Bioenergetics: Gibbs Energy Changes (ΔG and ΔG°')

The Big Picture of Gibbs Energy: Gibbs energy changes serve as the primary tool for predicting whether chemical reactions are thermodynamically favorable under two distinct sets of conditions:
- Standard Conditions ( $\Delta G^{\circ \prime}$ ): Defined as the biochemical standard state.
- Actual Conditions ( $\Delta G$ ): Specifically refers to conditions as they exist within a live cell, including varying temperatures and metabolite concentrations.
Concentration and Steady State: In a biological system, the concentration of reactants and products at steady state are used to calculate the actual Gibbs energy change for a reaction using the equilibrium relationship:
- $\Delta G = \Delta G^{\circ \prime} + RT \ln(Q)$

Thermodynamic Favorability: Gibbs Energy ( $G$ ) determines if a reaction is favorable or unfavorable in the direction written under a specific pressure and temperature.
The Enthalpy-Entropy Relationship: The change in Gibbs Energy ( $\Delta G$ ) is defined by the relationship between enthalpy ( $\Delta H$ ), entropy ( $\Delta S$ ), and the absolute temperature in Kelvin ( $T$ ):
- $\Delta G = \Delta H - T\Delta S$
Biochemical Standard Conditions ( $\Delta G^{\circ \prime}$ ): While standard chemical conditions often differ, the biochemical standard change in Gibbs energy ( $\Delta G^{\circ \prime}$ ) is adjusted to mimic the environment of a living cell. These specific constant conditions are:
- Pressure: $1\, \text{atm}$
- Temperature: $298\, \text{K}$
- pH: $7.0$
- Water Concentration: $[H_2O] = 55.5\, \text{M}$

Defining the Reaction: For a general reaction mechanism: $A + B \rightleftharpoons C + D$ .
The Equilibrium Constant ( $K_{eq}$ ): This is defined by the concentrations of reactants and products when the reaction has reached equilibrium:
- $K_{eq} = \frac{[C]_{eq} [D]_{eq}}{[A]_{eq} [B]_{eq}}$
Predicting Spontaneity and Reaction Direction:
- If K_{eq} > 1, the reaction proceeds spontaneously from left to right as written to form products C and D. In this case, \Delta G^{\circ \prime} < 0 and the reaction is classified as exergonic.
- If K_{eq} < 1, the reaction favors the formation of reactants A and B (moving right to left). In this case, \Delta G^{\circ \prime} > 0 and the reaction is classified as endergonic.

Mass Action Ratio ( $Q$ ): In actual cellular conditions, initial concentrations often differ from the standard $1\, \text{M}$ . Willard Gibbs defined the actual free energy change using the mass-action ratio ( $Q$ ), which is the ratio of initial concentrations of products over reactants:
- $Q = \frac{[C]_i [D]_i}{[A]_i [B]_i}$
- The subscript "i" denotes initial concentrations, and all values are expressed in units of Molarity ( $\text{M}$ , or moles/liter).
Physical Interpretation of $\Delta G^{\circ \prime}$ : This constant represents the amount of energy (exergonic or endergonic) required to move from a state where all reactants and products are initially at $1\, \text{M}$ to a state where the system has reached equilibrium.
- At equilibrium, the actual change in Gibbs energy $\Delta G = 0$ and the mass action ratio $Q$ becomes equal to $K_{eq}$ .
Historical Context: $\Delta G^{\circ \prime}$ values for most metabolic reactions were determined in the 1960s and consolidated into reference tables in the 1970s. These standardized values continue to be used in modern biochemistry.

Problem Statement: Calculate the $\Delta G$ value at $25^\circ \text{C}$ for the conversion of 2-phosphoglycerate (2PG) to phosphoenolpyruvate (PEP) by the enzyme enolase.
Given Parameters:
- $T = 25^\circ \text{C} = 298\, \text{K}$
- $R = 8.31 \times 10^{-3}\, \text{kJ/mol} \cdot \text{K}$
- $\Delta G^{\circ \prime} = +1.70\, \text{kJ/mol}$
- Steady state concentrations: $[2PG] = 0.06\, \text{M}$ , $[PEP] = 400\, \mu \text{M} = 4 \times 10^{-4}\, \text{M}$ .
Calculation Steps:
- $\Delta G = (+1.70\, \text{kJ/mol}) + (8.31 \times 10^{-3}\, \text{kJ/mol} \cdot \text{K}) \cdot (298\, \text{K}) \cdot \ln \left(\frac{4 \times 10^{-4}\, \text{M}}{6 \times 10^{-2}\, \text{M}}\right)$
- $\Delta G = (+1.70\, \text{kJ/mol}) + (2.47638) \cdot (-5.01)$
- $\Delta G = (+1.70\, \text{kJ/mol}) + (-12.41\, \text{kJ/mol})$
- Result: $\Delta G = -10.71\, \text{kJ/mol}$

Problem Statement: Calculate the $\Delta G$ value at $37^\circ \text{C}$ for the conversion of fructose-1,6-bisphosphate (F-1,6-BP) to glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone-phosphate (DHAP) by the enzyme aldolase.
Given Parameters:
- $T = 37^\circ \text{C} = 310\, \text{K}$
- $\Delta G^{\circ \prime} = +23.80\, \text{kJ/mol}$
- Steady state concentrations: $[F\text{-}1,6\text{-BP}] = 0.01\, \text{M}$ , $[GAP] = 0.02\, \text{mM} = 2 \times 10^{-5}\, \text{M}$ , $[DHAP] = 0.02\, \text{mM} = 2 \times 10^{-5}\, \text{M}$ .
Calculation Steps:
- $Q = \frac{(2 \times 10^{-5}\, \text{M}) \cdot (2 \times 10^{-5}\, \text{M})}{1 \times 10^{-2}\, \text{M}} = 4 \times 10^{-8}\text{?}$
- $\Delta G = +23.8\, \text{kJ/mol} + (8.31 \times 10^{-3}\, \text{kJ/mol} \cdot \text{K}) \cdot (310\, \text{K}) \cdot \ln \left(\frac{(2 \times 10^{-5}) \cdot (2 \times 10^{-5})}{1 \times 10^{-2}}\right)$
- $\Delta G = +23.80\, \text{kJ/mol} + (2.576) \cdot (-17.03)$
- $\Delta G = +23.80\, \text{kJ/mol} + (-43.87\, \text{kJ/mol})$
- Result: $\Delta G = -20.07\, \text{kJ/mol}$

Reaction Spontaneity: $\Delta G$ and $\Delta G^{\circ \prime}$ are the predictors of spontaneity.
Thermodynamic Connectivity: The formula $\Delta G = \Delta H - T\Delta S$ explicitly connects enthalpy, entropy, and absolute temperature.
Sign Conventions:
- \Delta G < 0: Exergonic (favorable).
- \Delta G > 0: Endergonic (unfavorable).
Thermodynamic vs. Kinetic Limits: $\Delta G$ indicates the thermodynamic favorability of a reaction; it does not indicate the rate of the reaction (kinetics).
Standard Relationship to Equilibrium: $\Delta G^{\circ \prime}$ is fundamentally related to the equilibrium constant via the equation: $\Delta G^{\circ \prime} = -RT \ln(K_{eq})$ .
Actual Energy Calculations: Actual Gibbs energy changes must account for real-time concentrations through the mass action ratio ( $Q$ ): $\Delta G = \Delta G^{\circ \prime} + RT \ln(Q)$ .