1
γ=0.5 is like saying the concentrations of H+ and A- are less. LeChatelier says this will shift the equilibrium to the right.
Electrostatic attraction makes ∆G more negative; The reaction goes further right than expected from ideal behavior. Actual concentration is 0.2M for each ion.
2
ATP + H2O ADP + H+ + Pi
3139.997)(5.79 −=−+=−×++=
−
−
+
+
×
+∆=+∆=∆
×+=+∆=∆
++=+∆=∆
)1)(1()1()1(1)(
)1)(1()101(1)(
log3.2GlnGG
)1)(1()101(1)(]ATP][X[][P]ADP][H[
ln-31lnGG
1]ATP][X[][P]ADP][H[
ln9lnGG
7
00
'000'
7OH
i
'00'
OH
i
00
2
2
RTQQRT
RTQQRT
RTQQRT
∆G0 = +9 kJ/mol [H+]=1M but, ∆G0’ = -31 kJ/mol [H+]=10-7 M, pH7
3
77
101)1)(1(
)101][]][ −
−+
×=×
==(1)(
ATP][X[PADP][HQOH
i0'
2
The famed “high energy phosphate bond” yields little more free energy than a hydrogen bond!
What is K when ∆G0’ = -31 kJ/mol ?
K = 10 ∆G0’/5.7 = 10-31/5.7= 2.7 x 105 i.e., enough but not too much.
4
LeChatlier: adding heat (raising temperature) shifts equilibrium (changes K) in the direction that uses the heat (lowers temperature). Endothermic (∆H = +) use heat and lower the T therefore shift to the right (K increases) Exothermic (∆H = negative) gives off heat and raise the T; therefore shift to the left (K decreases)
5
Saying the same thing with ∆G0
You will do this for the Lab experiment on the dissolving of butanol in water.
7
But, 5’ A-C is accidently = 5’ C-A and 5’ A-G is accidently = 5’ G-A
Independent of which pairs first!
9
Table 4.4
Initiation Fees ∆Ginit =+8.1 kJ/mol ∆Hinit =+0.8 kJ/mol
Independent of site of initiation and direction of zipping
10
5’...-A-A-...3’ 3’...-T -T-...5’
180 degree rotation 5’...-A-A-...3’
3’...-T -T-...5’
5’...-A-T-...3’ 3’...-T-A-...5’
180 degree rotation 5’...-A-T-...3’
3’...-T-A-...5’
NOT 5’...-T-A-...3’ 3’...-A-T-...5’
11
Independent of site of initiation and direction of zipping Ultimately , 1 init and ALL pairing are made regardless of starting point and direction.