van’thoff equation tue, 3oct17 - montana state university
TRANSCRIPT
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Temperature dependence of K depends onlyon ∆H0 by the Van’t Hoff equation
Van’t Hoff equation
−
∆−=
12
0
1
2 11lnTTR
HKK
−
∆−=
−
∆−=∴
12
0
12
12
0
1
2
11exp
11exp
TTRHKK
TTRH
KK
361 Lec 19Tue, 3oct17
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∆G0 = -RTlog (K)
K(T) = e-∆G0/RT
K(T) = e-∆G0/RT = e-(∆H0-T∆S0)/RT
Do NOT assume ∆G0 is independent of T !!!
Temperature dependence of K depends onlyon ∆H0 by the Van’t Hoff equation
WARNING!
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361 Lec 19Tue, 1oct13
∆G0’ for forming a base pair
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Table 4.4
Initiation Fees
∆Ginit =+8.1 kJ/mol
∆Hinit =+0.8 kJ/mol
Independent ofsite of initiationand direction ofzipping
Because, ultimately, 1 initiatation and ALL pairing are made, regardless of starting point and direction.
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All numbers in table are for adding a base pair in the
direction 5’ 3’
Doesn’t matter which strand
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But, 5’ A-C is accidently= 5’ C-A
and 5’ A-G is accidently= 5’ G-A
Independent of which pairs first!
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Also there are ACCIDENTALEquivalencies
Here A followed by C
is NOT expected to besame as C followed by A
It just happens to be.
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5’...-A-A-...3’3’...-T -T-...5’
180 degreerotation 5’...-A-A-...3’
3’...-T -T-...5’
5’...-A-T-...3’3’...-T-A-...5’
180 degreerotation 5’...-A-T-...3’
3’...-T-A-...5’
NOT 5’...-T-A-...3’3’...-A-T-...5’
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5’...-A-T-A-G-C-A...3’3’...-T-A-T-C-G-T...5’
∆G0 init +8.1 - 3.7 -2.4 -5.4 -9.3 -6.0 = -18.7∆H0 init +0.8 -30.2 -30.2 -32.7 -41.0 -35.6 = -168.9
Example:
This is for FORMING the double strandHomework S2 is for melting the double strand
1416)310(3145.8
700,18exp
exp0
=
−−=
∆−=
K
RTGK
11
12
HA --> A- + H+
Same math as dissociation of weak acid
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1 nM
nM
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Like Prob. 14 a (see Example 4.15 p. 133 ) 7 unknowns
need 7 equations
7th equation: Sum of the 4 His species = 0.10 mol
1st equation[H+]=10-pH
2nd equation
3rd equation
4th equation5th equation
6th equation
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Temporarily
Temporarily
Strategy:Find 3 ratios
First work withthe largest ratio
First work withthis FIRST
2 equations 2 unknowns
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Now, go back and solve for the small concentrations that were temporarily ignored