BCH 261 Lecture Notes - Lecture 6: Boltzmann Constant, Jmol, Gas Constant

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30 Nov 2020
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Lecture 6

  
= free E change under standard conditions
Through evolution, [ATP] > [ADP], thus >>>
for ATP = 32.2 kJ/mol
= free E change currently happening in rxn under given conditions
@ eq’m =0
 ,
R (gas constant) = 8.315 J/mol K OR 0.008315
kJ/mol K
K (Boltzmann constant) = 1.381 x 10^-23 J/K
1 cal = 4.84 J
= 
 
 ,
G & H = J/mol & S = J/mol K
If ∆G is negative; thus, transfer from region1 to region 2 is
favorable (this is the situation described in the “initial state”).
Ex. sugar & fat metabolism releases E that’s used to make
macromolecules
If ∆G is positive; thus, transfer from region 1 to region 2 would not
be favourable (but transfer in the opposite direction would be
favored)
Many biological process are thermodynamically
unfavourable
Ex. millions of nucleosides make 1 RNA molecule;
making proteins, moving mucles
If ∆G is zero; thus, there is no net driving force in either direction for
the transfer of A. The system is at equilibrium (this is the situation
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described in the “final state”)
Example: Calculate the standard free-energy change of the reaction
catalyzed by the enzyme phosphoglucomutase : Glucose 1-phosphate
<-> glucose 6-phosphate given that, starting with 20 mM glucose 1-
phosphate and no glucose 6-phosphate, the final equilibrium mixture at
25⁰C and pH 7.0 contains 1.0 mM glucose 1-phosphate and 19 mM
glucose 6-phosphate. Does the reaction in the direction of glucose 6-
phosphate formation proceed with a loss or a gain of free energy?
Example: Calculate the fraction of fructose 6-phosphate in the reaction:
fructose 1-phosphate <-> fructose 6-phosphate given when
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ΔH∘=−1kJ/mol and ΔS∘=5×10−4kJ/(mol⋅K) and the reaction occurs at
body temperature (37∘C)?
Adenosine triphosphate (ATP) is the energetic “currency” in cells
3 phosphoryl groups are referred to as the alpha (α), beta (β), and, for the
terminal phosphate, gamma (γ)
Phosphoanhydride bonds within ATP and ADP
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