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lec4_whc - Last Time-Ch 7 Phase Changes Reaction Enthalpy...

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Last Time--Ch. 7. Á Phase Changes Á Reaction Enthalpy H f = Σ nH° f (products) - Σ nH° f (reactants) H f positive Endothermic H f negative Exothermic

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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 fus = 6.01 kJ/mol C sp (Water) = 4.184 J/g°C C sp (Vapor) = 1.84 J/g°C vap = 40.7 kJ/mol
Step 1--heating of ice from -25 to 0 o C J = 2.09 J/g°C x 25 o C x 18 g of Ice = 940 J

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Step 2-- melting ice at 0 o C J = fus = 6.01 kJ/mol x 1 mol H 2 O = 6010 J
Step 3- heating water from 0 o C to 100 o C J = 4.184 J/g°C x 18 g x 100 °C = 7531.2 Joules

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Step 4 - boiling water at 0 o C = 40.7 kJ/mol x 1 mol = 40,700 Joules
Step 5- heating water vapor from 100 o C to 125 o C J = 1.84 J/g°C x 18 g x 25 o C = 828 Joules

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Standard Enthalpy of Formation, H f ° values can be tabulated for compounds by defining a reference state where H ° f 0 kJ mol -1 . H 2 (g) + ½O 2 (g) H 2 O(l) → f (H 2 O (l) ) = -285.83 kJ/mol The enthalpies of formation of the reactants in this case are zero because these are the reference states for these elements: f (O 2 (g)) 0.0 kJ/mol f (H 2 (g)) 0.0 kJ/mol
Hess’ Law allows you to calculate H f values for reactions that can't be directly measured. A C B D ? H f (AC) H f (DB) H f (CD)

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Hess’s Law ½N 2 (g) + O 2 (g) NO 2 (g) H° = ½N 2 (g) + ½O 2 (g) NO(g) H° = +90.25 kJ NO(g) + ½O 2 (g) NO 2 (g) H° = -57.07 kJ +33.18 kJ
New Material ch. 12: Liquids, Solids and Intermolecular Forces • Surface Forces and wetting • Viscosity • Vapor Pressures • Simple phase diagrams

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ch. 12: Surface Forces • Surface tension and wetting • Viscosity
Liquid Structure • Surface tension - – net inward force felt by molecules at a liquids surface.

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lec4_whc - Last Time-Ch 7 Phase Changes Reaction Enthalpy...

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