CHEM 1111notes

CHEM 1111notes - f (reactants) Calculate the ∆H° rxn...

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10/6/2008 Chapter 6 Notes Standard States: 1) Gases at 1 atm pressure 2) Aqueous solutions are at 1M concentration 3) Pure substances are in their most stable form at 1 atm and the temperature of interest (usually 25° Celsius) Water’s standard state is liquid Carbon’s standard state is graphite ∆H° rxn : change in enthalpy for a reaction with all reactants and products in their standard states. Standard Heat of Formation (∆H° f ): The enthalpy changes when 1 mole of a compound forms from its elements in their standard states. ( C(graphite) +2H 2 (g) CH 4 (g) ∆H° f (CH 4 (g)) ½ N 2 (g) + 3/2 H 2 (g) NH 3 (g) ∆H° f (NH 3 (g)) A few notes of ∆H° f : 1) ∆H° f for any element in its standard state is zero O 2 (g) O 2 (g) ∆H° f = 0 kJ/mol 2) Most ∆H° f are negative ∆H° f can be used to get ∆H° rxn : ∆H° rxn =Σn∆H° f (products) -Σn∆H°
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Unformatted text preview: f (reactants) Calculate the ∆H° rxn for: N 2 O 4 (g) 2NO 2 (g) ∆H° rxn =2∆H° f (NO 2 (g)) - ∆H° f (N 2 O 4 (g)) =2(+33.2 kJ/mol) – (+9.16 kJ/mol) =57.2 kJ ∆H° f (N2O4(g)) N2 (g) + 2O 2 (g) N 2 O 4 (g) ∆H° f =+33.2 kJ/mol ∆H° f (NO2(g)) ½ N 2 (g) + O 2 (g) NO 2 (g) ∆H° f =+9.16 kJ/mol N2O4 (g) N 2 (g) + 2O 2 (g) ∆H° f =9.16 kJ/mol N 2 (g) + 2O 2 (g) 2NO 2 (g) ∆H° f = 2(33.2) N 2 O 4 (g) 2NO 2 (g)-9.16 + 2(33.2) = 57.2 kJ/mol Chapter 7: Quantum Theory & Atomic Structure Review of Light: Frequency (v): number of cycles (waves) per second (hertz(Hz)s-1 ) Wavelength: the distance a wave travels in one cycle (m, nm) Speed of Light (C): distance traveled per unit time (m/s) (3.00 x 10 8 m/s)...
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This note was uploaded on 04/26/2009 for the course CHEM 1111 taught by Professor Robertparson during the Fall '08 term at Colorado.

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CHEM 1111notes - f (reactants) Calculate the ∆H° rxn...

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