4.6 Gibbs Free Energy, Equilibrium & Bioenergetic Reaction Coupling

4.6 Gibbs Free Energy, Equilibrium & Bioenergetic Reaction Coupling

5. Gibbs Free Energy ($\Delta G$) and Spontaneity Criteria

Gibbs free energy combines enthalpy and entropy into a single thermodynamic criterion that predicts spontaneity at constant temperature and pressure:

$$\Delta G = \Delta H - T \Delta S$$

• Criterion for Spontaneity:
  • $\Delta G < 0$: The process is exergonic and spontaneous in the forward direction.
  • $\Delta G = 0$: The system has reached dynamic equilibrium.
  • $\Delta G > 0$: The process is endergonic and non-spontaneous in the forward direction (the reverse process is spontaneous).

Temperature Dependence of Spontaneity

Because temperature in Kelvin ($T$) is strictly positive, the interplay between the signs of $\Delta H$ and $\Delta S$ produces four distinct spontaneity regimes:

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6. Standard Free Energy, Non-Standard States, and Equilibrium

Standard Free Energy and the Equilibrium Constant ($K_{\text{eq}}$)

Standard free energy change ($\Delta G^\circ$) describes a reaction where all reactants and products are present at standard conditions ($1\text{ atm}$ partial pressure for gases, $1.0\text{ M}$ concentration for dissolved solutes, $298\text{ K}$):

$$\Delta G^\circ = -RT \ln K_{\text{eq}}$$

Where $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$ and $T$ is temperature in Kelvin.

• If $K_{\text{eq}} > 1 \implies \ln K_{\text{eq}} > 0 \implies \Delta G^\circ < 0$ (products are thermodynamically favored at equilibrium).
• If $K_{\text{eq}} = 1 \implies \ln K_{\text{eq}} = 0 \implies \Delta G^\circ = 0$.
• If $K_{\text{eq}} < 1 \implies \ln K_{\text{eq}} < 0 \implies \Delta G^\circ > 0$ (reactants are thermodynamically favored at equilibrium).

Actual Non-Standard Free Energy ($\Delta G$)

Living cells never operate at $1.0\text{ M}$ standard solute concentrations. The actual, instantaneous free energy change depends on the current reaction quotient ($Q$):

$$\Delta G = \Delta G^\circ + RT \ln Q$$

Substituting $\Delta G^\circ = -RT \ln K_{\text{eq}}$ yields the fundamental relationship:

$$\Delta G = -RT \ln K_{\text{eq}} + RT \ln Q = RT \ln\left(\frac{Q}{K_{\text{eq}}}\right)$$

• If $Q < K_{\text{eq}} \implies Q/K_{\text{eq}} < 1 \implies \ln(Q/K_{\text{eq}}) < 0 \implies \Delta G < 0$: Reaction proceeds spontaneously forward.
• If $Q = K_{\text{eq}} \implies \Delta G = 0$: System is at equilibrium.
• If $Q > K_{\text{eq}} \implies Q/K_{\text{eq}} > 1 \implies \ln(Q/K_{\text{eq}}) > 0 \implies \Delta G > 0$: Reaction proceeds spontaneously in reverse.

The Biochemical Standard State ($\Delta G^{\circ\prime}$)

Because standard state $[\text{H}^+] = 1.0\text{ M}$ corresponds to $\text{pH} = 0$ (which would instantly denature all enzymes), biochemists define the biochemical standard state (designated by the prime symbol, $\Delta G^{\circ\prime}$):

• $\text{pH} = 7.0$ ($[\text{H}^+] = 10^{-7}\text{ M}$)
• $[\text{H}_2\text{O}] = 55.5\text{ M}$
• $[\text{Mg}^{2+}] = 1.0\text{ mM}$ (essential divalent cofactor for nucleotide coordination)
• Temperature $= 298.15\text{ K}$ ($25^\circ\text{C}$) or $310.15\text{ K}$ ($37^\circ\text{C}$)

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7. Bioenergetics: ATP Hydrolysis and Reaction Coupling

A central organizing principle of cellular metabolism is reaction coupling: an endergonic, thermodynamically unfavorable reaction ($\Delta G_1 > 0$) is linked to a strongly exergonic reaction ($\Delta G_2 \ll 0$) via a shared chemical intermediate. Because Gibbs free energy is an additive state function, the net free energy change is negative:

$$\Delta G_{\text{net}} = \Delta G_1 + \Delta G_2 < 0$$

Thermodynamic Basis of ATP Hydrolysis

Adenosine triphosphate (ATP) functions as the universal chemical energy currency. Hydrolysis of its terminal phosphoanhydride bond releases substantial free energy:

$$\text{ATP} + \text{H}_2\text{O} \longrightarrow \text{ADP} + \text{P}_i \quad (\Delta G^{\circ\prime} \approx -30.5\text{ kJ/mol} = -7.3\text{ kcal/mol})$$

(Under typical intracellular physiological conditions with high ATP/ADP ratios, the actual $\Delta G \approx -50\text{ to } -55\text{ kJ/mol}$).

Three Molecular Factors Driving High Phosphoryl Transfer Potential:

1. Electrostatic Repulsion Relief: At physiological $\text{pH} = 7.4$, ATP carries four localized negative charges ($\text{ATP}^{4-}$). Hydrolysis separates these clustered like charges onto $\text{ADP}^{3-}$ and $\text{HPO}_4^{2-}$, relieving electrostatic strain.
2. Resonance Stabilization: Inorganic phosphate ($\text{P}_i$) possesses four equivalent resonance structures with complete delocalization of negative charge over the entire tetrahedral ion. In contrast, the terminal phosphate in ATP is restricted in resonance because oxygen electrons compete for conjugation with adjacent phosphorus atoms.
3. Solvation Stabilization: The reaction products ($\text{ADP}$ and $\text{P}_i$) bind more surrounding water molecules, releasing greater hydration enthalpy compared to the intact ATP reactant.

Worked Glycolytic Coupling Example

The first step of glycolysis (phosphorylation of glucose catalyzed by hexokinase) is endergonic when uncoupled:

$$\begin{array}{rll} \text{Glucose} + \text{P}_i \longrightarrow \text{Glucose-6-Phosphate} + \text{H}_2\text{O} & \Delta G^{\circ\prime} = +13.8\text{ kJ/mol} & (\text{Unfavorable}) \\ \text{ATP} + \text{H}_2\text{O} \longrightarrow \text{ADP} + \text{P}_i & \Delta G^{\circ\prime} = -30.5\text{ kJ/mol} & (\text{Exergonic Drive}) \\ \hline \text{Glucose} + \text{ATP} \longrightarrow \text{Glucose-6-Phosphate} + \text{ADP} & \Delta G^{\circ\prime} = -16.7\text{ kJ/mol} & (\text{Spontaneous!}) \end{array}$$

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8. Worked Numerical Examples

Example 1: Enthalpy Calculation via Hess's Law and Bond Energies

Calculate the standard enthalpy of combustion for propane gas ($\text{C}_3\text{H}_8$) to form carbon dioxide and liquid water:

$$\text{C}_3\text{H}_{8(g)} + 5\text{O}_{2(g)} \longrightarrow 3\text{CO}_{2(g)} + 4\text{H}_2\text{O}_{(l)}$$

Given standard formation enthalpies: $\Delta H_f^\circ[\text{C}_3\text{H}_{8(g)}] = -104.7\text{ kJ/mol}$, $\Delta H_f^\circ[\text{CO}_{2(g)}] = -393.5\text{ kJ/mol}$, and $\Delta H_f^\circ[\text{H}_2\text{O}_{(l)}] = -285.8\text{ kJ/mol}$.

Solution:

1. Apply Hess's Law formula ($\Delta H_{\text{rxn}}^\circ = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants})$): $$\Delta H_{\text{rxn}}^\circ = [3(-393.5) + 4(-285.8)] - [1(-104.7) + 5(0)]$$
2. Multiply terms: $$\text{Products} = -1180.5 + (-1143.2) = -2323.7\text{ kJ/mol}$$ $$\text{Reactants} = -104.7\text{ kJ/mol}$$
3. Calculate net reaction enthalpy: $$\Delta H_{\text{rxn}}^\circ = -2323.7 - (-104.7) = -2323.7 + 104.7 = -2219.0\text{ kJ/mol}$$ Combustion of one mole of propane releases $2219\text{ kJ}$ of thermal energy.

Example 2: Non-Standard Free Energy and Metabolic Flux

Consider the interconversion of glucose-1-phosphate (G1P) and glucose-6-phosphate (G6P) catalyzed by phosphoglucomutase: $\text{G1P} \rightleftharpoons \text{G6P}$. The standard free energy change is $\Delta G^{\circ\prime} = -7.3\text{ kJ/mol}$ at $37^\circ\text{C}$ ($310\text{ K}$). In a red blood cell, the steady-state concentrations are $[\text{G1P}] = 0.050\text{ mM}$ and $[\text{G6P}] = 1.00\text{ mM}$. Calculate the non-standard free energy change ($\Delta G$) and determine if net glycolysis flux proceeds forward.

Solution:

1. Calculate the actual reaction quotient ($Q$): $$Q = \frac{[\text{G6P}]}{[\text{G1P}]} = \frac{1.00\text{ mM}}{0.050\text{ mM}} = 20$$
2. Calculate $RT \ln Q$ (using $R = 8.314 \times 10^{-3}\text{ kJ}/(\text{mol}\cdot\text{K})$ and $T = 310\text{ K}$): $$RT \ln Q = (8.314 \times 10^{-3}\text{ kJ/mol}\cdot\text{K})(310\text{ K}) \ln(20)$$ $$\ln(20) \approx 2.996 \implies RT \ln Q = (2.577\text{ kJ/mol})(2.996) \approx +7.72\text{ kJ/mol}$$
3. Calculate actual $\Delta G$: $$\Delta G = \Delta G^{\circ\prime} + RT \ln Q = -7.30\text{ kJ/mol} + 7.72\text{ kJ/mol} = +0.42\text{ kJ/mol}$$ Because $\Delta G > 0$ under these physiological concentrations, the reaction is near thermodynamic equilibrium (slightly favoring reverse flux), demonstrating that metabolic pathways are dynamically tuned by product-reactant concentration ratios.