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Phase Transitions H = Cice x T x mass Hfusion= 6.02 kJ/mol x mass

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Presentation on theme: "Phase Transitions H = Cice x T x mass Hfusion= 6.02 kJ/mol x mass"— Presentation transcript:

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

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

3 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
+ 3/2 O2(g)  Al2O3(s) H = kJ/mol _______________________________ __________ Fe2O3(s)  2Fe(s) + 3/2 O2(g) 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) H = kJ/mol + 2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s) -854 kJ/mol 2 ( ) +15 kJ/mol 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 Hof  NH2CH2COOH N2 + H2 + Cgr 2 + O2 Write the equation for which
Hrxn = Hof for NH2CH2COOH 1 mol product, from elements in their standard states  NH2CH2COOH N2 + H2 + Cgr 2 + O2 1/2 5/2

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

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

9 2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
Hrxn = nHof products - nHof reactants Hrxn = [Hof Al2O3(s) + 2 Hof Fe(l)] - [Hof Fe2O3(s) + Hof Al(s)] 2 [(-1676) + (15)] 2 - [(-822) + 0]kJ Hrxn = = -824 kJ

10 Bond Energies chemical reactions = bond breakage and bond formation
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 kJ O=O 495 kJ C=O 799 kJ O-H kJ - bonds formed Hrxn = [ (C-H) 4 + (O=O)] 2 - [ (C=O) 2 + (O-H)] 4 = -824 kJ Hrxn= Hof products - Hof reactants =- 802 kJ


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