A + B ⇌ C + D

The free energy change for the reaction at any given point can be calculated by the equation:

ΔG = ΔG° + RTln([C][D]/[A][B])

R = Gas constant (8.314 J/K.mol)

T = Temperature (K)

ΔG° = Free energy change under standard conditions.

Standard conditions

1. Standard states:

The standard state does not strictly include a temperature. However, standard states are typically measures at 298K and we assume this.

Standard states are denoted as ° e.g. ΔG°

Biochemists have adopted 2 further standards:

• pH of 7.0

• H₂O concentration defined as 1

When these additional standard are also applied, the free energy change is denoted as ΔG°’ (i.e. with the ‘prime’ symbol).

2. Standard change:

The reaction must be defined in terms of a balanced chemical equation

e.g. H₂(g) + ½O₂(g) → H₂O(l)

The standard change is for the complete reaction.

Many reactions will not go to completion and it is not always possible to set up a reaction where the reactants and products are in their standard states (e.g. using 1M solutions).

∆G=∆G°+RTln ([C][D])/([A][B])

The first term (ΔG°) can be thought of as a reference value for the reaction with reactants and products at 1M concentrations, temperature at 298K, and all of A and B converting into C and D.

The second term ****∆G=∆G°+RTln ([C][D])/([A][B]) represents the extra free energy (positive or negative) that is generated when the reaction is carried out at different concentrations and/or temperature.

The ΔG value calculated is the Gibbs free energy difference for the reaction in its current state compared to the equilibrium position. ΔG will change as the reaction proceeds towards equilibrium as the concentrations of the reactants and products change.

∆G=∆G°+RTln ([C][D])/([A][B]) at equilibrium

At equilibrium, ΔG = 0 and [C]eq[D]eq/[A]eq[B]eq = Keq (the equilibrium constant).

Therefore if we substitute in these terms:

0 = ∆G° + RTlnKeq

∆G° = -RTlnKeq

This is an important equation as it links the equilibrium constant directly with the Gibbs free energy under standard conditions.

Further it shows that the equilibrium position is determined by ∆G° and temperature.

Le Châtelier’s Principle with the equation for ΔG

Consider the reaction:

Glucose-6-phosphate ⇌ Fructose-6-phosphate

ΔG° = +1632 J/mol Keq = 0.517

It is at equilibrium when G6P = 6.59 mM and F6P = 3.41 mM

\small {\Delta \text G = \Delta \text G \degree + \text{RT}\ln \frac{[\text{F6P}]}{[\text{G6P}]} = 1632 + (8.314 \times 298 \times \ln \frac{3.41}{6.59}) = 0\text J / \text{mol}}

Reaction at equilibrium.

1. If 1mM more of G6P is added:

\small {\Delta \text G = \Delta \text G \degree + \text{RT}\ln \frac{[\text{F6P}]}{[\text{G6P}]} = 1632 + (8.314 \times 298 \times \ln \frac{3.41}{7.59}) = -350\text J / \text{mol}}

Reaction moves right to make more F6P.

2. If 1mM more of F6P is added:

\small {\Delta \text G = \Delta \text G \degree + \text{RT}\ln \frac{[\text{F6P}]}{[\text{G6P}]} = 1632 + (8.314 \times 298 \times \ln \frac{4.41}{6.59}) = +637\text J / \text{mol}}

Reaction moves left to make more G6P.

More on thermodynamics and catalysis

• Keq and ΔG are unaffected by the presence of an enzyme or catalyst.

• Enzymes increase the rate of reaction by stabilising the transition state (TS) and therefore reducing the energy barrier between the reactants/products and the TS.

• Reaction rate α activation energy (Ea).

• Reaction rates cannot be predicted from ΔG, ΔH and ΔS which only describe energy differences between the reactants and products.

Question 1:

In glycolysis, dihydroxyacetone phosphate (DHAP) is isomerised into glyceraldehyde-3-phosphate (G3P):

ΔG° = +7.5kJ/mol

1. Calculate the equilibrium constant at 37°C.

   Keq = [G3P]/[DHAP]

   ∆G° = -RTlnKeq

   7500 = -(8.314 × 310)lnKeq

   7500/-(8.314 × 310) = lnKeq = -2.910…

   e⁻²∙⁹¹⁰∙∙∙ = 0.054 = Keq

2. In the cell, depletion of G3P makes the reaction proceed to the right. Calculate ΔG if the concentration of G3P is always kept at 1% of the concentration of DHAP?

   7500 + (8.314 × 310) ln(0.01) = -4369.1 J/mol

3. How would the concentration of G3P be kept low?

   Concentration gradient to cause G3P to be carried out of the cell.