110c-lecture10 - Midterm 1 Wed April 21st 21 and 22.1-22.7...

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Midterm 1 Wed April 21 st . Chapters 19, 20, 21, and 22.1-22.7. 3” x 5” index card Review session? Problem Set #3 Due Wed April 28th.
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Chapter 19 Summary First Law of Thermodynamics: U = q + w or dU = δ q+ δ w dT T C q U T T V ) ( 2 1 = = = = 2 1 V V V dV nRT w q V = 0, w=0 T = 0, U=0 q=0 = = 2 1 T T dT nR w U Enthalpy (H) vs. Internal Energy (U): = = + = 2 1 V V V q PdV q w q U Const=V: PV U H + = P P q H = Const=P: (PV work) P P q VdP PdV PdV q dH δ = + + = () RT n U nRT U PV U H gas + + = + = ) ( Heat Capacity (C P vs. C V measured by calorimetry): ) ( T C T q T U V V V = = Const=V: ( U = q V ) ) ( T C T q T H P P P = = Const=P: ( H = q P ) Temperature Dependence of U and H: dT T C T U T U dV V U dT T U dU T T V T V = + = 2 1 1 2 ) ( ) ( ) ( dT T C T H T H dP P H dT T H dH T T P T P = + = 2 1 1 2 ) ( ) ( ) (
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Thermochemistry --- Summary 3 Different ways to calculate reaction enthalpy change ( r H): aA + bB cC+ dD ± r H 0 = -[ a f H A 0 + b f H B 0 ] + [ c f H C 0 + d f H D 0 ] Use heats of formation ( f H º) Pathway: reactants stable elements products ± r H 0 = [ a c H A 0 + b c H B 0 ] - [ c c H C 0 + d c H D 0 ] Use heats of combustion ( c H º) Pathway: reactants CO 2 + H 2 O products ± r H 0 = [ a BDE(A) + b BDE(B) ] -[ c BDE(C) + d BDE(D)] Use bond dissociation energy (BDE) Pathway: reactants atoms products
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Chapter 20 Summary 2 nd Law of Thermodynamics ( T q dS T q dS rev δ = ). Entropy (S) of an isolated system increases ( dS > 0) during a spontaneous process. + = + = = = 2 1 2 1 2 1 2 1 V V T T V V V T T V rev rev V nRdV T dT C T PdV T dT C T w T dU T q dS 1 2 1 2 1 1 2 2 ln ln ) , ( ) , ( V V nR T T C V T S V T S V + = (PV work) Boltzmann Equation of Entropy (S = k B lnW). W = number of microstates accessible to system (W V N , ideal gas) k B = Boltzmann constant = 1.38 x 10 -23 J K -1 = R/N A S total = S 1 + S 2 = k B (ln W 1 + ln W 2 ) where W total = W 1 *W 2 Entropy of Mixing ( mix S). {} { } 2 2 1 1 ln ln y R y y R y S mix = 2 1 1 1 n n n y + = 2 1 2 2 n n n y + = Entropy of Phase Transition ( fus S). ( ) ( ) l O H s O H 2 2 ⎯→ At T=273 K K mol J K mol J T H T q S fus fus rev fus + = = = = / 996 . 21 15 . 273 / 6008
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Chapter 21 Summary 3 rd Law of thermodynamics: the entropy (S) of a perfect crystal is 0 at 0 K (W=1). S(0) = k B lnW = k B ln(1) = 0 Single ground state (W=1) S(0) = k B lnW = k B ln(2 N ) = Nk B ln(2) 2-fold degeneracy (W=2 N ). 3 rd Law Entropy vs. Temperature by Calorimetry: ' ' ) ' ( ' ' ) ' ( ' ' ) ' ( ) 0 ( ) ( 0 dT T T C T H dT T T C T H T dT T C S T S T vap T g P vap vap fus T vap T fus T l P fus fus s P + + ∫∫ + + = () 3 T T C P (Debye T 3 Law) at low temperatures (<10 K). 3 ) ( ) 3 ( ' ' ' ' ' ' ' ) ' ( ) 10 ( 3 0 2 0 3 0 T C T c dT T c dT T T c dT T T C K T S P T T T P = = = = = < Debye Law for Entropy at T<10K: [ ] ) 38 . 17 . 17 . ( !
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This note was uploaded on 06/22/2010 for the course CHEM 21360 taught by Professor Ame during the Spring '09 term at East Los Angeles College.

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110c-lecture10 - Midterm 1 Wed April 21st 21 and 22.1-22.7...

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