2. 1 THERMOCHEMISTRY Thermodynamics The study of Heat and Work and State Functions

3. 2 Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — lightlight electricalelectrical kinetic and potentialkinetic and potential

4. 3 Potential & Kinetic Energy Potential energy — energy a motionless body has by virtue of its position.

5. 4 Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy Potential Energy on the Atomic Scale NaCl — composed of Na + and Cl - ions.

6. 5 Positive and negative particles (ions) attract one another. Two atoms can bond As the particles attract they have a lower potential energy Potential Energy on the Atomic Scale

7. 6 Potential & Kinetic Energy Kinetic energy — energy of motion Translation Translation

8. 7 Potential & Kinetic Energy Kinetic energy — energy of motion.

9. 8 Internal Energy (E) PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U) Int. E of a chemical system depends onInt. E of a chemical system depends on number of particlesnumber of particles type of particlestype of particles temperaturetemperature

10. 9 ThermodynamicsThermodynamics Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions. Heat transfers until thermal equilibrium is established.

11. 10 Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction.The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE.If PE of products is less than reactants, the difference must be released as KE.

12. 11 Energy Change in Chemical Processes PE of system dropped. KE increased. Therefore, you often feel a T increase.

13. 12 Heat Transfer No Change in State q transferred = (sp. ht.)(mass)(∆T)

14. 13 Heat Transfer with Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water q = (heat of fusion)(mass)

15. 14 Heat Transfer and Changes of State Requires energy (heat). This is the reason a)you cool down after swimming b)you use water to put out a fire. + energy Liquid ---> Vapor

16. 15 Heat water Evaporate water Melt ice Heating/Cooling Curve for Water Note that T is constant as ice melts

17. 16 Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/gK Heat of vaporization = 2260 J/g Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/gK Heat of vaporization = 2260 J/g What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 o C? Heat & Changes of State +333 J/g +2260 J/g

18. 17 What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 o C? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 10 5 J q = (500. g)(333 J/g) = 1.67 x 10 5 J 2.To raise water from 0 o C to 100 o C q = (500. g)(4.2 J/gK)(100 - 0)K = 2.1 x 10 5 J q = (500. g)(4.2 J/gK)(100 - 0)K = 2.1 x 10 5 J 3.To evaporate water at 100 o C q = (500. g)(2260 J/g) = 1.13 x 10 6 J q = (500. g)(2260 J/g) = 1.13 x 10 6 J 4. Total heat energy = 1.51 x 10 6 J = 1510 kJ Heat & Changes of State

19. 18 Heat Energy Transfer in a Physical Process CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C) Heat transfers from surroundings to system in endothermic process.

20. 19 Heat Energy Transfer in a Physical Process CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C)CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C) A regular array of molecules in a solid -----> gas phase molecules.A regular array of molecules in a solid -----> gas phase molecules. Gas molecules have higher kinetic energy.Gas molecules have higher kinetic energy.

21. 20 Energy Level Diagram for Heat Energy Transfer Transfer ∆E = E(final) - E(initial) = E(gas) - E(solid) CO 2 solid CO 2 gas

22. 21 Heat Energy Transfer in Physical Change Gas molecules have higher kinetic energy.Gas molecules have higher kinetic energy. Also, WORK is done by the system in pushing aside the atmosphere.Also, WORK is done by the system in pushing aside the atmosphere. CO 2 (s, -78 o C) ---> CO 2 (g, -78 o C) Two things have happened!

23. 22 FIRST LAW OF THERMODYNAMICS ∆E = q + w heat energy transferred energychange work done by the system Energy is conserved!

24. 23 ENTHALPYENTHALPY Most chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial Heat transferred at constant P = q p q p = ∆H where H = enthalpy Heat transferred at constant P = q p q p = ∆H where H = enthalpy

25. 24 If H final < H initial then ∆H is negative Process is EXOTHERMIC If H final < H initial then ∆H is negative Process is EXOTHERMIC If H final > H initial then ∆H is positive Process is ENDOTHERMIC If H final > H initial then ∆H is positive Process is ENDOTHERMIC ENTHALPYENTHALPY ∆H = H final - H initial

26. 25 Consider the formation of water H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) + 241.8 kJ USING ENTHALPY Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

27. 26 Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. USING ENTHALPY H 2 + O 2 gas Liquid H 2 OH 2 O vapor

28. 27 Making H 2 O from H 2 involves two steps. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g) + 242 kJ H 2 O(g) ---> H 2 O(liq) + 44 kJ ----------------------------------------------------------------------- H 2 (g) + 1/2 O 2 (g) --> H 2 O(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. USING ENTHALPY

29. 28 Hess’s Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed. Figure 6.18, page 227

30. 29 Hess’s Law & Energy Level Diagrams Forming CO 2 can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed. Figure 6.18, page 227

31. 30 This equation is valid because ∆H is a STATE FUNCTIONThis equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not how it got there.These depend only on the state of the system and not how it got there. V, T, P, energy — and your bank account!V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.  ∆H along one path =  ∆H along another path  ∆H along another path  ∆H along one path =  ∆H along another path  ∆H along another path

32. 31 Standard Enthalpy Values Most ∆H values are labeled ∆H o Measured under standard conditions P = 1 atm Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas

33. 32 Enthalpy Values H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H˚ = -242 kJ 2 H 2 (g) + O 2 (g) --> 2 H 2 O(g) ∆H˚ = -484 kJ H 2 O(g) ---> H 2 (g) + 1/2 O 2 (g) ∆H˚ = +242 kJ H 2 (g) + 1/2 O 2 (g) --> H 2 O(liquid) ∆H˚ = -286 kJ Depend on how the reaction is written and on phases of reactants and products

34. 33 Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆H f o = standard molar enthalpy of formation ∆H f o = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table Appendix A.2

35. 34 ∆H f o, standard molar enthalpy of formation H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f o (H 2 O, g)= -241.8 kJ/mol By definition, ∆H f o = 0 for elements in their standard states.

36. 35 Using Standard Enthalpy Values Use ∆H˚’s to calculate enthalpy change for H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) (product is called “water gas”)

37. 36 Using Standard Enthalpy Values H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) From reference books we find H 2 (g) + 1/2 O 2 (g) --> H 2 O(g)H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f ˚ of H 2 O vapor = - 242 kJ/mol ∆H f ˚ of H 2 O vapor = - 242 kJ/mol C(s) + 1/2 O 2 (g) --> CO(g)C(s) + 1/2 O 2 (g) --> CO(g) ∆H f ˚ of CO = - 111 kJ/mol ∆H f ˚ of CO = - 111 kJ/mol

38. 37 Using Standard Enthalpy Values H 2 O(g) --> H 2 (g) + 1/2 O 2 (g) ∆H o = +242 kJ C(s) + 1/2 O 2 (g) --> CO(g)∆H o = -111 kJ C(s) + 1/2 O 2 (g) --> CO(g) ∆H o = -111 kJ -------------------------------------------------------------------------------- To convert 1 mol of water to 1 mol each of H 2 and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic. H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) ∆H o net = +131 kJ ∆H o net = +131 kJ H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) ∆H o net = +131 kJ ∆H o net = +131 kJ

39. 38 Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known, Calculate ∆H of reaction? ∆H o rxn =  ∆H f o (products) -  ∆H f o (reactants) -  ∆H f o (reactants) ∆H o rxn =  ∆H f o (products) -  ∆H f o (reactants) -  ∆H f o (reactants) Remember that ∆ always = final – initial

40. 39 Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆H o rxn for CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react)

41. 40 Using Standard Enthalpy Values ∆H o rxn = ∆H f o (CO 2 ) + 2 ∆H f o (H 2 O) - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} = (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} - {0 + (-201.5 kJ)} ∆H o rxn = -675.6 kJ per mol of methanol CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react)

42. 41 Calorimetry Some heat from reaction warms water q water = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” q bomb = (heat capacity, J/K)(∆T) Total heat evolved = q total = q water + q bomb

43. 42 Calculate heat of combustion of octane. C 8 H 18 + 25/2 O 2 --> 8 CO 2 + 9 H 2 O Burn 1.00 g of octaneBurn 1.00 g of octane Temp rises from 25.00 to 33.20 o CTemp rises from 25.00 to 33.20 o C Calorimeter contains 1200 g waterCalorimeter contains 1200 g water Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K Measuring Heats of Reaction CALORIMETRY

44. 43 Step 1Calc. heat transferred from reaction to water. q = (4.184 J/gK)(1200 g)(8.20 K) = 41,170 J Step 2Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J = (837 J/K)(8.20 K) = 6860 J Step 3Total heat evolved 41,170 J + 6860 J = 48,030 J 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ Measuring Heats of Reaction CALORIMETRY