lec3 - H f (O 2 ) = 0 kJ/mol H f (H 2 O) = -285.83 kJ/mol H...

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1 Last Time First Law of Thermodynamics Heat Capacity P-V Work Enthalpy of Phase Changes Phase Changes • Require energy solid liquid gas • Energy “relaxes” molecule – molecule interactions • Before and after phase change T varies • During phase change T is constant • Total q from C p , C sp , Δ vap and Δ fus U U Example - Phase Changes Calculate the total heat needed to change 18 g (1 mol) of ice at - 25 °C to vapor at 125 °C. C sp (Ice) = 2.09 J/g°C Δ fus = 6.01 kJ/mol C sp (Water) = 4.184 J/g°C C sp (Vapor) = 1.84 J/g°C Δ vap = 40.7 kJ/mol
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2 Reaction Enthalpy Reactants Products Δ H f = Σ nH° f (products) - Σ nH° f (reactants) Δ H f positive Endothermic Δ H f negative Exothermic Δ f Standard enthalpy of formation Example - Combustion Is the combustion of propane C 3 H 8 (g) endothermic or exothermic?
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Unformatted text preview: H f (O 2 ) = 0 kJ/mol H f (H 2 O) = -285.83 kJ/mol H f (CO 2 ) = -393.51 kJ/mol H f (C 3 H 8 ) = -2220 kJ/mol Endothermic or Exothermic? Consider: Ba(OH) 2 H 2 O(s) + 2NH 4 NO 3 (s) Ba(NO 3 ) 2 + 2NH 3 (g) + 10H 2 O(l) H 2 SO 4 , Sugar, KClO 3 , salts, accelerant Hess Law Reaction enthalpies are additive H f (reverse) = - H f (forward) H condensation = H liquid- H vapor Balance and cancel known equations 3 Example - Hess Law Calculate H f for NO 2 (g) from N 2 (g) and O 2 (g). N 2 (g) + O 2 (g) 2NO(g) H f = 180 kJ 2NO(g) + O 2 (g) 2NO 2 (g) H f = -112 kJ _________________________________ N 2 (g) + 2O 2 (g) 2NO 2 (g) H f = 68 kJ...
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lec3 - H f (O 2 ) = 0 kJ/mol H f (H 2 O) = -285.83 kJ/mol H...

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