A biochemical reaction can be written as: Reactant A + Reactant B <-> Product AB

Question

A biochemical reaction can be written as: Reactant A + Reactant B <-> Product AB

To understand reactions, we often consider them at equilibrium (although within a cell, many reactions never reach equilibrium). We define the equilibrium constant, Keq, as the ratio of product concentrations to reactant concentrations at equilibrium:

Keq, = [AB]/[A][B]

If at equilibrium the reactants are at very low concentrations (less than one) and the product is at a concentration higher than one, will Keq be large or small?

The free energy change of a reaction, DG, indicates whether the reaction will proceed spontaneously. If we define a set of standard conditions for a reaction, we can determine a standard free energy change, DGo, that is specific for particular molecules in the reaction. This is defined by:

DG0 = -1.42 log Keq

     2. Determine the standard free energy changes for the following 4 cases:

[AB] =   1 ;     [A] = 10 ; [B] = 100 ; DG0 =

[AB] = 20;     [A] = 20 ; [B] = 10   ; DG0 =

[AB] = 5000; [A] = 1   ; [B] = 5    ; DG0 =

[AB] = 1000; [A] = 10 ; [B] = 100 ; DG0 = 0

     3. Consider the answers in 2. Which situations result in a negative DG0 ? Why?

Explanation / Answer

1) Keq, = [AB]/[A][B] if [A] and [B] are <1 and [AB]>1

then [A][B]<<1

therefore,Keq >>1 much larger

2)DG0 = -1.42 log Keq

so [AB] =   1 ;     [A] = 10 ; [B] = 100 ; DG0 =-1.42 log 1/10*100 =-1.42 log 10^-3= -1.42 * -3=4.26

[AB] = 20;     [A] = 20 ; [B] = 10   ; DG0 =-1.42 * log 20/20*10=-1.42 * log 0.1=-1.42*-1=1.42

[AB] = 5000; [A] = 1   ; [B] = 5    ; DG0 =-1.42 * log 5000/1*5=-1.42 * log 1000=-1.42* 3=-4.26

[AB] = 1000; [A] = 10 ; [B] = 100 ; DG0 = -1.42 * log 1000/10*100=-1.42 * log 1=-1.42* 0=0

3)[AB] = 5000; [A] = 1   ; [B] = 5    ; DG0 =-4.26 has negative DGo

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