1 / 33

q = (10.0g  2.09J/g o C  15.0 o C) + (10.0g  333J/g)

EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0 o C to steam at 127.0 o C ? q overall = q ice + q fusion + q water + q boil + q steam. q = (10.0g  2.09J/g o C  15.0 o C) + (10.0g  333J/g) + (10.0g  4.18J/g o C  100.0 o C) + (10.0g  2260J/g)

ghita
Download Presentation

q = (10.0g  2.09J/g o C  15.0 o C) + (10.0g  333J/g)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC?qoverall = qice + qfusion + qwater + qboil + qsteam q = (10.0g  2.09J/goC  15.0oC) + (10.0g  333J/g) + (10.0g  4.18J/goC  100.0oC) + (10.0g  2260J/g) + (10.0g  2.03J/goC  27.0oC) q = (314 + 3.33×103 + 4.18×103 + 2.26×104 + 548)J = 30.9 kJ

  2. Heat Flow in Reactions exothermic –reaction that gives off energy q < 0 isolated system E=0 heat released by reaction raises the temperature of the solvent constant T, heat is released to the surroundings endothermic – reaction that absorbs energy q > 0

  3. Expansion Type Work w = -PDV system does work DV = Vfinal - Vinitial P V P Vinitial qp = +2kJ

  4. 130 10 0 0 Do 250 J of work to compress a gas, 180 J of heat are released by the gasWhat is E for the gas? • 430 J • 70 J • -70 J • -180 J • -250 J

  5. Enthalpy H DE = q + w at constant V, wexpansion = 0 DE = qv at constant P, wexpansion = -PDV DE = qp - PDV DefineH=E + (PV) = E + PV at constant P Hence DH = qp

  6. Enthalpy Enthalpy heat at constant pressure or the heat of reaction qp = DH = Hproducts - Hreactants Exothermic Reaction DH = (Hproducts - Hreactants) < 0 2 H2(g) + O2(g) 2 H2O(l)DH < 0 Endothermic Reaction DH = (Hproducts - Hreactants) > 0 2 H2O(l) 2 H2(g) + O2(g)DH > 0

  7. State Functions • H and E along with P, T, V (or P, T, V) and many others are state functions. They are the same no matter what path we take for the change. • q and w are not state functions, they depend on which path we take between two points. q initial E=Efinal-Einitialq and w can be anything w q E w final

  8. Path Independent Energy Changes

  9. 130 10 0 0 Which day would you like OWL quizzes due (4 AM) • Monday • Tuesday • Wednesday • Thursday • Friday

  10. Stepwise Energy Changesin Reactions

  11. Laws of Thermochemistry 1. The magnitude of DH is directly proportional to the amount of reaction. H is for 1 mole of reaction as written 2 H2(g) + O2(g) 2 H2O(l)DH = -571.6 kJ H2(g) + ½ O2(g) H2O(l)DH = -285.8 kJ Can have ½ mole O2 just not ½ molecule

  12. Laws of Thermochemistry 2. DH for a reaction is equal in magnitude but opposite in sign to DH for the reverse reaction. H2(g) + ½ O2(g) H2O(l)DH = -285.8 kJ H2O(l) H2(g) + ½ O2(g)DH = +285.8 kJ

  13. Laws of Thermochemistry 3. The value of H for the reaction is the same whether it occurs directly or in a series of steps. DHoverall = DH1 + DH2 + DH3 + · · · also called Hess’ Law

  14. Enthalpy Diagram H2(g) + ½ O2(g)  H2O(l) DH = -285.8 kJ H2O(l)  H2O(g) DH = +44.0 kJ H2(g) + ½ O2(g)  H2O(g) DH = -241.8 kJ

  15. 130 10 0 0 Given 3 CO + 3/2 O2 3 CO2 H = -849 kJWhat is H for CO2 CO + ½ O2 ? • -283 kJ • +283 kJ • +849 kJ • -2547 kJ • +2547 kJ

  16. Energy and Stoichiometry • Since H is per mole of reaction we can relate heat to amount of reaction • Given C2H6 + 7/2 O2  2 CO2 + 3 H2O H=-1559.7 kJ • If 632.5 kJ are released to surroundings what mass of H2O is formed? • 632.5 kJ released means H = -632.5 kJ for this much H2O

  17. Bomb Calorimeter measure qv qrxn + qcal = 0 qrxn = -qcal qrxn = - ccalT Erxn = qrxn/moles rxn Erxn≈ Hrxn H = E + (PV) H = E + RTngas @298K RT = 2.5 kJ/mol

  18. “Coffee Cup” Calorimeter qp Photo by George Lisensky

  19. Measuring H • When 25.0 mL of 1.0 M H2SO4 are added to 50.0 mL of 1.0 M KOH, both initially at 24.6C the temperature rises to 33.9C. What is H for H2SO4 + 2 KOH  K2SO4 + 2 H2O ? (Assume d = 1.00 g/mL, c = 4.18 J/g.C) • qsoln = mcT • m = (25.0 + 50.0)mL×1.00g/mL = 75.0 g

  20. Measuring H cont. • q=mcT • qsoln = 75.0 g × 4.18 J/g.C × (33.9-24.6)C • qsoln = 2916 J • qrxn + qsoln = 0 • qrxn = -2916 J • Hrxn = qrxn/moles rxn

  21. Measuring H cont • How many moles rxn? • 1 mol rxn / 1 mol H2SO4 • 1 mol rxn / 2 mol KOH Stoichiometric mixture so 0.025 mol rxn

  22. Measuring H cont • Hrxn = qrxn/moles rxn • Hrxn = -2916 J / 0.025 mol rxn • Hrxn = -116622 J / mol rxn • Hrxn = -117 kJ • His per mole of reaction as written

  23. 130 10 0 0 If excess Al is added to 50 mL of 0.250 M H2SO4 how many moles of the following reaction occur?2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2 • 0.0125 mol • 0.0375 mol • 0.025 mol • 0.00625 mol • 0.00417 mol

  24. Hess’s Law • Can find H for an unknown, or hard to measure, reaction by summing measured H values of known reactions.

  25. EXAMPLE H for formation of CO cannot readily be measured since a mixture of CO and CO2 is always formed. C (s) + ½ O2 (g)  CO (g) H = ? C (s) + O2 (g)  CO2 (g) H1 = -393.5 kJ CO (g) + ½ O2 (g)  CO2 (g) H2 = -283.0 kJ C (s) + ½ O2 (g)  CO (g) H = H1 - H2 H = H1 - H2 = -393.5 – (-283.0) = -110.5 kJ

  26. Standard Enthalpy of Formation the enthalpy associated with the formation of 1 mol of a substance from its constituent elements under standard state conditions at the specified temperature For an element this is a null reaction O2 (g)  O2 (g) H = 0 Hf = 0 for all elements in their standard states

  27. 130 10 0 0 For which one of these reactions is ΔHºrxn = ΔHºf? • N2(g) + 3 H2(g)  2 NH3(g) • C(graphite) + 2 H2(g)  CH4(g) • C(diamond) + O2(g)  CO2(g) • CO(g) + ½O2(g)  CO2(g) • H2(g) + Cl2(g) 2 HCl(g)

  28. Calculation of DHo DHo = Smols  DHfoproducts – Smols  DHforeactants We can always convert products and reactants to the elements. Hess’s law says H is the same whether we go directly from reactants to products or go via elements

  29. ExampleWhat is the value of DHrxn for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g) from Appendix J Text C6H6(l)DHfo = + 49.0 kJ/mol O2(g) DHfo = 0 CO2(g) DHfo = - 393.5 H2O(g)DHfo = - 241.8 D Hrxn = [S mols  D Hfo]product –[S mols  D Hfo]reactants

  30. ExampleWhat is the value of DHrx for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)from Appendix J TextC6H6(l)DHfo = + 49.0 kJ/mol; O2(g) DHfo = 0CO2(g) DHfo = - 393.5; H2O(g)DHfo = - 241.8D Hrxn = [S mols D Hfo]product - [S mols  D Hfo]reactants D Hrxn = [12(- 393.5) + 6(- 241.8)]product - [2(+ 49.0 ) + 15(0)]reactants kJ/mol = - 6.2708  103 kJ

  31. Fossil Fuels coal petroleum natural gas

  32. Energy Resources in the U.S.

  33. Caloric Value of Some Foods

More Related