Warm up

1. Warm up Complete the following reactions • A warm solution of ammonium chloride is combined with a warm solution of sodium carbonate • Sulfur trioxide is bubbled through water • Potassium carbonate is heated

2. Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered in part by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer (Redox) in a battery

3. First, Some Terms Exothermic – process that releases energy to the environment Endothermic – process that absorbs energy from the environment Kinetic Energy – energy of motion (thermal, mechanical, electrical, sound) Potential Energy – energy that results from an object’s position (chemical, gravitational, electrostatic)

4. Chemical Reactivity But energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings (exothermic) are product-favored. So, let us consider heat transfer in chemical processes.

5. Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Heat transfers from surroundings to system in endothermic process.

6. Heat Energy Transfer in a Physical Process • CO2 (s, -78 oC) ---> CO2 (g, -78 oC) • A regular array of molecules in a solid -----> gas phase molecules. • Gas molecules have higher kinetic energy.

7. CO2 gas ∆E = E(final) - E(initial) = E(gas) - E(solid) CO2 solid Energy Level Diagram for Heat Energy Transfer

8. Heat Energy Transfer in Physical Change • Gas molecules have higher kinetic energy. • Also, WORKis done by the system in pushing aside the atmosphere. CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Two things have happened!

9. heat energy transferred work done by the system energy change FIRST LAW OF THERMODYNAMICS ∆E = q + w Energy is conserved!

10. heat transfer in (endothermic), +q heat transfer out (exothermic), -q w transfer in (+w) w transfer out (-w) SYSTEM ∆E = q + w

11. What the….? • So, I’ll skip the derivation but… • “H” is a measure of the heat content of a system • When a reaction is carried out at constant pressure H = q ≈ ∆E • And if no work is allowed to be done… • ∆E = ∆H • What does this mean? • When measuring energy changes of a system, conditions are set so that the energy change is due completely to heat loss or gain. • Therefore ∆H = Hfinal - Hinitial

12. ENTHALPY ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC

13. Warm-up • Write a heat of formation reaction for methane gas (CH4 (g)). • You dissolve sodium bicarbonate in water and when you touch the container it feels cold to you. If the system is the dissolving of sodium carbonate, is this process exothermic or endothermic?

14. USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

15. USING ENTHALPY Making liquid H2O from H2 + O2 involves twoexothermic steps. H2 + O2 gas H2O vapor Liquid H2O

16. USING ENTHALPY Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJ H2O(g) ---> H2O(liq) + 44 kJ ----------------------------------------------------------------------- H2(g) + 1/2 O2(g) --> H2O(liq) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

17. Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Figure 6.18, page 227

18. Hess’s Law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Figure 6.18, page 227

19. ∆H along one path = ∆H along another path • This equation is valid because ∆H is a STATE FUNCTION • These depend only on the state of the system and not how it got there. • V, T, P, energy — and your bank account! • Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.

20. Hess’s Law • The enthalpy change for a chemical reaction leading from reactants to products is the same no matter which route is chosen for the reaction. http://michele.usc.edu/105a/thermochemistry/hess.html

21. Hess’s Law • Addition of chemical equations can lead to a net equation. • If the chemical equation is reversed, the sign of ΔH changes. • If coefficients are multiplied, multiply ΔH by the same amount.

22. Hess’s Law Example #1 Before pipelines were built to deliver natural gas, individual towns and cities contained plants that produced a fuel known as town gas by passing steam over red-hot charcoal. C(s) + H2O(g) ==> CO(g) + H2(g) Calculate ∆H for this reaction from the following information. C(s) + 1/2 O2(g) ==> CO(g) ∆H = -110.53 kJ/mol H2(g) + 1/2 O2(g) ==> H2O(g) ∆H = -241.82 kJ/mol

23. Hess’s Law Example #2 • Use Hess's law to calculate ∆H for the reaction MgO(s) + CO2(g) ==> MgCO3(s) Mg(s) + 1/2 O2(g) ==> MgO(s) ∆H= -601.70 kJ/mol C(s) + O2(g) ==> CO2(g) ∆H = -393.51 kJ/mol Mg(s) + C(s) + 3/2 O2(g) ==> MgCO3(s) ∆H = -1095.8 kJ/mol

24. HW: • Topic Hess’s Law • Answer questions 35-40 on p. 273

25. Standard Enthalpy Values Most ∆H values are labeled ∆Ho Measured under standard conditions P = 1 bar Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas

26. Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 2 H2(g) + O2(g) --> 2 H2O(g) ∆H˚ = -484 kJ H2O(g) ---> H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ H2(g) + 1/2 O2(g) --> H2O(liquid) ∆H˚ = -286 kJ

27. Problem How much heat is needed to vaporize one mole of H2O (liquid)? H2(g) + ½ O2(g) H2O(l) H = - 285.9 kJ H2(g) + ½ O2(g) H2O(g) H = - 242.0 kJ

28. Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆Hfo = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L

29. ∆Hfo, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) ∆Hfo (H2O, g)= -241.8 kJ/mol By definition, ∆Hfo= 0 for elements in their standard states.

30. (product is called “water gas”) Using Standard Enthalpy Values Use ∆H˚’s to calculate enthalpy change for H2O(g) + C(graphite) --> H2(g) + CO(g)

31. Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find • H2(g) + 1/2 O2(g) --> H2O(g) ∆Hf˚ of H2O vapor = - 242 kJ/mol • C(s) + 1/2 O2(g) --> CO(g) ∆Hf˚ of CO = - 111 kJ/mol

32. Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ C(s) + 1/2 O2(g) --> CO(g)∆Ho = -111 kJ -------------------------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) ∆Honet = +131 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic.

33. Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known, Calculate ∆H of reaction? ∆Horxn =  ∆Hfo(products) -  ∆Hfo(reactants) Remember that ∆ always = final – initial

34. Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆Horxn for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn =  ∆Hfo(prod) -  ∆Hfo(react)

35. Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn =  ∆Hfo(prod) -  ∆Hfo(react) ∆Horxn = ∆Hfo(CO2) + 2 ∆Hfo(H2O) - {3/2 ∆Hfo(O2) + ∆Hfo(CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} ∆Horxn = -675.6 kJ per mol of methanol

36. Problem Consider the equations for these reactions: ½ N2 + ½ O2 NO ΔHf = +90.4 kJ NO + ½ O2 NO2ΔHf= -56.5 kJ All reactants are gaseous. What is the heat of formation, ΔHf, for NO2 gas?

37. Problem • A table of heats (enthalpies) of formation is given. Compound ΔHf (kJ·mol–1) • H2S(g) –21 • O2(g) 0 • H2O(l) –285 • SO2(g) –297 What is the heat (enthalpy) of reaction, ΔH, for this equation? 2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g)

38. Book problems: • Topic Heat of formation • Answer questions 43-48 on p. 274

39. HW • Webassign Username is student ID and password is atom is due by Friday… yep tomorrow.

40. Calorimetry introduction • Specific heat (an intensive property) • Heat released by a metal demo • BEST DEMO EVER… • Specific heat work • Heat of reaction for neutralization

41. CALORIMETRY Measuring Heats of Reaction Constant Volume “Bomb” Calorimeter • Burn combustible sample. • Measure heat evolved in a reaction. • Derive ∆E for reaction.

42. Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Calorimetry Total heat evolved = qtotal = qwater + qbomb

43. Measuring Heats of ReactionCALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200 g water • Heat capacity of bomb = 837 J/K

44. Measuring Heats of ReactionCALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ

45. Measuring Heats of ReactionCALORIMETRY • Is this the ΔH for this reaction? • Why or why not? • Right now our units are – 48.0kJ/g but we need a value that represents kJ/mole • Can you figure out how to use dimensional analysis (canceling of units) to give us the correct units? • Raise your hand when you have the answer • Answer: - 5480kJ/mole

46. HW • From book • #19-33 (there are different types of problems here, at a minimum, do some of each type)