![Page 1: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/1.jpg)
1
Termodynamik for Biokemikere Jan H. Jensen
Københavns Universitet
1. Ligevægt og ligevægtskonstanten 2. Enthalpi og entropi 3. Enthalpi og entropi for an ideal gas og van’t Hoff ligningen 4. Måling af enthalpi og entropi ændringer vha kalorimetri 5. Enthalpi og entropi for en ideal opløsning 6. Hydrofobisitet og entropi 7. Kemisk akIvitet og ikke-‐ideale opløsninger 8. Termodynamikens tre love og Boltzmannfordelingen
Playlist med alle videoer hPps://www.youtube.com/playlist?list=PLVxAq6ZYPp3154Tp_dmz9GOoo7g_3rQBH
![Page 2: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/2.jpg)
2
Indhold 1. Equilibrium and the equilibrium constant 1.1. Equilibrium constant (K): more reactant or product eeer equilibrium? 1.2. Standard free energy (ΔGo): a molecular understanding of K 1.3. 6 kJ/mol changes K by about an order of magnitude at 25 oC 1.4. How do you measure K? 1.5. Le Chatelier’s Principle 2. Enthalpi og entropi 2.1. EnergiIlstande (ingen slides) 2.2. Enthalpien (H) handler om energi 2.3. Standard dannelsesenthalpi 2.4. Bindingsenergier 2.5. Entropi (S) handler om muligheder 3. Enthalpi og entropi for en ideal gas og van’t Hoff ligningen 3.1. EnergiIlstande (ingen slides) 3.2. Enthalpibidrag for en ideal gas 3.3. Entropibidrag for en ideal gas 3.4. KonformaIonel entropi 3.5. Hvordan måler man standard enthalpi og entropi ændringer? van’t Hoff ligningen
![Page 3: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/3.jpg)
3
Indhold 4. Måling af enthalpi og entropi ændringer vha kalorimetri 4.1. Kalorimetri (ingen slides) 4.2. Hvordan måler man ΔHo? Kalorimetry 4.3. Hvorfor er varmekapaciteten et maximum når ΔGo(Tm) = 0? 4.4. Eksempel: Protein (polymer) foldning 4.5. Udfoldning ved høj temperatur sker pga entropi 5. Enthalpi og entropi for an ideal opløsning 5.1. Fri energibidrag for en ideal opløsning 5.2. Solveringsfrienergi: det polære bidrag 5.3. Solvent screening 5.4. Den ikke-‐polære solveringsfrienergi 5.5. Den hydrofobe effekt 6. Hydrofobicitet og entropi 6.1. Solventen bidrager Il entropiændringen 6.2. Entropien sIger når hydrofobe molekyler bindes 6.3. Hvordan måler man hydrofobicitet 6.4. Ligandbinding Il enzymet carbonic anhydrase
![Page 4: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/4.jpg)
4
Indhold 7. Kemisk akLvitet og ikke-‐ideale opløsninger 7.1. Ligevægtskonstanten har ingen enheder 7.2. AkIvitet for en opløsning 7.3. Den simple Debye-‐Hückel ligning 7.4. Brug af den simple Debye-‐Hückel ligning 8. Termodynamikens tre love og Boltzmannfordelingen 8.1. Termodynamikens tre love 8.2. Entropi og sandsynlighed – del 1 8.3. Entropi og sandsynlighed – del 2 8.4. Boltzmannfordelingen 8.5. Boltzmannfordelingen giver ligninger for fri energi-‐ bidrag for en ideal gas
![Page 5: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/5.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.,Ligevægt,og,ligevægtskonstanten,
2.$Enthalpi$og$entropi$3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.$Hydrofobisitet$og$entropi$7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 7: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/7.jpg)
!"#$%&'"(&)*+),-"'./&%0)
!:;0'0<+0;,(
-P#QRR'N7G*(G*%>7*%:R'*>"="%R)
E)
![Page 8: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/8.jpg)
!"#$%&'"(&)*+),-"'./&%0)
!:;0'0<+0;,(1-$"7&$7(W!XQ)'*%")%"$G&$(&)*%)#%*>CG&)$Y"%)#:;0'0<+0;,Z)W"BC.=.H%.C'Q)(*)G-$(:").()G*(G"(&%$D*(X)
K =P[ ]R[ ]
X[ ] = concentration of X
K>1! [P] > [R]! more product than reactant
R! P
[-$&)./)\6]^)+*%)$)4)F)/*=CD*()*+),6?,__6Z)
`7 47KE)a)4T1I)F))
27 T7TTE4J)F)
,7 T7I4;)F)
!7 47TTT)F)
CH3COOH! CH3COO! + H+ K = 1.74 "10!5
I)
![Page 10: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/10.jpg)
!"#$%&'"(&)*+),-"'./&%0)
=7&$.&+.(2+##(#$#+>/)Wf"*XQ)$)'*="GC=$%)C(>"%/&$(>.(:)*+)!#
K = e!"Go /RT
!Go = Go (P) "Go (R)
$)./)&-"):$/)G*(/&$(&)L7?4E)5R'*=)g)
*)h)/&$(>$%>)/&$&")
[-$&)./)f"*)+*%)&-./)%"$GD*()$&);I)*,Z))
`7 14;7I)e5R'*=)
27 T744)e5R'*=)
,7 ;K7;)e5R'*=)
!7 ;JiTT)e5R'*=)
CH3COOH! CH3COO! + H+ K = 1.74 "10!5
i)
![Page 11: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/11.jpg)
!"#$%&'"(&)*+),-"'./&%0)
K = e!"Go /RT #
"Go = !RT ln(K )
$)./)&-"):$/)G*(/&$(&)L7?4E)5R'*=)g)
CH3COOH! CH3COO! + H+ K = 1.74 "10!5
!Go = ("8.314)(298.15)ln(1.74 #10"5 )= 2.72 #104 J/mol = 27.2 kJ/mol
44)
=7&$.&+.(2+##(#$#+>/)Wf"*XQ)$)'*="GC=$%)C(>"%/&$(>.(:)*+)!#
[-$&)./)f"*)+*%)&-./)%"$GD*()$&);I)*,Z))
![Page 13: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/13.jpg)
!"#$%&'"(&)*+),-"'./&%0)
J)e5R'*=)G-$(:"/)!)H0)$H*C&)$()*%>"%)*+)'$:(.&C>")$&);I)*,)
K = e!"Go /RT
= e!"Go /(0.008314 #298.15)
= e!"Go /2.48
= 10!"Go /2.48#ln(10)
= 10!"Go /5.7
$ 10!"Go /6
4?)
![Page 14: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/14.jpg)
!"#$%&'"(&)*+),-"'./&%0)
K ! 10"#Go /6
K ! 10"27 /6
! 10"4.5
K between 10"4 and 10"5
CH3COOH! CH3COO! + H+
K = 1.74 "10!5 # $Go = 27.2 kJ/mol
U+)f"*)./)1;;7I)e5R'*=)$&);I)*,)N-$&)./)!Z)
`7 47;4)a)4T1;)
27 ;7;K))
,7 L7KI)a)4T?)
!7 I7ii)a)4TI)) 4E)
J)e5R'*=)G-$(:"/)!)H0)$H*C&)$()*%>"%)*+)'$:(.&C>")$&);I)*,)
![Page 16: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/16.jpg)
!"#$%&'"(&)*+),-"'./&%0)
?-@(.-(/-;(,#&";+#(!A(
HA! A! +H+ K = [A
! ][H+ ][HA]
fA! =
[A! ][HA]+ [A! ]
=1
[H+ ]K
+1=
K[H+ ]+ K
" fA! =
12" K = [H+ ]
A!
! log K( ) = 6.3
4K)
U'$:")$>$#&">)+%*'Q))-P#QRRD(0C=%7G*'R0lI##"))
![Page 17: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/17.jpg)
!"#$%&'"(&)*+),-"'./&%0)
HA! A! +H+ K = [A
! ][H+ ][HA]
A!
! log K( ) = 6.3
U+)m=*:W!XhJ7?)N-$&)./)f"*)$&);I)*,Z))
`7 I7?)e5R'*=)
27 ?J7T)e5R'*=)
,7 4IT7;)e5R'*=)
!7 J??7;)e5R'*=) 4L)
?-@(.-(/-;(,#&";+#(!A(
![Page 19: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/19.jpg)
!"#$%&'"(&)*+),-"'./&%0)
B#(6*&7#'0#+C"(D+0$10E'#(
%&#'()*+,-#./0&1,#2*#0#+3+2,(#'&#,45'6'78'5(#8,+562+#'�#+/'9#'&#,45'6'78'5(#2/02#.*5&2,80.2+#2/,#'()*+,-#./0&1,#
K =[B][A]
=1.0 ! x0.25 + x
= 2
x = 0.17 " [A] = 0.42 og [B] = 0.83 " A! B
A! B K =
[B][A]
= 2
<a$'#="7Q)$&)"BC.=.H%.C')\`^)h)T7;I)F)*:)\2^)h)T7IT)F))
U+)U)$>>)'*%")2o)/*)&-$&)\2^)h)47T)F)H"+*%")"BC.=.H%.C'o))[-$&)$%")\`^)$(>)\2^)$Y"%)"BC.=.H%.C'Z)
)
;4)
![Page 20: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/20.jpg)
!"#$%&'"(&)*+),-"'./&%0)
A! B+ C
`Y"%)"BC.=.H%.C').()$BC"*C/)/*=CD*()U)="&)-$=+)&-")N$&"%)"9$#*%$&"))
[-$&)N.==)-$##"()$GG*%>.(:)&*)c"),-$&"=."%@/)#%.(G.#="Z)
A. Equilibrium shifts towards products: A! B+ C
B. Equilibrium shifts towards reactant: A" B+ C
C. There is no change in equilibrium
;;)
![Page 21: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/21.jpg)
!"#$%&'"(&)*+),-"'./&%0)
A! B+ C<9$#*%$D*().(G%"$/"/)&-")G*(G"(&%$D*(/)
)M-")/-.Y).()"BC.=.H%.C')&*N$%>/)%"$G&$(&)>"G%"$/"/)&-")(C'H"%)*+)#$%DG="/)
$(>)&-"%"+*%")&-")G*(G"(&%$D*()
B#(6*&7#'0#+C"(D+0$10E(
K =[B][C]
[A]evaporation! "!!!
2[B]( ) 2[C]( )2[A]
> K equilibrium! "!!! K =[B #] [C #]
[A #]2[B] > [B #] , 2[C] > [C #] , 2[A] < [A #]
;E)
%&#'()*+,-#./0&1,#2*#0#+3+2,(#'&#,45'6'78'5(#8,+562+#'�#+/'9#'&#,45'6'78'5(#2/02#.*5&2,80.2+#2/,#'()*+,-#./0&1,#
![Page 22: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/22.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.,Enthalpi,og,entropi,
3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.$Hydrofobisitet$og$entropi$7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 25: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/25.jpg)
!Go = !H o " T!So
H2CO3(aq)! "!# !! H2O(l ) + CO2(g) !ngas = 1
Enthalpien*(H)*handler*om*energi*
!H o = !U + po!V
!H o = ændringen i enthapi når trykket er po = 1 bar!U = ændringen i den indre energi!V = ændringen i volumen
po!V " !ngasRT!ngas = ændring i antal mol af gas molekyler
RT = 2.5 kJ/mol ved 25 oC
![Page 26: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/26.jpg)
!H o = !H Molekyle + !H o,Translation + !H Rotation + !H Vibration
Enthalpien*(H)*s3ger*når*bindinger*brydes*
H2! "!# !! 2H
!H o = 460.2 + 6.3" 2.5 " 26.4 = 437.6 kJ/mol
!H Molekyle = energien for elektroner og kerner, såsom bindingsenergier
H2O ! ! !HOH! "!# !! 2H2O
"H o = 20.5 + 6.3+ 3.8 #17.6 = 13.0 kJ/mol
ΔHo&er&posiFv&primært&fordi&det&kræver&energi&at&bryde&&kovalente&bindinger&og&hydrogenbindinger&&
![Page 27: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/27.jpg)
!H o = !H Molekyle + !H o,Translation + !H Rotation + !H Vibration
Enthalpien*(H)*s3ger*når*bindinger*brydes*
!H Molekyle = energien for elektroner og kerner, såsom bindingsenergier
H2O ! ! !HOH! "!# !! 2H2O
"H o = 20.5 + 6.3+ 3.8 #17.6 = 13.0 kJ/mol
ΔHo&er&posiFv&primært&fordi&det&kræver&energi&at&bryde&&kovalente&bindinger&og&hydrogenbindinger&&
![Page 28: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/28.jpg)
!H o < 0 exotermisk A! "!# !! B + varme
!H o > 0 endotermisk A+varme! "!# !! B
Exoterme&og&endoterme&reakFoner&
Er&de+e&en&endotem&eller&exoterm&process&
![Page 30: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/30.jpg)
Standard*dannelsesenthalpi*(standard&enthalpi&of&formaFon,&heat&of&formaFon)&
elementer i standardtilstand! "!# !! molekyle !H fo
2C(s,grafit ) + 3H2(g)! "!# !! C2H6(g) !H f
o (C2H6 )
!H fo (C2H6 ) = H o (C2H6 ) " 2H o (C) " 3H o (H2 )
C2H4(g) + H2(g)! "!# !! C2H6(g) !H
o
!H o = H o (C2H6 ) " H o (C2H4 ) " H o (H2 )= !H f
o (C2H6 ) " !H fo (C2H4 ) " !H f
o (H2 )0
$ %& '&
![Page 31: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/31.jpg)
![Page 32: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/32.jpg)
C2H4(g) + H2(g)! "!# !! C2H6(g) !H
o = ?
Brug&eksperimentelle&dannelsesenthalpier&(fra&Google)&&og&Molecule&Calculator&Fl&at&udregne&ΔHo&for&denne&reakFon&&
A. 103.2&(eksp)&og&112.9&(MolCalc)&kJ/mol&
B. 53.2&(eksp)&og&49.3&(MolCalc)&kJ/mol&
C. Z33.2&(eksp)&og&Z77.2&(MolCalc)&kJ/mol&
D. Z135.9&(eksp)&og&&Z145.3&&kJ/mol&
![Page 34: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/34.jpg)
Bindingsenergier*(Bond&energies)&
C2H4(g) + H2(g)! "!# !! C2H6(g) !H
o = ?
!H o " 611+ 436 # 347 + 2 $ 414( ) = #128 kJ/mol
h+p://chemwiki.ucdavis.edu/TheoreFcal_Chemistry/Chemical_Bonding/General_Principles/Bond_Energies&
![Page 35: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/35.jpg)
Brug&bindingenergier&Fl&at&esFmere&ΔHo&for&denne&reakFon&
A. 152&kJ/mol&
B. 67&kJ/mol&
C. Z166&kJ/mol&
D. Z267&kJ/mol&
![Page 37: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/37.jpg)
Entropien*(S)*handler*om*muligheder*
S = k ln(W )
k = RNA
Boltzmanns konstant, NA = Avogadros tal
W = antal måder man kan lave den samme tilstand
AA! "!# !! A + A
!S = k ln WA+A( ) " k ln WAA( ) > 0
WAA = 6 WA+A = 15
!
2&parFkler&har&mere&entropi&end&1&
![Page 38: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/38.jpg)
Entropien*(S)*s3ger*når*bindinger*brydes*
H2! "!# !! 2H
!So = 11.6 +101.1"12.8 " 0.0 = 98.9 J/molK
H2O ! ! !HOH! "!# !! 2H2O
"So = #17.3+136.2 + 9.3# 66.0 = 79.4 J/molK
ΔSo&er&posiFv&primært&fordi&2&parFkler&har&mere&entropi&end&1&
!So = !SKonformation + !So,Translation + !SRotation + !SVibration
Mere&bevægelsesfrihed&=&større&entropi&
![Page 39: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/39.jpg)
Hvad&er&ΔSo&sandsynligvis&for&denne&process?&
A. !So > 0B. !So = 0C. !So < 0
![Page 40: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/40.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.$Enthalpi$og$entropi$3.,Enthalpi,og,entropi,for,an,ideal,gas,og,van’t,Hoff,ligningen,
4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.$Hydrofobisitet$og$entropi$7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 43: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/43.jpg)
ΔGo = ΔH o − TΔSo
ΔH o = ΔH Molekyle + ΔH o,Translation + ΔH Rotation + ΔH Vibration
H Molekyle = ingen simpel ligning
H o,Translation = 32 nRT + poV = 5
2 nRT
H Rotation = 32 nRT (nRT liniært molekyle)
H Vibration = nNAhc ν i12 +
1eNAhc νi /RT −1
⎛⎝⎜
⎞⎠⎟i=1
3Nat −X
∑
Enthalpibidrag-for-en-ideal-gas-
ν = bølgetal ( ≈ frekvens) i cm-1 Nat = antal atomer i molekyleth = Plancks konstant X = 6, 5 (liniær)c = lysets hastighed i cm/s NAhc = 11.96 J cm
![Page 44: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/44.jpg)
Hvad&er&Hrot&for&et&vand&molekyle&ved&25&oC&og&hvor&mange&forskellige&vibraOoner&bidrager&Ol&Hvib?&
A.&&2.5&kJ/mol&og&4&vibraOoner&&&B. 3.7&kJ/mol&og&3&vibraOoner&
C. 2.5&kJ/mol&og&3&vibraOoner&
D. 3.7&kJ/mol&og&2&vibraOoner&&
![Page 46: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/46.jpg)
ΔSo = ΔSKonformation + ΔSo,Translation + ΔSRotation + ΔSVibrationΔGo = ΔH o − TΔSo
Entropibidrag-for-en-ideal-gas-
SKonf = nR ln(gKonf ) gKonf = antal konformationer med samme energi (udartning)
So,Trans = nR ln2πm( )3/2 kTe( )5 /2
h3po
⎛
⎝⎜⎞
⎠⎟= nR ln aM 3/2T 5 /2( ) a = 0.3117 mol3/2
g3/2K5/2
SRot = nR ln 8π 2keTh2
⎛⎝⎜
⎞⎠⎟
3/2
π I1I2I3
⎡
⎣⎢⎢
⎤
⎦⎥⎥
(ikke liniær) I = intertimoment
SVib = nR NAhc ν i
RT eNAhc νi /RT −1( ) − ln 1− e−NAhc νi /RT( )⎛
⎝⎜
⎞
⎠⎟
i=1
3Nat −X
∑
ν = bølgetal ( ≈ frekvens) i cm-1 NAhc = 11.96 J cm
![Page 47: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/47.jpg)
Hvad&er&STrans&for&et&acetylen&molekyle&ved&25&oC&og&hvor&mange&forskellige&vibraOoner&bidrager&Ol&SVib?&
A.&&63.2&J/molK&og&7&vibraOoner&&&B. 149.4&J/molK&og&6&vibraOoner&&
C. 63.2&J/molK&og&6&vibraOoner&&
D. 149.4&J/molK&og&7&vibraOoner&&
![Page 49: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/49.jpg)
R A-
R B-
��� � �� ��������� � (ΔHo) � �� ���� �� � �� �-
R5A-&
Hvis �� ������ � ��� �, � ��� �, �� � ��� � �� �� ….&
A&&&&mere&RFA&end&RFB&&&&&&&&&&&&&&&&&&&&C&&&&&lige&meget&RFA&og&RFB-&B&&&&mere&RFB&end&RFA&&&&&&&&&&&&&&&&&&&&&&&&&&&&&D&&&&&ved&ikke&&
Se&bort&fra&translaOon,&rotaOon,&og&vibraOon&
Entropi-og-udartning-
![Page 50: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/50.jpg)
SR−AKonf = R ln(4)
SKonf = nR ln(gKonf ) gKonf = antal konformationer med samme energi (udartning)
Konforma;onel-entropi-og-udartning-
A& B&R& R&
SR−BKonf = R ln(1) = 0
1 234
3 412
2 341
4 123
SR−AKonf = −R fi ln fi( )
i=1
4
∑≤ R ln(4)
f4 = brøkdel i denne konformation
![Page 51: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/51.jpg)
S = k ln(W ) = k ln gNA( ) = R ln(g) hvis all tilstande har samme energi
S ≈ R ln(g) ⇒ gprodukt
greaktant
≈ eΔS /R ≈ 10ΔS /20
H2O ⋅ ⋅ ⋅HOH 2H2O
ΔSo = −17.3+136.2 + 9.3− 66.0 = 79.4 J/molK
ΔSo = ΔSKonformation + ΔSo,Translation + ΔSRotation + ΔSVibration
g2H2O
Konf
gH2O⋅⋅⋅HOHKonf = ?
A. 1/8&&&&&&&&&&C.&4&
B. ½&&&&&&&&&&&&&D.&8&&
![Page 52: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/52.jpg)
H2O ⋅ ⋅ ⋅HOH 2H2O
ΔSo = −17.3+136.2 + 9.3− 66.0 = 79.4 J/molK
ΔSo = ΔSKonformation + ΔSo,Translation + ΔSRotation + ΔSVibration
g2H2OKonf
gH2O⋅⋅⋅HOHKonf =
18
g2H2OTrans
gH2O⋅⋅⋅HOHTrans ≈ 107
g2H2ORot
gH2O⋅⋅⋅HOHRot ≈ 5
g2H2OVib
gH2O⋅⋅⋅HOHVib ≈
11000
gAA = 6 gA+A = 15
2 × 3 vibrationer12 vibrationer
10006 ≈ 3 (næsten) ens tilstande per ekstra vibration
![Page 54: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/54.jpg)
ΔGo = ΔH o − TΔSo
−RT ln K( ) = ΔH o − TΔSo
ln K( ) = −ΔH o
R1T
⎛⎝⎜
⎞⎠⎟+ΔSo
R
Man&går&ud&fra&at&ΔHo&og&ΔSo&er&uacængig&af&temperatur&&
Hvordan-måler-man-ΔHo-og-ΔSo?:-van’t-Hoff-ligningen-&
Måling&af&K&ved&forskellige&temperaturer&giver&ΔHo&og&ΔSo?&&&
ln K( )1TΔSo
R
hældning = −ΔH o
RLav&T&
Høj&T&
![Page 55: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/55.jpg)
Her&vises&et&van’t&Hoff&plot&for&en&reakOon&&
Er&reakOonen&endoterm&eller&exoterm?&&&&
ln K( )1T
![Page 56: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/56.jpg)
A B
Hvis&reakOonen&er&endoterm&hvad&sker&der&med&ligevægtskonstanten&når&temperaturen&sOger?&
A. K&falder&
B. K&sOger&
C. K&er&uændret&
![Page 57: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/57.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.$Enthalpi$og$entropi$3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.,Måling,af,enthalpi,og,entropi,ændringer,vha,kalorimetri,
5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.$Hydrofobisitet$og$entropi$7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 60: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/60.jpg)
ΔH o = Q
= mCpvandΔT
Varmekapaciteten&ved&konstant&tryk&
h+p://www.youtube.com/watch?v=EAgbknIDKNo&
Hvordan(måler(man(ΔHo?:(calorimetri(
&
Måling&af&temperatur&ændring&giver&ΔHo&
Cp =∂H o
∂T⎛⎝⎜
⎞⎠⎟ p
![Page 61: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/61.jpg)
K&måles&med&andre&metoder&og&giver&ΔSo&
ΔSo =ΔH o − ΔGo
T
=ΔH o + RT ln K( )
T
Hvordan(måler(man(ΔHo?:(calorimetri(
&
Måling&af&temperatur&ændring&giver&ΔHo&
ΔH o = Q
= mCpvandΔT
![Page 62: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/62.jpg)
Hvordan(måler(man(ΔHo(og(ΔSo?:(differen5al(scanning(calorimetry(
&
Måling&af&varmekapacitetsændring&giver&ΔHo&
Rent&vand&
ΔH o Tm( ) = arealet under kurven
“Smeltepunkt”&
ΔCp
![Page 63: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/63.jpg)
K(Tm ) = 1⇒ ΔGo (Tm ) = 0
ΔSo (Tm ) =ΔH o (Tm )
Tm
ΔH o (T ) = ΔH o (Tm ) + ΔCp T − Tm( )
ΔSo (T ) = ΔSo (Tm ) + ΔCp lnTTm
⎛⎝⎜
⎞⎠⎟
![Page 64: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/64.jpg)
Hvad&er&enhederne&for&varmekapaciteten?&
A. J/mol&
B. J/molK&
C. K/mol&J&
D. mol/J&
![Page 66: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/66.jpg)
10&
Hvorfor(er(varmekapaciteten(et(maximum(når(ΔGo(Tm)(=(0?((
I&de+e&eksempel:&ΔV&=&0&så&U&bruges&istedet&for&Ho&
Når CV =∂U∂T
⎛⎝⎜
⎞⎠⎟V
er størst, så er ∂T∂U
⎛⎝⎜
⎞⎠⎟V
mindst
Dvs&den&]lførte&energi&bruges&]l&at&bryde&bindinger&i&stedet&for&at&hæve&T''
Når&(alle)&bindinger&brydes&spontant&er&Goreaktant(=(Go
produkt&
∂T∂U
⎛⎝⎜
⎞⎠⎟V
≈ 0
![Page 67: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/67.jpg)
_1&
_0.8&
_0.6&
_0.4&
_0.2&
0&
0.2&
0.4&
0& 500& 1000& 1500& 2000& 2500& 3000&
U((eV)(
T((K)(
11&
The&simula]on&computes&T&as&a&func]on&of&U&Here&we&switch&the&axes&
![Page 68: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/68.jpg)
_1&
_0.8&
_0.6&
_0.4&
_0.2&
0&
0.2&
0.4&
0& 500& 1000& 1500& 2000& 2500& 3000&
U((eV)(
T((K)(
12&
The&data&is&noisy&primarily&because&of&the&fast&hea]ng&rate&We&smooth&it&by&fihng&it&to&a&polynomial&(blue&curve)&
![Page 69: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/69.jpg)
0&
0.0002&
0.0004&
0.0006&
0.0008&
0.001&
0.0012&
0.0014&
0.0016&
_1&
_0.8&
_0.6&
_0.4&
_0.2&
0&
0.2&
0.4&
0& 500& 1000& 1500& 2000& 2500& 3000&
Cv((eV/K)(
U((eV)(
T((K)(
13&
From&the&smoothed&data&(blue&curve)&we&can&compute&the&&heat&capacity&(red&curve,&right&y_axis)&
CV =∂U∂T
⎛⎝⎜
⎞⎠⎟V
≈U(T2 ) −U(T1)
T2 − T1
![Page 70: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/70.jpg)
Hvilken&]lstand&har&den&højeste&
varmekapacitet?&
A(
B(
C(
![Page 71: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/71.jpg)
0&
0.0002&
0.0004&
0.0006&
0.0008&
0.001&
0.0012&
0.0014&
0.0016&
_1&
_0.8&
_0.6&
_0.4&
_0.2&
0&
0.2&
0.4&
0& 500& 1000& 1500& 2000& 2500& 3000&
Cv((eV/K)(
U((eV)(
T((K)(
16&
![Page 73: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/73.jpg)
Protein((polymer)(foldning(eksempel(
&
Termisk&denaturering&af&proteinet&Barnase&
Tm&
![Page 74: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/74.jpg)
Varme(+(
F U K =[U][F]
= e−ΔGo /RT
fU =[U]
[F]+ [U]=
e−ΔGo /RT
1+ e−ΔGo /RT
fF =1
1+ e−ΔGo /RT
ΔH o ≈ ΔH Molekyle ΔSo ≈ ΔSKonf = R ln(4) − R ln 1( )
ΔGo ≈ ΔH Molekyle − T R ln 4( )( )
Foldet( Udfoldet&
En(meget(simpel(protein(model(
![Page 75: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/75.jpg)
Tmax&~&99&K&Tm&~&216&K&
For&små&systemer,&hvor&ΔSo&er&lille,&er&varmekapaciteten&ikke&et&maximum&ved&smeltetemperaturen&og&Cp&kurven&er&
bred&&
&Tm&≠&Tmax'fF
fU
Cp
H molekyle = 2.5 kJ/mol
![Page 76: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/76.jpg)
gU&=&100&&&
&gU&=&10.000&
For&store&systemer,&hvor&ΔSo&er&stor,&er&varmekapaciteten&i&et&maximum&ved&smeltetemperaturen&og&Cp&kurven&er&skarp&
&&Tm&≈&Tmax'
fF
fUCp
![Page 77: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/77.jpg)
Varmekapaciteten(er(et(maximum(når(halvdelen(af(proteinerne(er(udfoldet(
&
Tm,(ΔHo(og(ΔSo,(kan(måles(spektroskopisk(via(van’t(Hoff(methoden((
C: θ230nm = θ230nmU fU +θ230nm
F fF ∝ fU
Tm&
fF fU
Cp
∝ fU
θ230nm → fU (T )→ K(T )→ ΔH o ,ΔSo
fU(T ) =K(T )
1+ K(T )
![Page 78: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/78.jpg)
En(lille(ændring(i(ΔHo(kan(have(en(stor(effekt(på(stabiliteten(&
fF fU
Cp
Ca&100%&folded&protein&ved&25&oC& Ca&50%&udfolded&protein&ved&25&oC&
ΔHo&reduceret&med&15&kJ/mol&≈&1&H_binding&
![Page 79: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/79.jpg)
Video&4.4&
Ved&hvilken&pH&er&ΔSo&størst?&
DOI:&10.1021/bi00129a007&
![Page 81: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/81.jpg)
![Page 82: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/82.jpg)
Varme(+(
F U fU =e−ΔG
o /RT
1+ e−ΔGo /RT
ΔSo ≈ R ln(4) ⇒ fU =4e−ΔH
o /RT
1+ 4e−ΔHo /RT
= 4 fU ,mikro
fU ,mikro < fF altid ⇒ udfoldning (fU > fF ) sker pga entropi
Foldet(Udfoldet&makro]lstand&
Udfoldning(ved(høje(temperaturer(sker(pga(entropi(
mikro]lstand&
Image&adapted&from&Molecular'Driving'Forces'by&Dill&and&Bromberg&
![Page 83: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/83.jpg)
fF
fU
fU,mikro
Udfoldet&makro]lstand&
mikro]lstand&
Udfoldning(ved(høje(temperaturer(sker(pga(entropi(
![Page 84: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/84.jpg)
A. Udartningen&for&den&udfoldede&makro]lstand&er&4&
B. Udartningen&for&den&udfoldede&mikro]lstand&er&4&
C. Udartningen&for&den&foldede&makro]lstand&er&1&
D. Entropien&for&den&udfoldede&mikro]lstand&er&størrer&end&for&den&foldede&
Hvilken&påstand&er&ikke&sandt&(der&kan&godt&være&mere&end&én)&
![Page 85: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/85.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.$Enthalpi$og$entropi$3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.,Enthalpi,og,entropi,for,en,ideal,opløsning,
6.$Hydrofobisitet$og$entropi$7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 87: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/87.jpg)
Department)of)Chemistry)
ΔGo = ΔGMolecule/Conformation + ΔGo,Translation + ΔGRotation + ΔGVibration + ΔΔGSolvation
Free$energy$contribu.ons$for$an$ideal$solu.on$
ΔH o,Trans = 32 nRT + poΔV ≈ 3
2 nRT + ΔngasRT
So,Trans = nR ln2πm( )3/2 e5 /2
h3C o
⎛
⎝⎜⎞
⎠⎟= nR ln bM 3/2T 3/2( ) b = 3.7487 mol3/2
g3/2K3/2
Co)=)1.0)mol/L)
Ideal)soluEon)=)no)interacEons)between)
)))))))))))))))))))))))))))))solute)molecules)
H o,Trans ≈ 32 nRT
Solute:)the)molecule)that)is)dissolved)
)
Solvent)3)
![Page 88: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/88.jpg)
Department)of)Chemistry)
4)
ΔGo = ΔGMolecule/Conformation + ΔGsolutiono,Translation + ΔGRotation + ΔGVibration + ΔΔGSolvation
ΔGgaso + 5.5 A ⋅B A + B
ΔGgaso − 5.5 A + B A ⋅B
ΔGgaso A B
Ssolutiono,Trans = Sgas
o,Trans − R ln 0.31173.7487
T⎛⎝⎜
⎞⎠⎟= Sgas
o,Trans − 26.69 J/molK
H solutiono,Trans = Hgas
o,Trans − RT = Hgaso,Trans − 2.5 kJ/molK at$25$oC$
Gsolutiono,Trans = Hgas
o,Trans − 2.5( ) − T Sgaso,Trans − 0.02669( )
= Ggaso,Trans − 2.5 − 298.15(−0.02669)
= Ggaso,Trans + 5.5 kJ/mol
![Page 89: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/89.jpg)
Department)of)Chemistry)
5)
What)is)ΔH°correcEon
)at)50)°C)for)this)reacEon?)
A+B A ⋅B
A. O10.0)kJ/mol)
B. O2.7)kJ/mol)
C. 2.7)kJ/mol)
D. 10)kJ/mol))
ΔH o = ΔH Molecule + ΔH solutiono,Translation + ΔH Rotation + ΔH Vibration + ΔΔH Solvation
ΔHgaso + ΔH correction A+ B A ⋅B
![Page 91: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/91.jpg)
ΔGo = ΔGMolekyle/Konformation + ΔGo,Translation + ΔGRotation + ΔGVibration + ΔΔGSolvering
Solveringsfrienergi:$det$polære$bidrag$
+&
ΔGSolvering (A)
ΔGSolvering = ΔGpolærSolvering + ΔGikke polær
Solvering
A$ A$
![Page 92: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/92.jpg)
ΔGpolærSolvering ≈ −
694.7q2
R1− 1
ε⎛⎝⎜
⎞⎠⎟−60.25 ε −1( )µ2
2ε +1( )R3
Hvis&molekylet&er&kugleformet&og&solvent&beskrives&som&et&homogent&felt&med&dielectrisk&konstant&ε&
R#
ε#
≈&
q&=&ladningen&på&molekylet&&&&&&&&&&μ&=&dipolmomentet&(i&Debye)&&R&=&radius&a&kuglen&(i&Å)&&&&&&&&&&&&&&&&&ε&=&solventets&dielektriske&konstant&(1&<&ε#<&∞)&
694.7 kJ Åmol
og 60.25 kJ Å3
mol Debye2
![Page 93: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/93.jpg)
ΔGpolærSolvering ≈ −
694.7q2
R1− 1
ε⎛⎝⎜
⎞⎠⎟−60.25 ε −1( )µ2
2ε +1( )R3
Ud&fra&denne&ligning&hvad&er&sansynligvis&ikke&sandt?&(der&kan&være&mere&end&et&usandt&svar)&
A.&en&ion&har&en&større&solveringsenergi&end&et&neutralt&molekyle&&B.&alt&andet&lige,&et&lille&molekyle&har&en&mindre&solveringsenergi&end&et&stort&&C.&et&molekyle&har&en&størrer&solveringsenergi&i&et&opløsningsmiddel&med&stor&ε##D.&alt&andet&lige,&en&anion&har&en&større&solveringsenergi&&end&en&ka`on&&E.&et&neutralt&ikke&polært&molekyle&har&en&lille&solveringsenergi&&&&
solveringsenergi = ΔGpolærSolvering
![Page 95: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/95.jpg)
h+ps://www2.chemistry.msu.edu/faculty/reusch/vir+xtjml/enrgtop.htm&
+$ <$+&+&
+&
+&
+&
+& +&+&+&
+&
+&+&
i&
i&
i&
i&
i&i&i&
i&i&
i&
i&
i&
E =1389qAqB
εr
1389 kJ Åmol
ε# εvand = 80
εvakuum = 1r#
Solvent$“screening”$elektrosta`ske&vækselvirkninger&svækkes&af&solventen&
![Page 96: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/96.jpg)
I&hvilken&solvent&er&elektrosta`ske&vækselvirkninger&størst?&
A.&vand&&B.&acetone&&C.&cyclohexan&&D.&kloroform&
![Page 98: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/98.jpg)
ΔGSolvering = ΔGpolærSolvering + ΔGikke polær
Solvering
ΔGikke polærSolvering = ΔGkavitation
Solvering + ΔGvan der WaalsSolvering
ΔGkavitationSolvering = γ osSASA
γos&=&solventets&overfladespænding&&SASA&=&arealet&af&molekyle/solvent&overflade&#############(SASA&=&solvent&accessible&surface&area)&
Den$ikke<polære$solvaAonsenergi$
h+p://www.liv.ac.uk/researchintelligence/issue32/tension.htm&
![Page 99: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/99.jpg)
ΔGSolvering = ΔGpolærSolvering + ΔGikke polær
Solvering
ΔGikke polærSolvering = ΔGkavitation
Solvering + ΔGvan der WaalsSolvering
ΔGvan der WaalsSolvering ≈ −cSASA
c&=&empirisk&konstant&&
![Page 100: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/100.jpg)
ΔGikke polærSolvering = ΔGkavitation
Solvering + ΔGvan der WaalsSolvering
≈ γ osSASA − cSASA≈ γ ipSASA + b
γos&=&0.438&kJ/molÅ2&&&γip&=&0.0227&kJ/molÅ2&(b&=&3.85&kJ/mol)&&&
for&vand&ved&25&oC:&
DOI:&10.1063/1.4745084&&&&DOI:&10.1021/j100058a043&
![Page 101: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/101.jpg)
A$ B$
C$ D$
for&hvilken&`lstand&er&den&ikkeipolære&solva`onsenergi&mindst?&&
ΔGikke polærSolvering ≈ γ ipSASA + b
![Page 103: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/103.jpg)
Den$hydrofobe$effekt$Ikkeipolære&molekyler&opløst&i&vand&binder&stærkere&end&i&vakuum&
for&at&mindske&kontakten&med&solventen&(dvs&SASA)&
ΔGo ≈ ΔΔGikke polærSolvering ≈ γ ipΔSASA + b
SASAX+X SASAX⋅X
SASAX⋅X < SASAX+X ⇒ ΔGikke polærSolvering (X ⋅ X) < ΔGikke polær
Solvering (X + X)
Arealet&er&lavest&
![Page 104: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/104.jpg)
A$ B$ C$ D$
|ΔSASA|&størst&for&B$
Ud&fra&følgende&`lnærmelse,&hvilket&“molekyle”&bindes&sandsynligvis&stærkest&med&sig&selv?&
ΔGo ≈ ΔΔGikke polærSolvering ≈ γ ipΔSASA + b
![Page 105: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/105.jpg)
Olie&er&hydrofobisk&
h+p://youtu.be/D6aoJNqt1MQ&
![Page 106: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/106.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.$Enthalpi$og$entropi$3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.,Hydrofobisitet,og,entropi,
7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 108: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/108.jpg)
Hvad&er&&&&&&&&&&for&denne&simulaDon?&
A ΔSo = 0B ΔSo < 0C ΔSo > 0
ΔSo
h+p://youtu.be/zKNmBjqGijI&
![Page 109: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/109.jpg)
+&
+& +&
ΔSo > 0
2&parDkler& 1&parDkel&
2&parDkler& 5&parDkler&
Solventen(bidrager(/l(entropiændringen(
vandet&i&bindingslommen&er&anderledes&=&hydrofob&
![Page 110: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/110.jpg)
+& +&
h+p://youtu.be/1WkZznwmO0c&
![Page 111: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/111.jpg)
+& +&
h+p://youtu.be/ETMmH2trTpM&
ΔSo > 0
vandet&tæ+est&på&liganden&er&anderledes&=&hydrofob&
![Page 112: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/112.jpg)
ΔSo,Trans = ? ved 25 oC
(H2O)4'smelter'i'vand'
A.'2.4'J/molK''B.'35.2'J/molK''C.'99.2'J/molK''D.'337.0'J/molK'
![Page 113: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/113.jpg)
ΔSo,Trans = ? ved 25 oC
Sopløsningo,Trans = nR ln 3.7487M 3/2T 3/2( )
ΔSo,Trans = 4R ln 3.7487M 3/2T 3/2( ) − R ln 3.7487 4M( )3/2 T 3/2( )= 3R ln 3.7487M 3/2T 3/2( )
118.1
− 32 ln 4( )
17.3
= 337.0 J/molK
“isbjergsmodellen”'kun'kvalitaHv'ΔSo'kommer'sandsynligvis'mest'fra'ΔSvib'og'ΔSkonf''
'
(H2O)4'smelter'i'vand' A.'2.4'J/molK''B.'35.2'J/molK''C.'99.2'J/molK''D.'337.0'J/molK'
![Page 115: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/115.jpg)
h+p://youtu.be/ETMmH2trTpM&
ΔSo > 0 ΔH o ≈ +0ΔGo < 0
+&
Entropien(s/ger(når(hydrofobe(molekyler(bindes(
![Page 116: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/116.jpg)
ΔGikke polærSolvering = ΔGkavitation
Solvering + ΔGvan der WaalsSolvering
≈ γ osSASA − cSASA≈ γ ipSASA + b
ΔSikke polærSolvering =
∂ΔGikke polærSolvering
∂T
≈∂γ os
∂TSASA −
∂c∂T
SASA
≈∂γ os
∂TSASA
∂γ os
∂T= −0.00112 kJ/molÅ2K
Den&ikkePpolære&solveringsentropi&er&negaDv&
![Page 117: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/117.jpg)
+&
h+p://youtu.be/ETMmH2trTpM&
ΔSo > 0 ΔH o ≈ +0ΔGo < 0
ΔSo ≈ ΔΔSikke polærSolvering
≈∂γ os
∂T<0
ΔSASA<0
> 0
En&negaDv&ikkePpolær&solveringsentropi&betyder&at&entropien&sDger&når&hydrofobe&molekyler&bindes&
&
![Page 118: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/118.jpg)
benzen(benzen) benzen(aq)Hvad&er&&&&&&&&&&for&denne&proces?&
A ΔSo = 0B ΔSo < 0C ΔSo > 0
ΔSo
![Page 120: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/120.jpg)
Hvordan(måler(man(hydrofobisitet?(
solute(vand) solute(octanol)
log(Pwo ) = log[solute(octanol) ][solute(vand) ]
⎛
⎝⎜⎞
⎠⎟
log(P)&>&1&=&hydrofobisk&
ε&=&80&ε&=&10&
![Page 121: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/121.jpg)
Cl((O((C(((H( Hvilket&molekyle&har&det&laveste&log(Pwo)?&&
![Page 123: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/123.jpg)
P + L P ⋅L
For&hvilken&ligand&vil&ΔSo&være&mest&posiDv?&
Ligand&binding&Dl&enzymet&carbonic&anhydrase&(ved&25&oC)&
![Page 124: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/124.jpg)
ΔGo&&&&&&&&&&&&&&&P56.5&&&&&&&&&&&&&&&&&&&&&&&&&P54.4&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&P55.2&&&&&&&&&&&&&&&&&&&&&&&&&&&P55.6&kJ/mol&&ΔHo&&&&&&&&&&&&&&&P79.1&&&&&&&&&&&&&&&&&&&&&&&&&P68.2&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&P51.9&&&&&&&&&&&&&&&&&&&&&&&&&&&P35.1&kJ/mol&&PTΔSo&&&&&&&&&&&&&23.0&&&&&&&&&&&&&&&&&&&&&&&&&&&14.2&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&P2.9&&&&&&&&&&&&&&&&&&&&&&&&&&&P20.1&kJ/mol&&log(Pwo)&&&&&&&&&0.25&&&&&&&&&&&&&&&&&&&&&&&&&&1.33&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&P0.65&&&&&&&&&&&&&&&&&&&&&&&&&&&0.12&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
P + L P ⋅L
Binding&Dl&enzymet&carbonic&anhydrase&(ved&25&oC)&
DOI:&10.1002/anie.201301813&
![Page 125: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/125.jpg)
+& +&
+& +&
octanol&
log(Pwo)(måler(den(totale(hydrofobisitet(af(liganden((
![Page 126: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/126.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.$Enthalpi$og$entropi$3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.$Hydrofobisitet$og$entropi$7.,Kemisk,akLvitet,og,ikkeMideale,opløsninger,
8.$Termodynamikens$tre$love$og$Boltzmannfordelingen$
![Page 128: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/128.jpg)
A B+ C K =[B][C]
[A] ΔGo = −RT ln(K )
ΔGo = −J
mol K⎛⎝⎜
⎞⎠⎟
K( ) ln(K ) = J/mol
ΔGo = −
Jmol K
⎛⎝⎜
⎞⎠⎟
K( ) ln M ⋅MM
⎛⎝⎜
⎞⎠⎟= J/molln(M)
??
Ligevægtskonstanten-har-ingen-enheder-
![Page 129: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/129.jpg)
ΔGo = ΔGMolekyle/Konformation + ΔGo,Translation + ΔGRotation + ΔGVibration + ΔΔGSolvering
Fri-energibidrag-for-en-ideal-opløsning-
ΔH o,Trans = 32 nRT + poΔV ≈ 3
2 nRT + ΔngasRT
So,Trans = nR ln2πm( )3/2 e5 /2h3C o
⎛
⎝⎜⎞
⎠⎟Co&=&1.0&mol/L&
Ideal&opløsning&=&ingen&vækselvirkninger&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&mellem&solute&molekyler&
H o,Trans ≈ 32 nRT
Solute:&molekylet&der&er&opløst&&Solvent&
![Page 130: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/130.jpg)
STrans = R ln 2πm( )3/2 e5/2h3C
⎛
⎝⎜⎞
⎠⎟
= R ln 2πm( )3/2 e5/2
h3C o CC o
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
= R ln 2πm( )3/2 e5/2h3C o
⎛
⎝⎜⎞
⎠⎟− R ln C
C o⎛⎝⎜
⎞⎠⎟
= So,Trans − R ln CC o
⎛⎝⎜
⎞⎠⎟
S = So + R ln CC o
⎛⎝⎜
⎞⎠⎟⇒G = Go + RT ln C
C o⎛⎝⎜
⎞⎠⎟
STrans,&og&derfor&G,&er&en&funkKon&af&koncentraKon,&C&
![Page 131: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/131.jpg)
R P
G(X) = Go (X) + RT ln [X]C o
⎛⎝⎜
⎞⎠⎟
G(P) −G(R) = Go (P) −Go (R) + RT ln[P]
C o
[R]C o
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
ΔG = ΔGo + RT ln[P]
C o
[R]C o
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
![Page 132: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/132.jpg)
ΔG = 0⇒[P]
C o
[R]C o
= e−ΔGo /RT
vi skriver K =[P][R]
men mener K =[P]
C o
[R]C o
A B+ C
K =[B][C]
[A]
K-har-ingen-enheder,&men&bliver&Kt&angivet&som,&f.eks.&0.001&M&eller&1&mM&
ved&ligevægt&
ingen&enheder&
Ligevægtskonstanten-har-ingen-enheder-
R P
![Page 133: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/133.jpg)
G(X) = Go (X) + RT ln [X]C o
⎛⎝⎜
⎞⎠⎟
G(X) = Go (X) + RT ln pXpo
⎛⎝⎜
⎞⎠⎟
G(X) = Go (X) + RT ln [X]CXo
⎛⎝⎜
⎞⎠⎟
Co&=&1&M&ideal&opløsning&&
po&=&1&bar&ideal&gas&&
CXo&=&koncentraKon&af&ren&væske&eller&faststof&
opløsning&
gas&
faststof/væske&
H2O(l ) H2O(g)
K =
pH2Opo
⎛⎝⎜
⎞⎠⎟
[H2O]CH2Oo
⎛⎝⎜
⎞⎠⎟
1
=pH2O
po⎛⎝⎜
⎞⎠⎟ pH2O[H2O] = CH2O
o = 55.56 M
forskellige-standard7lstande-
uaMængig&af&fordampning&
![Page 134: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/134.jpg)
H2CO3(aq) H2O(l ) + CO2(g)
A. K =[HH2O ]pCO2
[H2CO3]
B. K =[HH2O ]
[H2CO3]
C. K =pCO2
[H2CO3]D. K = pCO2
Hvad&er&ligevægtskonstanten&for&denne&reakKon?&
![Page 136: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/136.jpg)
PbI2(s ) Pb(aq)2+ + 2I(aq)
−
K = [Pb2+ ][I− ]2
ΔGo = 46.1 kJ/mol ⇒ K = 5.93×10−9
⇒ [Pb2+ ] = 1.14 ×10−3 M
målt: [Pb2+ ] = 1.37 ×10−3 M
Mere&Pb2+&opløst&end&forventet!&
K = e−ΔGo /RT
20&oC&
![Page 137: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/137.jpg)
Vækselvirkninger&mellem&solute&molekyler&øger&koncentraKonen&af&fri&ioner&sammenlignet&med&ideal&opløsning&
![Page 138: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/138.jpg)
Vækselvirkninger&mellem&solute&molekyler&øger&koncentraKonen&af&fri&ioner&sammenlignet&med&ideal&opløsning&
+- +-
+-
9-
9-
![Page 139: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/139.jpg)
h+p://en.wikipedia.org/wiki/Debye%E2%80%93H%C3%BCckel_equaKon&
Vækselvirkninger&mellem&solute&molekyler&øger&koncentraKonen&af&fri&ioner&sammenlignet&med&ideal&opløsning&
![Page 140: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/140.jpg)
PbI2(s ) Pb(aq)2+ + 2I(aq)
−
K = aPb2+
aI−2
ΔGo = 46.1 kJ/mol ⇒ K = 5.93×10−9
⇒ aPb2+ = 1.14 ×10−3 M
målt: [Pb2+ ] = 1.37 ×10−3 M
Mere&Pb2+&opløst&end&forventet&
K = e−ΔGo /RT
aPb2+ = γ
Pb2+ [Pb2+ ]⇒γPb2+ =
1.14 ×10−3 M1.37 ×10−3 M
= 0.83
![Page 141: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/141.jpg)
Ak7vitet-(a)-for-en-ikke9ideal-opløsning-
R(aq) P(aq)
K =aPaR
= e−ΔGo /RT aP = γ P
[P]C o
ingen&enheder&
aP = 1⇒γ P = 1, [P]C o = 1
standard&Klstand&1&M&ideal&opløsning&
Kc =[P][R]
⇒ K =γ P
γ R
Kc ⇒ ΔGo = −RT ln K( )
ΔGo ⇒ K = e−ΔGo /RT ⇒ Kc =
γ R
γ P
K
γ&kaldes&akKvitetskoefficienten&
![Page 142: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/142.jpg)
A γNa+
≈ γCa2+
B γNa+
< γCa2+
C γNa+
> γCa2+
D ved ikke
Hvad&vil&du&forvente&(samme&koncentraKon)?&
![Page 143: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/143.jpg)
h+p://en.wikipedia.org/wiki/Debye%E2%80%93H%C3%BCckel_equaKon&
Vækselvirkninger&mellem&solute&molekyler&øger&koncentraKonen&af&fri&ioner&sammenlignet&med&ideal&opløsning&
![Page 145: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/145.jpg)
γ ± = 10− q+q− A I
I = 12 q+
2[+]+ q−2[−]( )
γ±&=&middel&akKvitetskoefficienten&'q+&=&ladning&af&kaKoner&&I&=&ionstyrke&&A&=&0.509&for&vandig&opløsning&ved&25&oC&&[+]&=&koncentraKon&af&kaKoner&
Den-simple-Debye9Hückel-ligning-(the&limited&DebyejHückel&law)&
KCl(s ) K(aq)+ + Cl(aq)
−
K = aK+aCl−
= γK+ [K+ ]γ
Cl−[Cl− ] = γ
K+γ Cl−( )[K+ ][Cl− ] = γ ±2[K+ ][Cl− ]
![Page 146: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/146.jpg)
γ ± = 10− q+q− A I
I = 12 q+
2[+]+ q−2[−]( )
Den-simple-Debye9Hückel-ligning-(the&limited&DebyejHückel&law)&
middel&akKvitetskoefficienten&for&0.001&M&CaCl2&
I = 12 q+
2[+]+ q−2[−]( )
= 12 +2( )2 0.001( ) + −1( )2 0.002( )⎡⎣ ⎤⎦
= 0.003
γ ± = γCa2+2 γ
Cl−( )1/3= 10− q+q− A I
= 10− +2⋅−1 0.509 0.003
= 0.88målt=&0.89&
![Page 147: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/147.jpg)
γ ± = 10− q+q− A I I = 1
2 q+2[+]+ q−
2[−]( )
Den-simple-Debye9Hückel-ligning-(the&limited&DebyejHückel&law)&
Ud&fra&denne&ligning&hvad&er&sansynligvis&ikke&sandt?&(der&kan&være&mere&end&et&usandt&svar)&
A.&&γ±&går&mod&1&når&ionkoncentraKonen&falder&&B.&alt&andet&lige&har&Zn2+&en&mindre&γ±&end&Brj&&&C.&γ±&er&0&for&neutrale&molekyler&&D.&0.01&M&CaCl2&har&en&størrer&γ±&end&0.01&M&CaSO4&&&E.&γ±&kan&være&størrer&end&1&&&
![Page 149: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/149.jpg)
PbI2(s ) Pb(aq)2+ + 2I(aq)
−
5.93×10−9 = [Pb2+ ][I− ]2
= x 2x( )2
= 4x3
⇒ [Pb2+ ] = 1.14 ×10−3 M
målt: [Pb2+ ] = 1.37 ×10−3 M
Mere&Pb2+&opløst&end&forventet!&
Den-simple-Debye9Hückel-ligning:-et-eksempel-
![Page 150: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/150.jpg)
PbI2(s ) Pb(aq)2+ + 2I(aq)
−
5.93×10−9 = γPb2+[Pb2+ ]γ
I−2 [I− ]2 = γ ±
3[Pb2+ ][I− ]2
⇒ [Pb2+ ] = 1.32 ×10−3 M
målt: [Pb2+ ] = 1.37 ×10−3 M
Mere&Pb2+&opløst&end&forventet!&
γ ± = 10−2 0.509( ) I I = 1
2 22[Pb2+ ]+ [I- ]( )
5.93×10−9 = 10−2 0.509( ) 3x( )3 4x3( )
![Page 151: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/151.jpg)
log γ ±( ) = −A q−q+ I1+ B I
+ CI
γ ± = 10− q+q− A I
Den-simple-Debye9Hückel-ligning-(the&limited&DebyejHückel&law)&
Den-udvidede-Debye9Hückel-ligning-(the&extended&DebyejHückel&law)&
I ≤ 0.01 M
log γ ±( ) = −0.509 q−q+ I
1+ I− 0.3IDavies:& I ≤ 0.1 M
![Page 152: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/152.jpg)
Termodynamik,for,Biokemikere,,
Jan$H.$Jensen$Københavns$Universitet$
1.$Ligevægt$og$ligevægtskonstanten$2.$Enthalpi$og$entropi$3.$Enthalpi$og$entropi$for$an$ideal$gas$og$van’t$Hoff$ligningen$4.$Måling$af$enthalpi$og$entropi$ændringer$vha$kalorimetri$5.$Enthalpi$og$entropi$for$en$ideal$opløsning$6.$Hydrofobisitet$og$entropi$7.$Kemisk$akIvitet$og$ikkeJideale$opløsninger$8.,Termodynamikens,tre,love,og,Boltzmannfordelingen,
![Page 154: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/154.jpg)
dU = dq + dw
qTkoldt
−q
Tvarmt> 0
ΔSunivers > 0
dS = dqrevT
dqrev&=&en&varmeoverførsel&der&så&lille&at&T&er&upåvirket&&&&&&&&&&&&&&T&er&upåvirket&=&reversible&process&=&ligevægt&
S 0 K( ) = 0
U&=%indre&energi%q&=&varme&&w&=&arbejde&
Termodynamikkens&3&love&
1.&
2.&
3.&
Varme&overførers&spontant'fra&varme&Ll&kolde&legemer&
![Page 155: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/155.jpg)
ΔSunivers ≥ 0dSsystem + dSbad ≥ 0
dSsystem +dqbadT
≥ 0
dSsystem −dqsystemT
≥ 0
dSsystem −dUsystem + pdVsystem( )
T≥ 0
0 ≥ dH system − TdSsystem0 ≥ dGsystem
2.'lov' >&for&spontan&process&=&ved&ligevægt&&
dU = dq + dw= dq − pdV
dq = dU + pdV
dGsystem ≡ dG = 0
universets'entropi's0ger,'og'systemets'fri'energi'falder,'for'en'spontant'process'
Ligevægt:& ⇒ K = e−ΔGo /RT
Dill&&&Bromberg&Molecular%Driving%Forces%
![Page 156: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/156.jpg)
Hvad&er&sandt?&
A.&G&er&uaUængig&af&koncentraLon,&Go&er&aUængig&af&koncentraLon&&B.&ΔG&er&fri&energi&ændring&for&blandingen&af&reaktant&og&produkt,&&&&&&ΔGo&er&fri&energi&forskellen&i&fri&energi&af&reaktant&eller&produkt&i&deres&standard&Llstande&&C.&ΔG&er&den&fri&energi&ændring&ved&et&tryk&andet&end&1&bar,&&&&&&ΔGo&er&den&fri&energi&ændring&for&et&tryk&=&1&bar,&&&D.&Go&den&lavest&mulige&værdi&af&G%
![Page 158: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/158.jpg)
A&B&
B&A&
A&B&
A&Ll&venstre&x&B&Ll&venstre&½&x&½&=&¼&
½&x&½&=&¼&½&x&½&=&¼&
½&x&½&=&¼&
![Page 159: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/159.jpg)
½&x&½&x&½&=&⅛&
½&x&½&x&½&=&⅛&
⅛& ⅛& ⅛&
A&B& C&
A&C& B&
B&C& A&
3&x&⅛&
![Page 161: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/161.jpg)
0&
0.1&
0.2&
0.3&
0.4&
0.5&
0.6&
0& 1& 2&
sand
synlighe
d'
Antal'par0kler'i'højre'side'
NH
p(NH )
p(0) = 1× 12( )2 p(2) = 1× 1
2( )2
p(1) = 2 × 12( )2
![Page 162: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/162.jpg)
0&
0.05&
0.1&
0.15&
0.2&
0.25&
0.3&
0.35&
0.4&
0& 1& 2& 3&
sand
synlighe
d'
antal'par0kler'i'højre'side'
NH
p(NH )
p(0) = 1× 12( )3 p(3) = 1× 1
2( )3
p(1) = 3× 12( )3 p(2) = 3× 1
2( )3
p(NH ) = 3!NH !(3− NH )!
12( )3 3!= 3 ⋅2 ⋅1
![Page 163: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/163.jpg)
0&
0.01&
0.02&
0.03&
0.04&
0.05&
0.06&
0.07&
0.08&
0.09&
0& 5& 10& 15& 20& 25& 30& 35& 40& 45& 50& 55& 60& 65& 70& 75& 80& 85& 90& 95& 100&
sand
synlighe
d'
antal'par0kler'i'højre'side'
p(NH ) =100!
NH !(100 − NH )!12( )100
=W 12( )100
NH
p(NH )
![Page 164: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/164.jpg)
dGsystem ≡ dH system − TdSsystem = −TdSsystem ≤ 0
Hvorfor'fylder'gassen'spontant'begge'beholdere?'
systemet&går&spontant&mod&den&Llstand&der&har&den&højeste&sandsynlighed&&systemet&går&spontant&mod&den&Llstand&der&har&den&højeste&mulLplicitet&(W)&&
systemet&går&spontant&mod&den&Llstand&der&har&den&højeste&entropi&&
S ∝W
p(NH ) =W (NH ) 12( )100
![Page 165: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/165.jpg)
S ∝W
S = k ln W( )
S'='k'ln(W)'
dS = dqrevT
S&har&J/K&enheder&
dS = dSA + dSBW =WAWB
ln WA( ) + ln WB( ) = ln WAWB( )k = R
NA
S&er&addiLv&
![Page 166: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/166.jpg)
Hvad&er&ΔS&for&denne&process?&(N&=&100)&
A.&1.11&x&10j18&J/K&&&B.&5.32&x&10j19&J/K&&C.&8.47&x&10j20&J/K&&D.&2.96&x&10j21&J/K&&E.&9.22&x&10j22&J/K&
![Page 167: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/167.jpg)
Hvad&er&ΔS&for&denne&process?&(N&=&100)&
ΔS = S(NH = 50) − S(NH = 0)
= k ln 100!50!( )2
⎛
⎝⎜⎞
⎠⎟− k ln 100!
0! 100!( )⎛⎝⎜
⎞⎠⎟
1
= 1.38 ×10−23 J/K( ) 66.8( )= 9.22 ×10−22 J/K≈ 100k ln(2)
W =100!
NH !(100 − NH )!
![Page 169: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/169.jpg)
S = k ln W( ) = k ln N !
N1!N2 !…Nt !⎛⎝⎜
⎞⎠⎟
ε1&=&0&ε2&&ε3&&
N1&=&antal&molekyler&&&&&&&&&med&energi&ε1&
ε2&&ε1&=&0&
ε2&&ε1&=&0&
′S = k ln N !( ) − ln N1 −1( )!( ) − ln N2 +1( )!( )⎡⎣ ⎤⎦S = k ln N !N1!N2
⎛⎝⎜
⎞⎠⎟
= k ln N !( ) − ln N1!( ) − ln N2 !( )⎡⎣ ⎤⎦
dU = ε2 − ε1 ⇒ N1 → N1 −1 and N2 → N2 +1
Boltzmann'fordelingen'
![Page 170: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/170.jpg)
dS = ′S − S
= k ln N1!N1 −1( )!
⎛
⎝⎜⎞
⎠⎟+ ln N2 !
N2 +1( )!⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
≈ k ln N1( ) + ln 1N2 +1
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
= k ln N1N2 +1
⎛⎝⎜
⎞⎠⎟
≈ k ln N1N2
⎛⎝⎜
⎞⎠⎟
ε2&&ε1&=&0&
ε2&&ε1&=&0&
dU = ε2 − ε1 ⇒ N1 → N1 −1 and N2 → N2 +1
ln x!( ) ≈ x ln(x) − xstor&x%
![Page 171: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/171.jpg)
ε2&&ε1&=&0&
ε2&&ε1&=&0&
dU = ε2 − ε1 = ε2 ⇒ N1 → N1 −1 and N2 → N2 +1
dS = dUT
k ln N1N2
⎛⎝⎜
⎞⎠⎟=ε2T
k ln N2
N1
⎛⎝⎜
⎞⎠⎟= −
ε2T
N2
N1= e−ε2 /kT
Ni = e−εi /kT N1
N1 e−εi /kTi∑ = N
N1N
=1e−εi /kT
i∑
Ni
N=
e−εi /kT
e−εi /kTi∑
ved&ligevægt:&
![Page 172: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/172.jpg)
Ni
N=
e−εi /kT
e−εi /kTi∑
pi =e−εi /kT
q
Boltzmann'fordelingen'den&fordeling&af&energi&blandt&molekyler&
der&har&den&laveste&fri&energi&
pi&=&sandsynligheden&for&at&et&molekyle&har&energi&εi&eoer&ligevægt&&&q&kaldes&Llstandssummen&(parLLon&funcLon)&
ε1&=&0&ε2&&ε3&&
Ni&=&antal&molekyler&&&&&&&&&med&energi&εi&
![Page 173: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/173.jpg)
ε1&=&0&ε2&&ε3&&Hvad&er&S&for&denne&fordeling?&
A.&1.91&x&10j20&J/K&&&B.&1.22&x&10j22&J/K&&C.&8.58&x&10j23&J/K&&D.&3.36&x&10j24&J/K&&E.&6.42&x&10j25&J/K&
![Page 175: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/175.jpg)
ΔSunivers ≥ 0 og S = k ln W( )⇒ pi =e−εi /kT
q
U =U(0) + N ε
ε = piεii∑ =
1q
εie−βεi
i∑ = −
1q
ddβ
e−βεii∑⎡
⎣⎢
⎤
⎦⎥ = −
1qdqdβ
Boltzmann'fordelingen'giver'ligningerne'for''fri'energibidrag'for'en'ideal'gas'
β = 1kT
S = k ln N !N1!N2 !…Nt !
⎛⎝⎜
⎞⎠⎟
= k ln Ni lnN − Ni lnNi( )i∑ = −k Ni ln
Ni
Ni∑
= −k Nii∑ ln e−βεi
q⎛⎝⎜
⎞⎠⎟= kβ Niεi
i∑ + Nk lnq
=UT
+ Nk lnq
ε1&=&0&ε2&&ε3&&
![Page 176: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/176.jpg)
Eksempel:'den'transla0onelle'indre'energi'
εnxnynzT = εnx
T + εnyT + εnz
T
εnxT = (nx
2 −1) h2
8mX 2 = (nx2 −1)ε nx = 1,2,3,....
Kvantemekanik&(parLkel&i&en&kasse&med&længde&X%og&masse%m)&
qXT = e−βεi
i=1
∞
∑ = e−(n2 −1)βε dn
1
∞
∫ ≈ e−n2βε dn
0
∞
∫ ≈2πmh2β
X
qT =2πmh2β
⎛⎝⎜
⎞⎠⎟
3/2
XYZ =2πmh2β
⎛⎝⎜
⎞⎠⎟
3/2
V
εT = −1q
dqdβ
⎛⎝⎜
⎞⎠⎟V
= 32 kT
UT =UT (0)
0 + N εT = 3
2 nRT
ε Trans
kT≈ 10−20
![Page 177: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/177.jpg)
ε Trans
kT≈ 10−20 ⇒U Trans =
32nRT ⇒ H Trans =U Trans + pV =
52nRT
εRot
kT≈ 0.01− 0.001⇒U Rot = H Rot =
32nRT
εVib
kT≈ 2 −10⇒UVib = H Vib = nNAhc ν i
12 +
1eNAhc νi /RT −1
⎛⎝⎜
⎞⎠⎟i=1
3Nat −X
∑
UVib =UVib (0)
≠0 + N εVib
Vibra0on'er'lidt'anderledes'
ε1&=&0&ε2&&ε3&&
![Page 178: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/178.jpg)
pi =e−εi /kT
q og S = U
T+ Nk lnq
STrans = nR ln2πm( )3/2 kTe( )5 /2
h3p⎛
⎝⎜⎞
⎠⎟
SRot = nR ln 8π 2keTh2
⎛⎝⎜
⎞⎠⎟
3/2
π I1I2I3
⎡
⎣⎢⎢
⎤
⎦⎥⎥
SVib = nR NAhc ν i
RT eNAhc νi /RT −1( ) − ln 1− e−NAhc νi /RT( )⎛
⎝⎜
⎞
⎠⎟
i=1
3Nat −6
∑
εiTrans ,εi
Vib , og εiVib fra kvantemekanik
![Page 179: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/179.jpg)
S = k ln W( ) = k ln N !
N1!N2 !…Ng !⎛
⎝⎜⎞
⎠⎟
Ni
N=
e−εi /kT
e−εi /kTi∑
ε1 = ε2 =… = εg⇒ N1 = N2 =… = Ng = N g
W =N !
N g( )!g =N e( )N
N g( ) N g( )g
=N e( )N
N e( ) 1 g( )⎛
⎝⎜
⎞
⎠⎟
N
= gN
Entropi'og'udartning'
S = k ln W( ) = Nk ln g( )
x!≈ x e( )x
![Page 180: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/180.jpg)
ε1&=&0&ε2&=&kT&&ε3&=&2kT&Hvad&er&U&for&denne&fordeling?&
A. U =9qkTe−1 + 2kTe−2( )
B. U =9qkTe + 2kTe2( )
C. U =9q
1+ kTe−1 + 2kTe−2( )D. U = kTe−1 + 2kTe−2( )
![Page 181: Thermodynamics for Biochemists: a YouTube textbook](https://reader035.vdocument.in/reader035/viewer/2022062503/58760d981a28ab306c8b49c1/html5/thumbnails/181.jpg)
ε1&=&0&ε2&=&kT&&ε3&=&2kT&Hvad&er&U&for&denne&fordeling?&
U =U(0)
0 + N ε
ε = piεi
i∑ = p1 ε1
0 + p2ε2 + p3ε3 = p2kT + p32kT
pi =e−εi /kT
q
p2 =e−kT /kT
q=e−1
q
p3 =e−2kT /kT
q=e−2
q q = e−εi /kTi∑
= 1+ e−1 + e−2
N ε =9qkTe−1 + 2kTe−2( )