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Phase Transitions

melting point. boiling point. 100 o C. 0 o C. Phase Transitions. H = C ice x T x mass. C ice =2.087 J/g o C. H fusion = 6.02 kJ/mol x mass. H = C water x T x mass. C water =4.184 J/g o C. H vap = 40.7 kJ/mol x mass. T. H = C steam x T x mass. C steam =1.996 J/g o C.

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Phase Transitions

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  1. melting point boiling point 100oC 0oC Phase Transitions H = Cice x T x mass Cice=2.087 J/goC Hfusion= 6.02 kJ/mol x mass H = Cwater x T x mass Cwater=4.184 J/goC Hvap = 40.7 kJ/mol x mass T H = Csteam x T x mass Csteam=1.996 J/goC steam water ice -20oC heat added

  2. Thermite reaction Al2O3(s) + 2Fe(l) 2Al(s) +Fe2O3(s) H is an Exothermic extensive, State function Hess’ Law

  3. 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l) 2Al(s) + 3/2 O2(g)  Al2O3(s) H = -1676 kJ/mol _______________________________ __________ Fe2O3(s)  2Fe(s) + 3/2 O2(g) 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) H= - 822 kJ/mol +  Al2O3(s) + 2Fe(s) -854 kJ/mol 2Al(s) + Fe2O3(s) 2 ( ) +15 kJ/mol 2 2 Fe(s)  Fe(l) _______________________________ __________ 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l) Hrxn = -824 kJ/mol

  4. Hess’ Law • Always end up with exactly the same reactants and products • If you reverse a reaction, reverse the sign of H • If you change the stoichiometry, change H

  5. Heats of formation, Hof H = heat lost or gained by a reaction “o” = standard conditions: all solutes 1M all gases 1 atm “f” = formation reaction: 1mol product from elements in standard states for elements in standard states, Hof= 0

  6. Write the equation for which Hrxn = Hof Hof for NH2CH2COOH from elements 1 mol product, in their standard states  NH2CH2COOH + O2 N2 + H2 + Cgr 2 1/2 5/2

  7. 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l) products reactants elements Hof 2 Al(s) Al2O3(s) 2 Al(s) 2 Fe(s) Fe2O3 2 Fe (l) 3/2 O2(g) Hof Al2O3(s) + 2 HofFe (l) Al(s) Fe2O3

  8. 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l) products reactants elements Hof 2 Al(s) Al2O3(s) 2 Al(s) 2 Fe(s) Fe2O3 2 Fe (l) 3/2 O2(g) Hof Al2O3(s) + 2 HofFe (l) - Hof - Hof Fe2O3 Al(s) - Hrxn = nHofproducts nHofreactants

  9. 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l) - nHofreactants Hrxn = nHofproducts [HofAl2O3(s) + 2 HofFe(l)] Hrxn= - [HofFe2O3(s) + HofAl(s)] 2 [(-1676) + (15)] 2 - [(-822) + 0]kJ Hrxn = = -824 kJ

  10. Bond Energies chemical reactions = bond breakage and bond formation bond energies positive energy required to break bond bond breakage a) endothermic b) exothermic (raise P.E.) bond formation exothermic (lower P.E.)

  11. Bond energies CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) Hrxn= bonds broken C-H 413 kJ O=O 495 kJ C=O 799 kJ O-H 467 kJ - bonds formed Hrxn= 4 [ (C-H) + (O=O)] 2 - [ (C=O) 2 + (O-H)] 4 = -824 kJ Hrxn= Hof products - Hof reactants =- 802 kJ

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