06_lecture_ppt-Kuila

# V p x v x d 3 f x d w f d 2 v 0 p v 0 w sys 0 work is

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V P x V = x d 3 = F x d = w F d 2 V > 0 -P V < 0 w sys < 0 Work is not a state function. w = w final - w initial initial final

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11 A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = - P V (a) V = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L•atm = 0 joules (b) V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 L•atm w = -14.1 L•atm x 101.3 J 1L•atm = -1430 J
12 Heat, q, is also not a state function. You can heat water with a Bunsen burner without doing any work on the water. Also, you can heat the water by other means

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13 Chemistry in Action: Making Snow U = q + w q = 0 (no heat exchange, adiabatic process ) w < 0, U < 0 U = C T T < 0, SNOW!
14 Enthalpy and the First Law of Thermodynamics U = q + w U = H - P V H = U + P V q = H and w = - P V At constant pressure:

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15 Enthalpy is a state function.
16 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure H products < H reactants H < 0 H products > H reactants H > 0

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17 Thermochemical Equations H 2 O ( s ) H 2 O ( l ) H = 6.01 kJ/mol Is H negative or positive? System absorbs heat Endothermic H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.
18 Thermochemical Equations CH 4 ( g ) + 2O 2 ( g) CO 2 ( g) + 2H 2 O ( l ) H = -890.4 kJ/mol Is H negative or positive? System gives off heat Exothermic H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.

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19 H 2 O ( s ) H 2 O ( l ) H = 6.01 kJ/mol The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of H changes H 2 O ( l ) H 2 O ( s ) H = - 6.01 kJ/mol If you multiply both sides of the equation by a factor n , then H must change by the same factor n . 2H 2 O ( s ) 2H 2 O ( l ) H = 2 x 6.01 = 12.0 kJ
20 H 2 O ( s ) H 2 O ( l ) H = 6.01 kJ/mol The physical states of all reactants and products must be specified in thermochemical equations.

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• Fall '07
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