore we can add up all of the components from the cycle which yield the overall formation reaction (from the elements).  H f NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

19. Formation of NaCl (s) Na (s) + ½ Cl 2(g)  H sub = 107 kJ Na (g) + ½ Cl 2(g)  H BDE = 122 kJ* Na (g) + Cl (g)  H I1 = 496 kJ Na + (g) + Cl (g) + e -  H EA = -349 kJ Na + (g) + Cl - (g)  H latt = -787 kJ NaCl (s)  H f ° = -411 kJ/mol  H f NaCl = 107 + 122 + 496 + (-349) + (-787) = -411 kJ/mol of NaCl

20.  H sub Mg (s)  Mg (g) = 146 kJ/mol  H I1 Mg (g)  Mg + (g) = 738 kJ/mol  H I2 Mg + (g)  Mg 2+ (g) = 1451 kJ/mol  H BDE F 2(g) = 159 kJ/mol of F 2  H EA F = -328 kJ/mol of F  H form MgF 2(s) = -1124 kJ/mol (this is a  H° f ) Determine the lattice energy of MgF 2(s)

21. Lattice Energy of MgF 2(s) Mg (s) + F 2(g)  H sub = 146 kJ Mg (g) + F 2(g)  H BDE = 159 kJ Mg (g) + 2F (g)  H I1 = 738 kJ Mg 2+ (g) + 2F (g) + 2e -  H EA = -656 kJ (2 x -328 kJ) Mg 2+ (g) + 2F - (g)  H latt = ? MgF 2(s)  H f ° = -1124 kJ/mol  H I2 = 1451 kJ Mg + (g) + 2F (g) + e -  H latt =  H f – (  H sub +  H BDE +  H I1 +  H I2 +  H EA ) = -2962 kJ/mol MgF 2(s)

22. Lets leave it here as far as new material...

23. Your Assignment (and no not if you choose to accept it, just accept it) 1)Using your notes and the textbook suggest possible reasons why some reactions are exothermic and some are endothermic (5.4.2) in terms of average bond energy/enthalpy. 2)The combustion of methane is represented by the equation CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l) + 890.3 kJ. a) what mass of CH 4(g) must be burned to give off 1.00 x 10 5 kJ of heat? b) how much heat is produced when 2.78 moles of CO 2(g) are generated? 3) Using standard enthalpies of formation from Appendix B in your textbook calculate the standard enthalpy change for the following reactions: a) NH 3(g) + HCl (g)  NH 4 Cl (s) b) 3C 2 H 2(g)  C 6 H 6(l) c) FeO (s) + CO (g)  Fe (s) + CO 2(g)

24. 4)When burning a Dorito you find that the temperature of 150 g of water in an aluminum (mass 12 g) can is raised by 64 K. What amount of energy was released by the Dorito? You may assume that no heat was lost to the surrounding and it was completely transferred to the can and water. 5)Use the following 2 reactions calculate the  H rxn for 2NO 2(g)  N 2 O 4(g). N 2(g) + 2O 2(g)  N 2 O 4(g) ;  H = 9.2 kJ and N 2(g) + 2O 2(g)  2NO 2(g) ;  H = 33.2 kJ 6)Calculate the enthalpy of reaction: BrCl (g)  Br (g) + Cl (g)  H rxn = ? Using the following data: Br 2(l)  Br 2(g)  H = 30.91 kJ Br 2(g)  2Br (g)  H = 192.9 kJ Cl 2(g)  2Cl (g)  H = 243.4 kJ Br 2(l) + Cl 2(g)  2BrCl (g)  H = 29.2 kJ 7) Question 9.33 from your textbook. Using a Born-Haber cycle for KF to calculate  H EA of fluorine.

25. Tying Up Some Loose Ends... Enthalpy Cycles, Calculation of  H rxn and Variations in Lattice Energy The brocolli must die!

26. Multiple Representations of Enthalpy So far we have looked at 2 ways of representing the energy/enthalpy term for a chemical reaction: 1) as a component in a thermochemical equation 2) as a term outside the equation, calculated from a formula such as q = mc  T 3) as an enthalpy diagram  a graphical way to show the change in enthalpy rather than relying solely on equations  this in NOT a commonly used method and it will be addressed purely as the presentation of a third option to enthalpy problems, I strongly recommend you use equations and Hess’ Law to solve enthalpy problems

27. What is an Enthalpy Diagram? A diagram that shows the overall and net reaction steps with their corresponding energy terms for a chemical reaction. Consider below and the combustion of methane.

28. What is an Enthalpy Cycle? Let’s consider an alternative route to finding the  H comb of methane. ΔH 1 = enthalpy change for bond breaking = 4 x E(C-H) + 2 x E(O=O) = 4 x 413 + 2 x 498 = + 2648 kJ mol -1 ΔH 2 = enthalpy change for bond making = - [ 2 x E(C=O) + 4 x E(O-H) ] = - [ 2 x 805 + 4 x 464 ] = - 3466 kJ mol-1  H rxn =  H 1 +  H 2 = -818 kJ/mol

29. What is an Enthalpy Cycle? A B C H2H2 H1H1 H3H3  H 1 =  H 2 +  H 3

30. Variations in Lattice Energy Sometimes when we calculate a lattice energy and then compare it to our experimental value, there is a difference. If this difference is large enough between the theoretical and experimental values this indicates that rather than the lattice being totally ionic, there is a significant degree of covalent character to each interaction.  Yes this is a return to the bonding continuum! Consider the following: NaClAgCl Theoretical Value (kJ/mol) 766770 Experimental Value (kJ/mol) 771905 AgCl has significant covalent character! Is there another way we could have identified this phenomenon?