Lecture 16 - Chemical Equilibrium stoichiometric...

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Unformatted text preview: Chemical Equilibrium stoichiometric coefficients A A + B B or C C + D D molecular species A A + B B - C C - D D = 0 j j X j =0 dG = -SdT + Vdp + where , <0 C D j dN j j A , B >0 dA = -SdT - pdV + j j dN j at T,V dA = j j dN j = j j j d at T,p dG= j j dN j reaction parameter: equilibrium A = 0 6 T, V dN j = j d j A is minimum with respect to ALL parameters 6 j j = 0 j ln Q 6 we know that j = -kT (canonical ensemble) 6N j Ideal non-interacting gases (with many more states than N h Boltzmann statistics) Bln 6 j Q= j q Nj j Nj ! ln q j - ln N j !) Nj = -kT ln j lnQ 6 = 6N j 6 6 qj j Nj ! N B = j qj j ln Nj ! Nj N 6N j = j (N j = ln qj Nj j j qj N = kT ln 0 = j j = - j kT ln N 6 q j j j j 6 j j j q N j 6 j j N q j =1 j equilibrium condition 6 j j N q j =1 j A B C D N N q q =1 q q N N A B C D C A A B C D ( qA ) C ( qc ) ( qD ) D B ( qB ) D = ( NC ) C A ( NA ) ( ND ) ( NB ) D B and dividing by volume q q c D ( ) C ( ) D ( pC ) C ( pD ) D V V + + + = C D Kc ( T ) = = kT C D A B A A B qA B q B A ) A ( B ) B ( ( pA ) ( pB ) V V for an ideal gas Nj pj = kT V K p ( T ) = ( kT ) j j Kc ( T ) Examples q Na2 V Associatio n 1- 2 K p ( T ) = ( kT ) 2 2 Na h Na2 q Na V qvibration qtranslation 6 4 4 7 4 4 8 qrotation 6 4 7 4 8 q } e lectronic 3 - vib 64 7 4 8 2 2T 2mNa kT T e De 1 2 e V 2 2 vib elec V - rot h 1 - e T 6 6 -1 K p ( T ) = ( kT ) qelectronic 2 3 2 2m kT 2 } 1 6 6 Na 6 6 V elec 2 V h 6 6 1 2 De = Do + h g Na elec =* 1 S1 Na elec = 2 2 2 mNa , mNa , rot , and vib can all be extracted from tables 2 We can predict Kp(T) at any T!! Isotope exchang e H2 + D2 2HD q HD V 2 -1-1 K p ( T ) = ( kT ) q H2 qD2 V V HD 2 2 K p (T ) = K c ( T ) = 2 - 3 2T 2 De ,HD e 1 2mHD kT T HD HD,elec e h2 V V - HD T 1- e 2 2 mH2 kT VT 2 2 h H2 3 e - H2 2T - 1- e H2 T H2 e 2 De ,H2 mD2 kT VT 2 3 h 2 2 D2 e - D2 2T - 1- e D2 T D2 e De ,D2 1 V 2 Fortunately MANY terms cancel... 2 kT 3 2 2 m 2 h K p (T ) = HD 3 m 2 kT H2 mD2 2 h 3 T HD T 2 2 ( e - HD H2 D2 T + 2T + 2T ) (- 1 e - H2 T (- 1 )( - ) 1 - 4 H2 D2 HD T e ) e - D2 T 2 2HD ( 2De,HD - De,H2 - De,D2 ) e H2 D2 2 4 H2 D2 K p (T ) = H mD 2 m 2 2 HD 2 mHD 3 ( e - 2HD +H2 + D2 2T ) (- 1 e - H2 T (- 1 )( - ) 1 - HD T e ) e - D2 T 2 2 HD ( 2De,HD - De,H2 - De,D2 ) e H2 D2 If we consider the B-O approximation,the 3 PES are all the same, the changes are in the masses, thus mH2 mD2 mHD H2 D2 HD kH2 = kD2 = kHD De,H= De,D= De,HD 2 2 and HD = H2 = D2 = 1 kH2 = kD2 = kHD D2 = H2 2 2 2 H2 H2 = D2 D2 = HD HD H2 D2 2 and mH 2 HD = H2 D = 2 H2 HD mH mD mH + mD and since H = mD 2 1 2 HD = D2 ,vib = HD,vib h D2 k = h H2 k mH mD = H2 ,vib h HD h H2 = = k k mH ( mH + mD ) 2mH mD = H2 ,vib 3 4 j ,rot h2 = 8 2kI j where Ij = re2 H2 3 HD,rot I H2 = = = H2 ,ro t I HD HD 4 considering 1 ? e 3 - D2 3 HD,rot I D2 and = = = D2 ,rot I HD HD 2 K p (T ) = j T 2 2 mHD 4 H2 D2 H mD HD m 2 2 3 ( e - 2HD +H2 +D2 2T (- ) 1 e - H2 T )( - ) - 1- e HD T 1 e - D2 T e ( 2De ,HD - De ,H2 - De ,D2 ) 2 3 = 4 2 4 2 - 3 3 e 2 4 2 H2 3 1 + +1 4 2 2T ( 2 -1-1) De 9 H2 ( e =4 - e 8 B 4 for m 3 + 2 +1 2T ) 2 ( m1 - m2 ) 2 m1 ( m1 - m2 ) K ( T) = 4 + 1 - 2 2 32m1 T 8m1 e in general polya tomi c 1 H2 + O2 H2O 2 K p ( T ) = ( kT ) 1-1- 1 2 q H 2O V q H2 qO2 V V De,H 2O e 1 2 - j q 2 H2O mH2O kT T T n e 2T T = 6 V h2 A B C j =1 - j - e T 1 3 2 1 2 1 2 q 2 H2 mH2 kT T e De ,H 2 = e 2 V h 2 H2 1 - e - T q O2 V 2 mO2 kT T = 2 2 h O2 3 2 3 2 - 2T e De ,O 2 3 e - - e T 1 - 2T 3 - g 1 D D h , e 2T e e = e o and H2O = 1 2 we can look at the T dependence of K considering that De = Do + - Do ,H2O kT Do ,O 2 Do ,H2O - Do ,H2 - Do ,O 2 Kp ( T ) 6 T - 1 2 T T T T e Do ,H2 kT 3 4 1 2 3 2 1 2 1 2 1 2 T T e 3 2 = e 2 kT 5 4 2 kT T T e T We can also evaluate chemical potentials (T,p) q ( T , p ) = -kT ln N 3 2 q mkT V 2 for a monoatomic ideal gas = 2 qnuclear qelectronic N h N 2 mkT kT ( T , p ) = -kT ln 2 - kT ln ( qnuclear qelectronic ) h p 3 2 2 mkT ( T , p ) = -kT ln kTqnuclear qelectronic kT ln ( p ) + 2 h 1 4 4 4 4 4 4 2 4 4 4 4 4 4 43 4 o ( T ) 3 2 ( T , p ) = o ( T ) + kT ln p which can be replaced in j p=e - o kT j j = 0 0 = C o ( T ) + kT ln pC + D o ( T ) + kT ln pD C D - A C D ( ( A o ( T ) + kT ln pA A B ) ( ) - ( B B o ) ( T ) + kT ln p ) B C o + D o - A o - B o = -kT ( C ln pC + D ln pD - B ln pB - A ln pA ) 14444 244444 4 3 o CC p D p D o = -kT ln A B pA pB 6 or K p ( T ) = e - o kT How important are the chosen units? 3 2 2 mkT o ( T , p ) = -kT ln 2 kTqnuclear qelectronic h 3 2 mkT = 2 kT h 2 oK kg 2 o kg kg m m K 2 o 2 o s K s K 2 3 6 kg m s 3 2 kg 4 m 5 s 3 = s 5 kg 3 m 6 3 2 m mkT kT kg = kg 6 1 = [ pressure ] = 2 2 h s 6 m s 2 m2 thermodynamics tables, [pressure] = atm p ( T ) atm K = For the zero energy, consider that D o o qe qv = qe qv e o { { shows how to scale it calculated as if the |g> of the molecule is =0 prod - react with respect to the energy of separated atoms o 2 mkT kT - o o o ( T , p ) = -kT ln 2 e qn qrot qelec qv + kT ln ( p ) 106 h 1.016 3 2 Do oo for one mol of an element at T=0 o K o q kT o o G - Eo = -RT ln V 1.10 x106 Free Gibbs energy per mol (1 atm), relative to that zero energy as T h 0 Go o Eo ...
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This note was uploaded on 05/29/2011 for the course CHM 6461 taught by Professor Bowers during the Spring '08 term at University of Florida.

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