8.1: Mass action

There is a lot of misleading information in the scientific literature about how to quantify the amount of energy released by a microorganism’s catabolic reaction (Amend and LaRowe, 2019). To help you avoid mistakes, we will start with the basics and first consider the concept of mass action before moving on to free energy calculations.

At equilibrium, the forward rate of a reaction equals the reverse reaction rate, so there is no net change in the reaction mixture. We can describe the equilibrium point of a reaction using the mass action equation. Consider the following generic reaction: $$a \ \text{A} + b \ \text{B} \longleftrightarrow c \ \text{C} + d \ \text{D}$$ where $\text{A}$, $\text{B}$, $\text{C}$, and $\text{D}$ represent reactants and products and $a$, $b$, $c$, and $d$ represent their respective stoichiometric coefficients. For this reaction, we can define a reaction quotient $(Q)$ as follows: $$Q = \frac{[\text{C}]^{c} [\text{D}]^{d}}{[\text{A}]^{a} [\text{B}]^{b}}$$

Here and throughout this book, brackets $[ \ ]$ indicate chemical activity for aqueous species and fugacity for gases, following Faure (1991). For a brief review of the concepts of activity and fugacity, see Box $8.1$. Also, when formulating the reaction quotient, we can often exclude pure solids and liquids if they are present in a reaction. The activity of pure solids and liquids is $1$ and the activity of dilute aqueous solutions is also nearly $1$.

When a reaction mixture is in equilibrium, the value of $Q$ is equal to $K$, the equilibrium constant for the reaction. Specifically, regardless of what values of activity and/or fugacity are inserted for each product and reactant, if the reaction mixture is at equilibrium, then the value of the quotient is constant at a given temperature and pressure and equal to $K$. This relationship is referred to as the law of mass action. If $Q$ is less than $K$, then the reaction needs to move forward to reach equilibrium. If the value of $Q$ is greater than $K$, then the reaction needs to go backward to reach equilibrium.

As an example, we will calculate the value of $Q$ for the following reaction, which describes microbial reduction of iron in hematite $\left(\text{Fe}_{2} \text{O}_{3}\right)$ coupled with dihydrogen oxidation: $$\text{H}_{2} \ (aq) + \text{Fe}_{2} \text{O}_{3} \ (s) + 4 \ \text{H}^{+} \longleftrightarrow 3 \ \text{H}_{2} \text{O} \ (l) + 2 \ \text{Fe}^{2+}$$

Using an activity of $1$ for water and hematite, the reaction quotient is: $$Q = \frac{\left[\text{Fe}^{2+}\right]^{2}}{\left[\text{H}_{2}\right] \left[\text{H}^{+}\right]^{4}}$$

It is acceptable to calculate Q as formulated in equation $\PageIndex{4}$. However, by putting the quotient in log form, we can make the calculation more convenient by removing exponents: $$\log Q = 2 \log \left[\text{Fe}^{2+}\right] - \log \left[\text{H}_{2}\right] - 4 \log \left[\text{H}^{+}\right]$$

If the environment where the reaction is occurring has a composition consistent with that in Table $8.1$, the $\log Q$ value is $23.68$. The $\log K$ at the temperature of the environment $\left(25^{\circ} \mathrm{C}\right)$ is equal to $29.18$ (Appendix A). Thus, the value of $Q$ is less than $K$ and we conclude that the reaction needs to move forward to reach equilibrium.

Table $8.1$: Hypothetical Conditions used for Example and Practice Calculations

Parameter	Concentration/Value	Activity*	Log activity
$\text{H}_{2} \ (aq)$	$10 \ \text{nM}$	$9.89 \times 10^{-9}$	$-8.005$
$\text{CH}_{3} \text{COO}^{-}$	$50 \ \mu \text{M}$	$4.51 \times 10^{-5}$	$-4.3654$
$\text{O}_{2} \ (aq)$	$100 \ \mu \text{M}$	$9.89 \times 10^{-5}$	$-4.005$
$\text{NO}_{3}^{-}$	$200 \ \mu \text{M}$	$1.83 \times 10^{4}$	$-3.7376$
$\text{SO}_{4}^{2-}$	$1 \ \text{mM}$	$6.69 \times 10^{-4}$	$-3.1746$
$\text{N}_{2} \ (aq)$	$0.5 \ \text{mM}$	$4.94 \times 10^{-4}$	$-3.3065$
$\text{NH}_{4}^{+}$	$100 \ \mu \text{M}$	$9.13 \times 10^{-5}$	$-4.0394$
$\text{Fe}^{2+}$	$100 \ \mu \text{M}$	$7.29 \times 10^{-5}$	$-4.1633$
$\text{H}_{2} \text{S} \ (aq)$	$5 \ \mu \text{M}$	$4.41 \times 10^{-6}$	$-5.3559$
$\text{CH}_{4} \ (aq)$	$100 \ \mu \text{M}$	$9.89 \times 10^{-5}$	$-4.005$
$\text{HCO}_{3}^{-}$	$2 \ \text{mM}$	$5.80 \times 10^{-4}$	$-3.2369$
$\text{H}^{+}$	$\text{pH} \ 6.00$	$1.00 \times 10^{-6}$	$-6$
*Activities are calculated using The Geochemist’s Workbench program SpecE8 with temperature at $25^{\circ}\mathrm{C}$ and solution ionic strength at 4.2 mmolal.

Practice $\PageIndex{1}$

Calculate the value of the reaction quotient for the following reactions under the hypothetical conditions listed in Table $8.1$.

$$\begin{align*} & \text{CH}_{3} \text{COO}^{-} + \text{H}^{+} + \text{SO}_{4}^{2-} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + \text{H}_{2} \text{S} \ (aq) \\ & \text{CH}_{3} \text{COO}^{-} + 8 \ \text{FeOOH} \ (s) + 15 \ \text{H}^{+} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + 12 \ \text{H}_{2} \text{O} \ (l) + 8 \ \text{Fe}^{2+} \end{align*}$$

Use an activity of 1 for solids and liquids. Answers are provided at the end of the chapter (Practice $8.1.1$).

Box $8.1$: Activity and Fugacity

The activity of an aqueous chemical species can be thought of as an effective concentration. In very dilute solutions, activity is numerically equivalent to concentration in molality (Bethke, 2008). However, as the concentrations of solutes increases in a solution, the difference between activity and concentration grows.

Activity differs from concentration in response to electrostatic interactions within solutions. Ions and polar molecules with like charges repulse one another whereas those with opposite charges attract. These interactions increase order within the solution by creating a loose structure. Some of the system’s energy is consumed in creating that order and the amount consumed is proportional to the extent of electrostatic interactions. In other words, as solute concentrations increase, electrostatic interactions increase and cause the difference between activity and concentration to grow. To account for these effects in our thermodynamic calculations, we use activities rather than concentrations.

Several approaches have been developed to calculate activities. Many are based on the Debye-Hückel equation, which calculates an activity coefficient $(\gamma)$ that is used to calculate the activity $(a)$ of chemical species $(i)$ as: $$a_{i} = \gamma_{i} C_{i} \nonumber$$ where $C$ is the concentration in molal units (mol solute/kg solvent) and activity and the activity coefficient are unitless. In an ideal solution, the activity coefficient is one and therefore activity and concentration are equal.

Fugacity is conceptually similar to activity except it is applied to gases rather than aqueous chemical species. Specifically, we can think of fugacity as the effective partial pressure of a gas. For gases at low partial pressures, fugacity is numerically equivalent to partial pressure, but as partial pressure increases, the difference between fugacity and partial pressure grows to account for greater deviation from ideal gas behavior. Fugacity $(f)$ for a gas species $(i)$ is related to partial pressure $(P)$ in bars according to the following equation: $$f_{i} = \gamma_{i} P_{i} \nonumber$$ where $\gamma$ is the fugacity coefficient, and fugacity and the fugacity coefficient are unitless.

For most parameters, analytical laboratories do not report results in terms of activity and fugacity. Therefore, in order to calculate the free energy change of a microbial reaction, we first need to calculate the activities and fugacities of all chemical species found in the reaction. These are the values that should typically be inserted into the reaction quotient, not concentrations and partial pressures. A convenient way to accomplish this task is with geochemical modeling software. Numerous geochemical modeling programs have been developed. Two commonly used options include The Geochemist’s Workbench (GWB) (Bethke, 2018) and PHREEQC (Parkhurst, 1995). Among other capabilities, these programs calculate chemical speciation and the activity or fugacity of those species within a given geochemical system. An example of how to carry out a speciation calculation with one of the GWB programs is available online (Bethke and Farrell, 2016).