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THERMOCHEMISTRY OR THERMODYNAMICS

THERMOCHEMISTRY OR THERMODYNAMICS. Chapter 6. Chemical Reactivity. What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS . We have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED .

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THERMOCHEMISTRY OR THERMODYNAMICS

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  1. THERMOCHEMISTRY OR THERMODYNAMICS Chapter6

  2. Chemical Reactivity • What drives chemical reactions? How do they occur? • The first is answered by THERMODYNAMICS and the second by KINETICS. • We 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 in a battery

  3. Energy and Chemistry • ENERGY is the capacity to do work or transfer heat. • HEAT is the form of energy that flows between 2 samples because of their difference in temperature. • Other forms of energy — • lightelectrical nuclear • kinetic potential

  4. Law of Conservation of Energy • Energy can be converted from one form to another but can neither be created nor destroyed. • (Euniverse is constant)

  5. The Two Types of Energy • Potential: due to position or composition - can be converted to work • PE = mgh • (m = mass, g = acceleration of gravity, and h = height) • Kinetic: due to motion of the object • KE = 1/2 mv2 • (m = mass, v = velocity)

  6. Kinetic and Potential Energy • Potential energy — energy a motionless body has by virtue of its position.

  7. Kinetic and Potential Energy Kinetic energy — energy of motion. • Translation • Rotation • Vibration

  8. Units of Energy • 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. • 1000 cal = 1 kilocalorie = 1 kcal • 1 kcal = 1 Calorie (a food “calorie”) • But we use the unit called the JOULE • 1 cal = 4.184 joules James Joule 1818-1889

  9. Temperature v. Heat • Temperature reflects random motions of particles, therefore related to kinetic energy of the system. • Heat involves a transfer of energy between 2 objects due to a temperature difference

  10. Extensive properties depends directly on the amount of substance present. mass volume heat heat capacity (C) Intensive properties is not related to the amount of substance. temperature concentration pressure specific heat (s) Extensive & Intensive Properties

  11. State Function • Depends only on the present state of the system - not how it arrived there. • It is independent of pathway. • Energy change is independent of the pathway (and, therefore, a state function), while work and heat are dependent on the pathway.

  12. System and Surroundings • System: That on which we focus attention • Surroundings: Everything else in the universe • Universe = System + Surroundings

  13. Exo and Endothermic • Heat exchange accompanies chemical reactions. • Exothermic: Heat flows out of the system (to the surroundings). • Endothermic: Heat flows into the system (from the surroundings).

  14. ENERGY DIAGRAMS Exothermic Endothermic

  15. Endo- and Exothermic Surroundings System heat qsystem > 0 w > 0 E goes up ENDOTHERMIC

  16. Endo- and Exothermic Surroundings Surroundings System System heat heat qsystem > 0 w > 0 qsystem < 0 w < 0 E(system) goes down E(system) goes up ENDOTHERMIC EXOTHERMIC

  17. First Law • First Law of Thermodynamics: The energy of the universe is constant.

  18. Enthalpy • DH = Hfinal - Hinitial • If Hfinal > Hinitial then DH is positive • Process is ENDOTHERMIC • If Hfinal < Hinitial then DH is negative • Process is EXOTHERMIC

  19. First Law • E = q + w • E = change in system’s internal energy • q = heat • w = work

  20. Piston moving a distance against a pressure does work.

  21. Work • w & V must have opposite signs, since work is being done by the system to expand the gas. • wsystem = -P V • 1 Latm = 101.3 J • 1 J = kgm2/s2 P = F/A F = PA w = F h w = PA h w = P V

  22. Enthalpy • Enthalpy = H = E + PV • E = HPV • H = E + PV • At constant pressure, qP = E + PV, where qP = H at constant pressure • H = energy flow as heat (at constant pressure)

  23. Some Heat Exchange Terms • specific heat capacity (s) heat capacity per gram = J/°C g or J/K g • molar heat capacity (s) heat capacity per mole = J/°C mol or J/K mol

  24. Specific Heat Capacity • Substance Spec. Heat (J/g•K) • H2O 4.184 • Al 0.902 • glass 0.84 Aluminum

  25. Ho = - qp qp = mst q = heat (J) m = mass (g) s = specific heat (j/gCo) t = “change” in temperature (Co) Ho = “change in” enthalpy (kJ) Simple Calorimeter

  26. Specific Heat Capacity • If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? • where DT = Tfinal - Tinitial heat gain/lost = q = m s DT

  27. Specific Heat Capacity • If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? • where DT = Tfinal - Tinitial • q = (0.902 J/g•K)(25.0 g)(37 - 310)K • q = - 6160 J heat gain/lost = q = m s DT

  28. Specific Heat Capacity • If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? • q = - 6160 J • Notice that the negative sign on q signals heat “lost by” or transferred out of Al.

  29. Bomb Calorimeter

  30. Heat Capacity E = qv & qv = -(C t + ms t) E = “change in” internal energy (J) qv = heat at constant volume (J) C = heat capacity (J/Co) t = “change”in temperature (Co)

  31. REMEMBER!!! • In regular calorimetry pressure is constant, but the volume will change so: • qp = -H • qp =E + p V • In bomb calorimetry, volume is constant so: • qv = E • since p V = zero.

  32. 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 • Hcm = -(Ct + mst) where Hc is heat of combustion.

  33. 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 = C 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

  34. Hess’s Law • Reactants  Products • The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

  35. Standard States • Compound • For a gas, pressure is exactly 1 atmosphere. • For a solution, concentration is exactly 1 molar. • Pure substance (liquid or solid), it is the pure liquid or solid. • Element • The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.

  36. Calculations via Hess’s Law • 1. If a reaction is reversed, H is also reversed. • N2(g) + O2(g)  2NO(g) H = 180 kJ • 2NO(g)  N2(g) + O2(g) H = 180 kJ • 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. • 6NO(g) 3N2(g) + 3O2(g) H = 540 kJ

  37. Using Enthalpy • Consider the decomposition of water • H2O(g) + 243 kJ ---> H2(g) + 1/2 O2(g) • Endothermic reaction — heat is a “reactant” • DH = + 243 kJ

  38. Using Enthalpy • Making H2 from H2O involves two steps. • H2O(l) + 44 kJ ---> H2O(g) • H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g) • ----------------------------------------------------------------- • H2O(l) + 286 kJ --> H2(g) + 1/2 O2(g) • Example of HESS’S LAW— • If a rxn. is the sum of 2 or more others, the net DH is the sum of the DH’s of the other rxns.

  39. Using Enthalpy • Calc. DH for S(s) + 3/2 O2(g) --> SO3(g) • S(s) + O2(g) --> SO2(g) -320.5 kJ • SO2(g) + 1/2 O2(g) --> SO3(g) -75.2 kJ • _______________________________________ • S(s) + 3/2 O2(g) --> SO3(g) -395.7 kJ

  40. energy S solid direct path DH = 1 +O 2 -320.5 kJ + 3/2 O 2 DH = SO gas 2 -395.7 kJ + 1/2 O 2 SO gas 3 DH = -75.2 kJ 2 DH along one path = DH along another path

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

  42. Standard Enthalpy Values • NIST (Nat’l Institute for Standards and Technology) gives values of • DHof = standard molar enthalpy of formation • This is the enthalpy change when 1 mol of compound is formed from elements under standard conditions. DHof is always stated in terms of moles of product formed. • See Appendix A21-A24.

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

  44. Using Standard Enthalpy Values • Use DH°’s to calculate enthalpy change for • H2O(g) + C(graphite) --> H2(g) + CO(g) • (product is called “water gas”)

  45. 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) • DH°f of H2O vapor = - 242 kJ/mol • C(s) + 1/2 O2(g) --> CO(g) • DH°f of CO = - 111 kJ/mol

  46. Using Standard Enthalpy Values • H2O(g) --> H2(g) + 1/2 O2(g) DHo = +242 kJ • C(s) + 1/2 O2(g) --> CO(g) DHo = -111 kJ • ----------------------------------------------------------------- • H2O(g) + C(graphite) --> H2(g) + CO(g) • DHonet = +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.

  47. Change in Enthalpy • Can be calculated from enthalpies of formation of reactants and products. • Hrxn° = npHf(products) nrHf(reactants) • H is an extensive property--kJ/mol • For the reaction: 2H2(g) + O2 (g) ---> 2H2O(g) • Enthalpy would be twice the H value for the combustion of hydrogen.

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

  49. Using Standard Enthalpy Values • CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) • DHorxn = DHof (prod) - DHof (react) • DHorxn = DHof (CO2) + 2 DHof (H2O) • - {3/2 DHof (O2) + DHof (CH3OH)} • = (-393.5 kJ) + 2 (-241.8 kJ) • - {0 + (-201.5 kJ)} • DHorxn = -675.6 kJ per mol of methanol • DHorxn is always in terms of moles of reactant.

  50. Pathway for the Combustion of Methane

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