8.2: Gibbs free energy

As described above, the difference between the reaction quotient \((Q)\) and the equilibrium constant \((K)\) tells us about the extent to which a reaction is out of equilibrium. By calculating the Gibbs free energy change of a reaction \(\left(\Delta G_{r}\right)\), we take this a step further and convert the extent of disequilibrium into an amount of energy.

For a reaction at equilibrium, \(Q = K\) and the free energy change is zero \(\left(\Delta G_{r} = 0\right)\). Exergonic reactions are those that need to move forward to reach equilibrium \((Q < K)\) and in doing so, they release energy and thus have a negative free energy change \(\left(\Delta G_{r} < 0\right)\) (Fig. \(8.1\)). Endergonic reactions are those that need to move backward to reach equilibrium \((Q > K)\). For an endergonic reaction to move forward, energy would need to be added to the reaction mixture \(\left(\Delta G_{r} > 0\right)\). Microorganisms catalyze both types of reactions as part of their metabolisms, but only exergonic reactions can supply chemotrophs with energy.

Figure \(8.1\): Variation in free energy during an exergonic and endergonic reaction according to transition-state theory.
https://commons.wikimedia.org/wiki/File:Gibbs_free_energy_changes.png

There are multiple ways to calculate the free energy change of a reaction. Here, we will calculate free energy changes by correcting the standard state free energy change of the reaction \(\left({\Delta G_{r}}^{\circ}\right)\) for the conditions of the environment: \[\Delta G_{r} = {\Delta G_{r}}^{\circ} + RT \ln Q\]where \(R\) is the gas constant \(\left(0.008314 \ \text{kJ mol}^{-1} \text{K}^{-1}\right)\), \(T\) is temperature (in Kelvin \((K)\)), and \(Q\) is the reaction quotient. We can also write the equation in terms of log base 10: \[\Delta G_{r} = {\Delta G_{r}}^{\circ} + 2.303 RT \log Q\]

The value of \(Q\) reflects the activities of reactants and products of microbial reactions within an environment. \(T\) accounts for the temperature of the environment of interest. Thus, we can see that changes in the composition and temperature of an environment can alter the energy yields of microbial reactions.

Before we can carry out this calculation, we also need the standard state free energy change for the reaction. We can calculate one based on equation \((\PageIndex{2})\). As noted above, when \(Q = K\), a reaction is at equilibrium and \(\Delta G_{r} = 0\). Inserting these values into equation \((\PageIndex{2})\), we get \[0 = {\Delta G_{r}}^{\circ} + 2.303 RT \log K\]

Solving for the standard state free energy change gives the following: \[{\Delta G_{r}}^{\circ} = -2.303 RT \log K\]

Therefore, given values of \(\log K\) at the temperature of interest, we can calculate standard state free energy changes for a reaction at that temperature. Equilibrium constant values for select microbial reactions are provided in Appendix A. The values listed are only appropriate for environments with temperatures of \(25^{\circ} \mathrm{C}\) \((298.15 \ \mathrm{K})\). However, the appendix also includes polynomial equations that can be used to estimate the log value of the equilibrium constants at different temperatures.

For an example calculation, let’s reconsider the reaction from our example reaction quotient calculation (reaction \((8.1.3)\)). Using equation \((\PageIndex{4})\) and a \(\log K\) value for the reaction at \(25^{\circ} \mathrm{C}\) \((29.1806)\), the standard state free energy change of the reaction is: \[{\Delta G_{r}}^{\circ} = -2.303 \left(0.008314 \ \frac{\text{kJ}}{\text{mol K}}\right) (298.15 \ \mathrm{K}) (29.1806) = −166.58 \ \text{kJ} / \text{mol}\]

Assuming conditions consistent with our hypothetical environment (Table \(8.1\)) the log of the reaction quotient is \(23.68\). Inserting this value and the standard state free energy into equation \((\PageIndex{2})\), we get: \[\Delta G_{r} = \left(-166.58 \ \frac{\text{kJ}}{\text{mol}}\right) + 2.303 \left(0.008314 \ \frac{\text{kJ}}{\text{mol K}}\right) (298.15 \ \mathrm{K}) (23.68)\]which gives a Gibbs free energy change of \(-31.4 \ \text{kJ}/\text{mol}\). Under these conditions, therefore, the reaction is thermodynamically favorable \(\left(\Delta G_{r} < 0\right)\) and has the potential to serve as an energy source for iron-reducing microorganisms.

Practice \(\PageIndex{1}\)

Based on reaction quotient values calculated for parts A and B of Practice \(8.1.1\), what are the free energy yields of those reactions? Use Appendix A to obtain \(\log K\) values for the reactions at \(25^{\circ} \mathrm{C}\).