Thermodynamics for Biochemists: a YouTube textbook

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

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

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

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

8.1.
Termodynamikens
tre
love

8.2.
Entropi
og
sandsynlighed
–
del
1

8.3.
Entropi
og
sandsynlighed
–
del
2

8.4.

8.5.
giver
ligninger
for
fri
energi-‐

bidrag
for
en
ideal
gas

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$Boltzmannfordelingen$

!"#$%&'"(&)*+),-"'./&%0)
;)
9.>"*)474)
-P#QRR0*C&C7H"R!STU42IO,1V)

!"#$%&'"(&)*+),-"'./&%0)
!:;0'0<+0;,(
-P#QRR'N7G*(G*%>7*%:R'*>"="%R)
E)

!"#$%&'"(&)*+),-"'./&%0)
!:;0'0<+0;,(1-$"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)

!"#$%&'"(&)*+),-"'./&%0)
L)
9.>"*)47;)
-P#QRR0*C&C7H"RbcI:I.+Nde8)

!"#$%&'"(&)*+),-"'./&%0)
=7&$.&+.(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)

!"#$%&'"(&)*+),-"'./&%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))

!"#$%&'"(&)*+),-"'./&%0)
4;)
9.>"*)47?)
-P#QRR0*C&C7H"R5:SjE$T[.LF)

!"#$%&'"(&)*+),-"'./&%0)
J)e5R'*=)G-$(:"/)!)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?)

!"#$%&'"(&)*+),-"'./&%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)*,)

!"#$%&'"(&)*+),-"'./&%0)
4J)
9.>"*)47E)
-P#QRR0*C&C7H"R$!"[k`BF5-V)

!"#$%&'"(&)*+),-"'./&%0)
?-@(.-(/-;(,#&";+#(!A(
HA ! A!
+ H+
K =
[A!
][H+
]
[HA]
fA! =
[A!
]
[HA]+ [A!
]
=
1
[H+
]
K
+1
=
K
[H+
]+ K
" fA! =
1
2
" K = [H+
]
A!
!log K( ) = 6.3
4K)
U'$:")$>$#&">)+%*'Q))
-P#QRRD(0C=%7G*'R0lI##"))

!"#$%&'"(&)*+),-"'./&%0)
HA ! A!
+ H+
K =
[A!
][H+
]
[HA]
A!
!log K( ) = 6.3
U+)m=*:W!XhJ7?)N-$&)./)f"*)$&);I)*,Z))
`7  I7?)e5R'*=)
27  ?J7T)e5R'*=)
,7  4IT7;)e5R'*=)
!7  J??7;)e5R'*=) 4L)
?-@(.-(/-;(,#&";+#(!A(

!"#$%&'"(&)*+),-"'./&%0)
;T)
9.>"*)47I)
-P#QRR0*C&C7H"R6,1*UJ,bnMN)

!"#$%&'"(&)*+),-"'./&%0)
B#(6*&7#'0#+C"(D+0$10E'#(
%&#'()*+,-#./0&1,#2*#0#+3+2,(#'&#,45'6'78'5(#8,+562+#
'&#0#+/'9#'&#,45'6'78'5(#2/02#.*5&2,80.2+#2/,#'()*+,-#./0&1,#
K =
[B]
[A]
=
1.0 ! x
0.25 + x
= 2
x = 0.17 " [A] = 0.42 og [B] = 0.83 " A ! B
A ! B K =
[B]
[A]
= 2
<a$'#="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)

!"#$%&'"(&)*+),-"'./&%0)
A ! B+ C
`Y"%)"BC.=.H%.C').()$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
;;)

!"#$%&'"(&)*+),-"'./&%0)
A ! B+ C
<9$#*%$D*().(G%"$/"/)&-")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+#
'&#0#+/'9#'&#,45'6'78'5(#2/02#.*5&2,80.2+#2/,#'()*+,-#./0&1,#

,
Jan$H.$Jensen$
1.$Ligevægt$og$ligevægtskonstanten$
2.,Enthalpi,og,entropi,

Video&2.1&
h+p://youtu.be/Qv2rKiNi2rQ&
(ingen&slides)&

Video&2.2&
h+ps://youtu.be/NGkGEoQ503A&

!Go
= !H o
" T !So
H2CO3(aq)
! "!# !! H2
O(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 o
C

!H o
= !H Molekyle
+ !H o,Translation
+ !H Rotation
+ !HVibration
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&&

!H o
= !H Molekyle
+ !H o,Translation
+ !H Rotation
+ !HVibration
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&&

!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&

Video&2.3&
h+ps://youtu.be/S03m6ADws9o&

Standard*dannelsesenthalpi*
(standard&enthalpi&of&formaFon,&heat&of&formaFon)&
elementer i standardtilstand! "!# !! molekyle !H f
o
2C(s,grafit) + 3H2(g)
! "!# !! C2H6(g) !H f
o
(C2H6 )
!H f
o
(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 f
o
(C2H4 ) " !H f
o
(H2 )
0
$ %& '&

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&

Video&2.4&
h+ps://youtu.be/7gDqFbMZhd0&

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&

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&

Video&2.5&
h+ps://youtu.be/dhUsMOH9dc&

Entropien*(S)*handler*om*muligheder*
S = k ln(W )
k =
R
NA
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&

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&

Hvad&er&ΔSo&sandsynligvis&for&denne&process?&
A. !So
> 0
B. !So
= 0
C. !So
< 0

,
Jan$H.$Jensen$
3.,Enthalpi,og,entropi,for,an,ideal,gas,og,van’t,Hoﬀ,ligningen,

Video&3.1&
h+ps://youtu.be/Qv2rKiNi2rQ&
(ingen&slides)&

Video&3.2&
h+ps://youtu.be/47MEHOwFxAg&

ΔGo
= ΔH o
− T ΔSo
ΔH o
= ΔH Molekyle
+ ΔH o,Translation
+ ΔH Rotation
+ ΔHVibration
H Molekyle
= ingen simpel ligning
H o,Translation
= 3
2 nRT + po
V = 5
2 nRT
H Rotation
= 3
2 nRT (nRT liniært molekyle)
HVibration
= nNAhc νi
1
2 +
1
eNAhc νi /RT
−1
⎛
⎝⎜
⎞
⎠⎟
i=1
3Nat − X
∑
Enthalpibidrag-for-en-ideal-gas-
ν = bølgetal ( ≈ frekvens) i cm-1
Nat = antal atomer i molekylet
h = Plancks konstant X = 6, 5 (liniær)
c = lysets hastighed i cm/s NAhc = 11.96 J cm

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&&

Video&3.3&
h+ps://youtu.be/wWx3b3MZidc&

ΔSo
= ΔSKonformation
+ ΔSo,Translation
+ ΔSRotation
+ ΔSVibration
ΔGo
= ΔH o
− T ΔSo
Entropibidrag-for-en-ideal-gas-
SKonf
= nRln(gKonf
) gKonf
= antal konformationer med samme energi
(udartning)
So,Trans
= nRln
2πm( )3/2
kTe( )5/2
h3
po
⎛
⎝
⎜
⎞
⎠
⎟ = nRln aM 3/2
T 5/2
( ) a = 0.3117
mol3/2
g3/2
K5/2
SRot
= nRln
8π2
keT
h2
⎛
⎝⎜
⎞
⎠⎟
3/2
πI1I2 I3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
(ikke liniær) I = intertimoment
SVib
= nR
NAhcνi
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

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&&

Video&3.4&
h+ps://youtu.be/tcvaf0iAxbo&

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-

SR− A
Konf
= Rln(4)
SKonf
= nRln(gKonf
) gKonf
= antal konformationer med samme energi
(udartning)
Konforma;onel-entropi-og-udartning-
A& B&R& R&
SR− B
Konf
= Rln(1) = 0
1
2
3
4
3
4
1
2
2
3
4
1
4
1
2
3
SR− A
Konf
= −R fi ln fi( )
i=1
4
∑
≤ Rln(4)
f4 = brøkdel i denne
konformation

S = k ln(W ) = k ln gNA
( )= Rln(g) hvis all tilstande har samme energi
S ≈ Rln(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⋅⋅⋅HOH
Konf
= ?
A.  1/8&&&&&&&&&&C.&4&
B.  ½&&&&&&&&&&&&&D.&8&
&

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⋅⋅⋅HOH
Konf
=
1
8
g2H2O
Trans
gH2O⋅⋅⋅HOH
Trans
≈ 107
g2H2O
Rot
gH2O⋅⋅⋅HOH
Rot
≈ 5
g2H2O
Vib
gH2O⋅⋅⋅HOH
Vib
≈
1
1000

gAA = 6 gA+ A = 15
2 × 3 vibrationer
12 vibrationer
10006
≈ 3 (næsten) ens tilstande per ekstra vibration

Video&3.5&
h+ps://youtu.be/ZOWBLWp32N4&

ΔGo
= ΔH o
− T ΔSo
−RT ln K( ) = ΔH o
− T ΔSo
ln K( ) =
−ΔH o
R
1
T
⎛
⎝⎜
⎞
⎠⎟ +
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( )
1
TΔSo
R
hældning =
−ΔH o
R
Lav&T&
Høj&T&

Her&vises&et&van’t&Hoﬀ&plot&for&en&reakOon&
&
Er&reakOonen&endoterm&eller&exoterm?&
&&&
ln K( )
1
T

A  B
Hvis&reakOonen&er&endoterm&hvad&sker&der&med&
ligevægtskonstanten&når&temperaturen&sOger?&
A.  K&falder&
B.  K&sOger&
C.  K&er&uændret&

,
Jan$H.$Jensen$
4.,Måling,af,enthalpi,og,entropi,ændringer,vha,kalorimetri,

Video&4.1&
h+ps://youtu.be/EAgbknIDKNo&
(ingen&slides)&

Video&4.2&
h+ps://youtu.be/OKamlXPmkvw&

ΔH o
= Q
= mCp
vand
ΔT
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

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
= mCp
vand
ΔT

Hvordan(måler(man(ΔHo(og(ΔSo?:(diﬀeren5al(scanning(calorimetry(
&
Måling&af&varmekapacitetsændring&giver&ΔHo&
Rent&vand&
ΔH o
Tm( )= arealet under kurven
“Smeltepunkt”&
ΔCp

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 ln
T
Tm
⎛
⎝⎜
⎞
⎠⎟

Hvad&er&enhederne&for&varmekapaciteten?&
A.  J/mol&
B.  J/molK&
C.  K/mol&J&
D.  mol/J&

Video&4.3&
h+ps://youtu.be/MB7akfMkReQ&

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&Go
reaktant(=(Go
produkt&
∂T
∂U
⎛
⎝⎜
⎞
⎠⎟
V
≈ 0

_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&

_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)&

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

Hvilken&]lstand&
har&den&højeste&
varmekapacitet?&
A(
B(
C(

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&

Video&4.4&
h+ps://youtu.be/C2GheQMeVnY&

Protein((polymer)(foldning(eksempel(
&
Termisk&denaturering&af&proteinet&Barnase&
Tm&

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
= Rln(4) − Rln 1( )
ΔGo
≈ ΔH Molekyle
− T Rln 4( )( )
Foldet( Udfoldet&
En(meget(simpel(protein(model(

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

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
fU
Cp

Varmekapaciteten(er(et(maximum(når(halvdelen(af(proteinerne(er(udfoldet(
&
Tm,(ΔHo(og(ΔSo,(kan(måles(spektroskopisk(via(van’t(Hoﬀ(methoden((
C: θ230nm = θ230nm
U
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 )

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&

Video&4.4&
Ved&hvilken&pH&er&ΔSo&størst?&
DOI:&10.1021/bi00129a007&

Video&4.5&
h+ps://youtu.be/jeZAbwNE3zM&

Varme(+(
F  U fU =
e−ΔGo
/RT
1+ e−ΔGo
/RT
ΔSo
≈ Rln(4) ⇒ fU =
4e−ΔH o
/RT
1+ 4e−ΔH o
/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&

fF
fU
fU,mikro
Udfoldet&makro]lstand&
mikro]lstand&
Udfoldning(ved(høje(temperaturer(sker(pga(entropi(

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)&

,
Jan$H.$Jensen$
5.,Enthalpi,og,entropi,for,en,ideal,opløsning,

Department)of)Chemistry)
Video)5.1)
h6ps://youtu.be/6i4tDj6f6uc)
2)

ΔGo
= ΔGMolecule/Conformation
+ ΔGo,Translation
+ ΔGRotation
+ ΔGVibration
+ ΔΔGSolvation
Free$energy$contribu.ons$for$an$ideal$solu.on$
ΔH o,Trans
= 3
2 nRT + po
ΔV ≈ 3
2 nRT + ΔngasRT
So,Trans
= nRln
2πm( )3/2
e5/2
h3
Co
⎛
⎝
⎜
⎞
⎠
⎟ = nRln bM 3/2
T 3/2
( ) b = 3.7487
mol3/2
g3/2
K3/2
Co)=)1.0)mol/L)
Ideal)soluEon)=)no)interacEons)between)
)))))))))))))))))))))))))))))solute)molecules)
H o,Trans
≈ 3
2 nRT
Solute:)the)molecule)that)is)dissolved)
)
Solvent) 3)

4)
ΔGo
= ΔGMolecule/Conformation
+ ΔGsolution
o,Translation
+ ΔGRotation
+ ΔGVibration
+ ΔΔGSolvation
ΔGgas
o
+ 5.5 A ⋅B  A + B
ΔGgas
o
− 5.5 A + B  A ⋅B
ΔGgas
o
A  B
Ssolution
o,Trans
= Sgas
o,Trans
− Rln
0.3117
3.7487
T
⎛
⎝⎜
⎞
⎠⎟ = Sgas
o,Trans
− 26.69 J/molK
Hsolution
o,Trans
= Hgas
o,Trans
− RT = Hgas
o,Trans
− 2.5 kJ/molK at$25$oC$
Gsolution
o,Trans
= Hgas
o,Trans
− 2.5( )− T Sgas
o,Trans
− 0.02669( )
= Ggas
o,Trans
− 2.5 − 298.15(−0.02669)
= Ggas
o,Trans
+ 5.5 kJ/mol

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
+ ΔHsolution
o,Translation
+ ΔH Rotation
+ ΔHVibration
+ ΔΔHSolvation
ΔHgas
o
+ ΔHcorrection A + B  A⋅B

Video&5.2&
h+ps://youtu.be/CyR7JiLw1O0&

ΔGo
= ΔGMolekyle/Konformation
+ ΔGo,Translation
+ ΔGRotation
+ ΔGVibration
+ ΔΔGSolvering
Solveringsfrienergi:$det$polære$bidrag$
+&
ΔGSolvering
(A)
ΔGSolvering
= ΔGpolær
Solvering
+ ΔGikke polær
Solvering
A$ A$

ΔGpolær
Solvering
≈ −
694.7q2
R
1−
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 Å
mol
og 60.25
kJ Å3
mol Debye2

ΔGpolær
Solvering
≈ −
694.7q2
R
1−
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 = ΔGpolær
Solvering

Video&5.3&
h+ps://youtu.be/SoHWl96_L68&
&

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 Å
mol
ε# εvand = 80
εvakuum = 1
r#
Solvent$“screening”$
elektrosta`ske&vækselvirkninger&svækkes&af&solventen&

I&hvilken&solvent&er&elektrosta`ske&vækselvirkninger&størst?&
A.&vand&
&
B.&acetone&
&
C.&cyclohexan&
&
D.&kloroform&

Video&5.4&
h+ps://youtu.be/C0R59ekRy2o&

ΔGSolvering
= ΔGpolær
Solvering
+ ΔGikke polær
Solvering
ΔGikke polær
Solvering
= ΔGkavitation
Solvering
+ ΔGvan der Waals
Solvering
ΔGkavitation
Solvering
= 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&

ΔGSolvering
= ΔGpolær
Solvering
+ ΔGikke polær
Solvering
ΔGikke polær
Solvering
= ΔGkavitation
Solvering
+ ΔGvan der Waals
Solvering
ΔGvan der Waals
Solvering
≈ −cSASA
c&=&empirisk&konstant&
&

ΔGikke polær
Solvering
= ΔGkavitation
Solvering
+ ΔGvan der Waals
Solvering
≈ 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&

A$ B$
C$ D$
for&hvilken&`lstand&er&den&ikkeipolære&solva`onsenergi&mindst?&&
ΔGikke polær
Solvering
≈ γ ipSASA + b

Video&5.5&
h+ps://youtu.be/xVmrgm5XobM&

Den$hydrofobe$eﬀekt$
Ikkeipolære&molekyler&opløst&i&vand&binder&stærkere&end&i&vakuum&
for&at&mindske&kontakten&med&solventen&(dvs&SASA)&
ΔGo
≈ ΔΔGikke polær
Solvering
≈ γ ipΔSASA + b

SASAX+X SASAX⋅X
SASAX⋅X < SASAX+ X ⇒ ΔGikke polær
Solvering
(X ⋅ X) < ΔGikke polær
Solvering
(X + X)
Arealet&er&lavest&

A$ B$ C$ D$

|ΔSASA|&størst&for&B$
Ud&fra&følgende&`lnærmelse,&hvilket&“molekyle”&bindes&sandsynligvis&stærkest&med&sig&selv?&
ΔGo
≈ ΔΔGikke polær
Solvering
≈ γ ipΔSASA + b

Olie&er&hydrofobisk&
h+p://youtu.be/D6aoJNqt1MQ&

,
Jan$H.$Jensen$
6.,Hydrofobisitet,og,entropi,

Video&6.1&
h+ps://youtu.be/KqSjPspb0EA&

Hvad&er&&&&&&&&&&for&denne&simulaDon?&
A ΔSo
= 0
B ΔSo
< 0
C ΔSo
> 0
ΔSo
h+p://youtu.be/zKNmBjqGijI&

+& 
+&  +&
ΔSo
> 0
2&parDkler& 1&parDkel&
2&parDkler& 5&parDkler&
Solventen(bidrager(/l(entropiændringen(
vandet&i&bindingslommen&er&anderledes&=&hydrofob&

+&  +&
h+p://youtu.be/1WkZznwmO0c&

+&  +&
h+p://youtu.be/ETMmH2trTpM&
ΔSo
> 0
vandet&tæ+est&på&liganden&er&anderledes&=&hydrofob&


ΔSo,Trans
= ? ved 25 o
C
(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'


ΔSo,Trans
= ? ved 25 o
C
Sopløsning
o,Trans
= nRln 3.7487M 3/2
T 3/2
( )
ΔSo,Trans
= 4Rln 3.7487M 3/2
T 3/2
( )− Rln 3.7487 4M( )3/2
T 3/2
( )
= 3Rln 3.7487M 3/2
T 3/2
( )
118.1
  
− 3
2 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'

Video&6.2&
h+ps://youtu.be/g7Fe4FGN9tE&

ΔSo
> 0
ΔH o
≈ +0
ΔGo
< 0
 +&
Entropien(s/ger(når(hydrofobe(molekyler(bindes(

ΔGikke polær
Solvering
= ΔGkavitation
Solvering
+ ΔGvan der Waals
Solvering
≈ γ osSASA − cSASA
≈ γ ipSASA + b
ΔSikke polær
Solvering
=
∂ΔGikke polær
Solvering
∂T
≈
∂γ os
∂T
SASA −
∂c
∂T
SASA
≈
∂γ os
∂T
SASA
∂γ os
∂T
= −0.00112 kJ/molÅ2
K
Den&ikkePpolære&solveringsentropi&er&negaDv&

 +&
ΔSo
> 0
ΔH o
≈ +0
ΔGo
< 0
ΔSo
≈ ΔΔSikke polær
Solvering
≈
∂γ os
∂T
<0

ΔSASA
<0

> 0
En&negaDv&ikkePpolær&solveringsentropi&betyder&
at&entropien&sDger&når&hydrofobe&molekyler&bindes&
&

benzen(benzen)  benzen(aq)
Hvad&er&&&&&&&&&&for&denne&proces?&
A ΔSo
= 0
B ΔSo
< 0
C ΔSo
> 0
ΔSo

Video&6.3&
h+ps://youtu.be/gAaV7bEWqOw&

Hvordan(måler(man(hydrofobisitet?(

solute(vand)  solute(octanol)
log(Pwo ) = log
[solute(octanol) ]
[solute(vand) ]
⎛
⎝
⎜
⎞
⎠
⎟
log(P)&>&1&=&hydrofobisk&
ε&=&80&ε&=&10&

Cl((O((C(((H( Hvilket&molekyle&har&det&laveste&log(Pwo)?&&

Video&6.4&
h+ps://youtu.be/7CVzdXnM1eM&

P + L  P ⋅L
For&hvilken&ligand&vil&ΔSo&være&mest&posiDv?&
Ligand&binding&Dl&enzymet&carbonic&anhydrase&(ved&25&oC)&

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&

+&  +&
+&  +&

octanol&
log(Pwo)(måler(den(totale(hydrofobisitet(af(liganden((

,
Jan$H.$Jensen$
7.,Kemisk,akLvitet,og,ikkeMideale,opløsninger,

Video&7.1&
h+ps://youtu.be/D_eGyby1whM&

A  B+ C K =
[B][C]
[A]
ΔGo
= −RT ln(K)
ΔGo
= −
J
mol K
⎛
⎝⎜
⎞
⎠⎟ K( )ln(K) = J/mol
ΔGo
= −
J
mol K
⎛
⎝⎜
⎞
⎠⎟ K( )ln
M ⋅M
M
⎛
⎝⎜
⎞
⎠⎟ = J/molln(M)
??

Ligevægtskonstanten-har-ingen-enheder-

ΔGo
= ΔGMolekyle/Konformation
+ ΔGo,Translation
+ ΔGRotation
+ ΔGVibration
+ ΔΔGSolvering
Fri-energibidrag-for-en-ideal-opløsning-
ΔH o,Trans
= 3
2 nRT + po
ΔV ≈ 3
2 nRT + ΔngasRT
So,Trans
= nRln
2πm( )3/2
e5/2
h3
Co
⎛
⎝
⎜
⎞
⎠
⎟ Co&=&1.0&mol/L&
Ideal&opløsning&=&ingen&vækselvirkninger&
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&mellem&solute&molekyler&
H o,Trans
≈ 3
2 nRT
Solute:&molekylet&der&er&opløst&
&
Solvent&

STrans
= Rln
2πm( )3/2
e5/2
h3
C
⎛
⎝
⎜
⎞
⎠
⎟
= Rln
2πm( )3/2
e5/2
h3
Co C
Co
⎛
⎝
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
= Rln
2πm( )3/2
e5/2
h3
Co
⎛
⎝
⎜
⎞
⎠
⎟ − Rln
C
Co
⎛
⎝⎜
⎞
⎠⎟
= So,Trans
− Rln
C
Co
⎛
⎝⎜
⎞
⎠⎟
S = So
+ Rln
C
Co
⎛
⎝⎜
⎞
⎠⎟ ⇒ G = Go
+ RT ln
C
Co
⎛
⎝⎜
⎞
⎠⎟
STrans,&og&derfor&G,&er&en&funkKon&af&koncentraKon,&C&

R  P
G(X) = Go
(X) + RT ln
[X]
Co
⎛
⎝⎜
⎞
⎠⎟
G(P) − G(R) = Go
(P) − Go
(R) + RT ln
[P]
Co
[R]
Co
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
ΔG = ΔGo
+ RT ln
[P]
Co
[R]
Co
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟

ΔG = 0 ⇒
[P]
Co
[R]
Co
= e−ΔGo
/RT
vi skriver K =
[P]
[R]
men mener K =
[P]
Co
[R]
Co
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

G(X) = Go
(X) + RT ln
[X]
Co
⎛
⎝⎜
⎞
⎠⎟
G(X) = Go
(X) + RT ln
pX
po
⎛
⎝⎜
⎞
⎠⎟
G(X) = Go
(X) + RT ln
[X]
CX
o
⎛
⎝⎜
⎞
⎠⎟
Co&=&1&M&ideal&opløsning&&
po&=&1&bar&ideal&gas&&
CX
o&=&koncentraKon&af&ren&væske&
eller&faststof&
opløsning&
gas&
faststof/væske&
H2O(l)  H2O(g)
K =
pH2O
po
⎛
⎝⎜
⎞
⎠⎟
[H2O]
CH2O
o
⎛
⎝⎜
⎞
⎠⎟
1
  
=
pH2O
po
⎛
⎝⎜
⎞
⎠⎟  pH2O[H2O] = CH2O
o
= 55.56 M
forskellige-standard7lstande-
uaMængig&af&fordampning&

H2CO3(aq)  H2
O(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?&

Video&7.2&
h+ps://youtu.be/Va5LJ8UX7sY&

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&

Vækselvirkninger&mellem&solute&molekyler&øger&koncentraKonen&af&fri&ioner&
sammenlignet&med&ideal&opløsning&

+- +-
+-
9-
9-

h+p://en.wikipedia.org/wiki/Debye%E2%80%93H%C3%BCckel_equaKon&

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+
] ⇒ γ Pb2+ =
1.14 ×10−3
M
1.37 ×10−3
M
= 0.83

Ak7vitet-(a)-for-en-ikke9ideal-opløsning-
R(aq)  P(aq)
K =
aP
aR
= e−ΔGo
/RT
aP = γ P
[P]
Co
ingen&
enheder&
aP = 1⇒ γ P = 1,
[P]
Co
= 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&

A γ Na+ ≈ γ Ca2+
B γ Na+ < γ Ca2+
C γ Na+ > γ Ca2+
D ved ikke
Hvad&vil&du&forvente&(samme&koncentraKon)?&

Video&7.3&
h+ps://youtu.be/LFqWtMudxJc&

γ ± = 10− q+ q− A I
I = 1
2 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− = γ K+ [K+
]γ Cl− [Cl−
] = γ K+ γ Cl−( )[K+
][Cl−
] = γ ±
2
[K+
][Cl−
]

γ ± = 10− q+ q− A I
I = 1
2 q+
2
[+]+ q−
2
[−]( )
middel&akKvitetskoeﬃcienten&for&0.001&M&CaCl2&
I = 1
2 q+
2
[+]+ q−
2
[−]( )
= 1
2 +2( )2
0.001( )+ −1( )2
0.002( )⎡
⎣
⎤
⎦
= 0.003
γ ± = γ Ca2+
2
γ Cl−( )
1/3
= 10− q+ q− A I
= 10− +2⋅−1 0.509 0.003
= 0.88
målt=&0.89&

γ ± = 10− q+ q− A I
I = 1
2 q+
2
[+]+ q−
2
[−]( )
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&&&

Video&7.4&
h+ps://youtu.be/MZGwsQ61T7M&

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
Den-simple-Debye9Hückel-ligning:-et-eksempel-

PbI2(s)  Pb(aq)
2+
+ 2I(aq)
−
5.93×10−9
= γ Pb2+ [Pb2+
]γ I−
2
[I−
]2
= γ ±
3
[Pb2+
][I−
]2
⇒ [Pb2+
] = 1.32 ×10−3
M
målt: [Pb2+
] = 1.37 ×10−3
M
γ ± = 10−2 0.509( ) I
I = 1
2 22
[Pb2+
]+ [I-
]( )
5.93×10−9
= 10−2 0.509( ) 3x
( )
3
4x3
( )

log γ ±( )= −
A q−q+ I
1+ B I
+ CI
γ ± = 10− q+ q− A I
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

,
Jan$H.$Jensen$
8.,Termodynamikens,tre,love,og,Boltzmannfordelingen,

Video&8.1&
h+ps://youtu.be/XeupZ3YwCaM&

dU = dq + dw
q
Tkoldt
−
q
Tvarmt
> 0
ΔSunivers > 0
dS =
dqrev
T
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&

ΔSunivers ≥ 0
dSsystem + dSbad ≥ 0
dSsystem +
dqbad
T
≥ 0
dSsystem −
dqsystem
T
≥ 0
dSsystem −
dUsystem + pdVsystem( )
T
≥ 0
0 ≥ dHsystem − TdSsystem
0 ≥ 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%

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%

Video&8.2&
h+ps://youtu.be/vVSDlReY2LQ&

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&

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&

Video&8.3&
h+ps://youtu.be/h2h3nNiQYxs&

0&
0.1&
0.2&
0.3&
0.4&
0.5&
0.6&
0& 1& 2&
sandsynlighed'
Antal'par0kler'i'højre'side'
NH
p(NH )
p(0) = 1× 1
2( )2
p(2) = 1× 1
2( )2
p(1) = 2 × 1
2( )2

0&
0.05&
0.1&
0.15&
0.2&
0.25&
0.3&
0.35&
0.4&
0& 1& 2& 3&
sandsynlighed'
antal'par0kler'i'højre'side'
NH
p(NH )
p(0) = 1× 1
2( )3
p(3) = 1× 1
2( )3
p(1) = 3× 1
2( )3
p(2) = 3× 1
2( )3
p(NH ) =
3!
NH !(3− NH )!
1
2( )3
3!= 3⋅2 ⋅1

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&
sandsynlighed'
antal'par0kler'i'højre'side'
p(NH ) =
100!
NH !(100 − NH )!
1
2( )100
= W 1
2( )100
NH
p(NH )

dGsystem ≡ dHsystem − 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 ) 1
2( )100

S ∝W
S = k ln W( )
S'='k'ln(W)'
dS =
dqrev
T
S&har&J/K&enheder&
dS = dSA + dSB
W = WAWB
ln WA( )+ ln WB( )= ln WAWB( )
k =
R
NA
S&er&addiLv&

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&

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 )!

Video&8.4&
h+ps://youtu.be/r_jHsUjXWHQ&

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'

dS = ′S − S
= k ln
N1!
N1 −1( )!
⎛
⎝
⎜
⎞
⎠
⎟ + ln
N2 !
N2 +1( )!
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
≈ k ln N1( )+ ln
1
N2 +1
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢
⎤
⎦
⎥
= k ln
N1
N2 +1
⎛
⎝⎜
⎞
⎠⎟
≈ k ln
N1
N2
⎛
⎝⎜
⎞
⎠⎟
ε2&&
ε1&=&0&
ε2&&
ε1&=&0&
dU = ε2 − ε1 ⇒ N1 → N1 −1 and N2 → N2 +1
ln x!( ) ≈ xln(x) − x
stor&x%

ε2&&
ε1&=&0&
ε2&&
ε1&=&0&
dU = ε2 − ε1 = ε2 ⇒ N1 → N1 −1 and N2 → N2 +1
dS =
dU
T
k ln
N1
N2
⎛
⎝⎜
⎞
⎠⎟ =
ε2
T
k ln
N2
N1
⎛
⎝⎜
⎞
⎠⎟ = −
ε2
T
N2
N1
= e−ε2 /kT
Ni = e−εi /kT
N1
N1 e−εi /kT
i
∑ = N
N1
N
=
1
e−εi /kT
i
∑
Ni
N
=
e−εi /kT
e−εi /kT
i
∑
ved&ligevægt:&

Ni
N
=
e−εi /kT
e−εi /kT
i
∑
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&

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&

Video&8.5&
h+ps://youtu.be/gne9xlUe5lQ&

ΔSunivers ≥ 0 og S = k ln W( )⇒ pi =
e−εi /kT
q
U = U(0) + N ε
ε = piεi
i
∑ =
1
q
εie−βεi
i
∑ = −
1
q
d
dβ
e−βεi
i
∑
⎡
⎣
⎢
⎤
⎦
⎥ = −
1
q
dq
dβ
Boltzmann'fordelingen'giver'ligningerne'for''
fri'energibidrag'for'en'ideal'gas'
β = 1
kT
S = k ln
N!
N1!N2 !…Nt !
⎛
⎝⎜
⎞
⎠⎟
= k ln Ni ln N − Ni ln Ni( )
i
∑ = −k Ni ln
Ni
Ni
∑
= −k Ni
i
∑ ln
e−βεi
q
⎛
⎝⎜
⎞
⎠⎟ = kβ Niεi
i
∑ + Nk lnq
=
U
T
+ Nk lnq
ε1&=&0&
ε2&&
ε3&&

Eksempel:'den'transla0onelle'indre'energi'
εnx nynz
T
= εnx
T
+ εny
T
+ εnz
T
εnx
T
= (nx
2
−1)
h2
8mX2
= (nx
2
−1)ε nx = 1,2,3,....
Kvantemekanik&
(parLkel&i&en&kasse&
med&længde&X%
og&masse%m)&
qX
T
= e−βεi
i=1
∞
∑ = e−(n2
−1)βε
dn
1
∞
∫ ≈ e−n2
βε
dn
0
∞
∫ ≈
2πm
h2
β
X
qT
=
2πm
h2
β
⎛
⎝⎜
⎞
⎠⎟
3/2
XYZ =
2πm
h2
β
⎛
⎝⎜
⎞
⎠⎟
3/2
V
εT
= −
1
q
dq
dβ
⎛
⎝⎜
⎞
⎠⎟
V
= 3
2 kT
UT
= UT
(0)
0
 + N εT
= 3
2 nRT
εTrans
kT
≈ 10−20

εTrans
kT
≈ 10−20
⇒UTrans
=
3
2
nRT ⇒ H Trans
= UTrans
+ pV =
5
2
nRT
εRot
kT
≈ 0.01− 0.001⇒URot
= H Rot
=
3
2
nRT
εVib
kT
≈ 2 −10 ⇒UVib
= HVib
= nNAhc νi
1
2 +
1
eNAhc νi /RT
−1
⎛
⎝⎜
⎞
⎠⎟
i=1
3Nat − X
∑
UVib
= UVib
(0)
≠0
 + N εVib
Vibra0on'er'lidt'anderledes'
ε1&=&0&
ε2&&
ε3&&

pi =
e−εi /kT
q
og S =
U
T
+ Nk lnq
STrans
= nRln
2πm( )3/2
kTe( )5/2
h3
p
⎛
⎝
⎜
⎞
⎠
⎟
SRot
= nRln
8π2
keT
h2
⎛
⎝⎜
⎞
⎠⎟
3/2
πI1I2 I3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
SVib
= nR
NAhcνi
RT eNAhc νi /RT
−1( )
− ln 1− e−NAhc νi /RT
( )
⎛
⎝
⎜
⎞
⎠
⎟
i=1
3Nat −6
∑
εi
Trans
,εi
Vib
, og εi
Vib
fra kvantemekanik

S = k ln W( ) = k ln
N!
N1!N2 !…Ng !
⎛
⎝
⎜
⎞
⎠
⎟
Ni
N
=
e−εi /kT
e−εi /kT
i
∑
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

ε1&=&0&
ε2&=&kT&&
ε3&=&2kT&Hvad&er&U&for&denne&fordeling?&
A. U =
9
q
kTe−1
+ 2kTe−2
( )
B. U =
9
q
kTe + 2kTe2
( )
C. U =
9
q
1+ kTe−1
+ 2kTe−2
( )
D. U = kTe−1
+ 2kTe−2
( )

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 + p3 2kT
pi =
e−εi /kT
q
p2 =
e−kT /kT
q
=
e−1
q
p3 =
e−2kT /kT
q
=
e−2
q q = e−εi /kT
i
∑
= 1+ e−1
+ e−2
N ε =
9
q
kTe−1
+ 2kTe−2
( )

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