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Chemical Equilibrium

LMA – Law of Mass Action

Consider the chemical reaction between educts A, B and products C, D:

(1.1)

a A + b B  ⟷  c C + d D

with a, b, c, and d as the stoichiometric coefficients. The left-right arrow tells us that the reaction proceeds in both directions: forwards and backwards. It’s quite common to replace the left-right arrow with an equal sign.

When equilibrium is reached, the forward and the backward rates become equal, and the amount of educts and products obeys the fundamental law of mass action:

(1.2)

\(K \ = \ \dfrac{\{C\}^c\{D\}^d}{\{A\}^a\{B\}^b}\)

(Law of Mass Action)

K is called the equilibrium constant. This law is so simple, yet universal and powerful. It describes aqueous speciation (including acid-base and redox reactions), solid phase and gas equilibria, ion exchange, surface complexation, and other phenomena. Thus, it’s not surprising that the mass-action law forms the backbone of many approaches in hydro­chemistry – including PhreeqC and aqion.

The ratio in 1.2 is often interpreted as a ratio of molar (mol/L) or molal (mol/kg) concentrations. This, however, only applies to infinitely diluted systems. In the case of real solutions, the concentrations should be replaced by activities. We emphasize this by using curly brackets {} for activities in 1.2, while square brackets [] are reserved for molar concentrations.

Equilibrium Constants

Our central quantity is the equilibrium constant K. Its value describes something like the degree of completeness of the reaction when equilibrium is established:

	K		K
	1		0		½ of the educts have changed into products (50 of the reaction is complete)
	10-10		-10		reaction is very incomplete
	1010		10		reaction is almost complete

Since K varies over several orders of magnitude, it’s more convenient to work with the decadic logarithm of K rather than with K itself. Equation (1.2) then becomes:

(2.1)

lg K  =  c · lg {C} + d · lg {D} – a · lg {A} – b · lg {B}

Note that outside the English-speaking world, the term “K” is used instead of K. Nevertheless, we will use K and K as synonyms.

Instead of K, the pK value is often used:

(2.2)

pK = – K

which parallels the definition of pH as – {H⁺}.

Simple Rules. The reaction formula in 1.1 determines the form of 1.2. Other reactions with more or with less educts and/or products lead to other formulas of K. Let’s consider the most simple reaction type A ⟷ B with K = {B}/{A} and what happens when we manipulate it:

(2.3a)		reaction	A ⟷ B		K		K
(2.3b)		reverse reaction	B ⟷ A		1/K		– K
(2.3c)		multiplication by 2	2 A ⟷ 2 B		K2		2 K
(2.3d)		division by 2	½ A ⟷ ½ B		K½		½ K
(2.3e)		two-step reaction	A ⟷ B    (with K1)		K1 · K2		K1 + K2
			B ⟷ C   (with K2)

Thermodynamic Data (log K Values)

Two things are necessary to perform an equilibrium calculation:

•  the model:	LMA algorithm	(e.g. PhreeqC, aqion)
•  thermodynamic data:	lg K’s	(e.g. wateq4f1)

The role of the thermodynamic database cannot be overestimated. Models are empty shells unless thermodynamic data breathes life into them.

Today, dozens of hydrochemistry models/software exist. The vast majority of them are LMA-based and of high numerical accuracy. In other words, when using the same input solution and the same thermodynamic dataset the calculations will lead to exactly the same results. In reality, however, each model comes with its own thermodynamic database, so the same input solution will produce different results (i.e. speciation, saturation indices, etc.).

Model comparisons show that the cause for different results almost always lies in the differences between the thermodynamic databases used. The typical question, “Which model/software is used?” is incomplete. One should always ask additionally: “Which thermodynamic database is used?”

Inside the Database. The thermodynamic database² comprises at least two items for each species: the reaction formula (stoichiometry) and the K value. Example: the 1^st dissociation step of sulfuric acid (at 25) is defined by the entry:

H+ + SO4-2 = HSO4-
log_k 1.988

Note 1. The K value is only meaningful in context with the chemical reaction formula (1.1). For example, if you reverse the order in 1.1 to cC + dD ⟷ aA + bB, then K changes its sign from K to – K.

Note 2. The K is temperature dependent. The conversion of K from standard temperature (25) to other temperatures is described here.

Stoichiometric Coefficients

As we have learned above, the K value and the reaction formula are inseparable things. It’s a good idea to always put both together and write:

(3.1)

a A + b B  ⟷  c C + d D

lg K

This line contains three pieces of information:

• the stoichiometric coefficients (a, b, c, d)
• the names of the constituents (A, B, C, D)
• the equilibrium constant (in form of K)

To emphasize the similarity with an algebraic equation, we replace the right-left arrow in the reaction formula by an equals sign and move all terms to the right side of the equation:

(3.2)

0  =  c C + d D – a A – b B

or, more generally:

(3.3)

0  =  Σi νi Ai

where A_i (= A, B, C, D) denotes the i^th chemical constituent and the small Greek letter ν_i symbolizes the corresponding stoichiometric coefficient with the convention: ν_i > 0 for products and ν_i < 0 for educts. Stoichiometric coefficients are often overlooked, but they play an important role (along with the sign convention for educts and products).

Conditional Equilibrium Constants

There is some relaxed version of the mass-action law based on concentrations alone (rather then activities). In contrast to 1.2, it relies on a so-called conditional or apparent equilibrium constant:

(4.1)

\(^cK \ =\ \dfrac{[C\,]^c \,[D\,]^d}{[A\,]^a \,[B\,]^b}\)

Since concentrations are converted to activities by {i} = γ_i [i], the conditional equilibrium constant ^cK is related to the thermodynamic equilibrium constant K by.

(4.2)

\(^cK \ =\ \left[ \dfrac{\gamma_c \,\gamma_d}{\gamma_a \, \gamma_b} \right]^{-1} \ K\)

For sufficiently dilute solutions, i.e. when activity coefficients are γ_i = 1, both equilibrium constants become equal: ^cK = K.

Again, the conditional equilibrium constant ^cK can be abbreviated by p^cK = – ^cK, which yields the relation:

(4.3)

pcK  =  pK – ( lg γa + lg γb – lg γc – lg γd )

log K as an “Energy”

There is a fundamental relationship between the equilibrium constant K and the Gibbs energy change:³

(5.1)

ΔG0  =  – RT ln K

where R = 8.314 J mol^-1K^-1 is the gas constant and T the absolute temperature in Kelvin. This equation can be re-arranged to K:⁴

(5.2)

\(\lg K = -\dfrac{\Delta G^{0}}{\ln 10 \,\cdot RT} \, = -\dfrac{\Delta G^{0}}{2.3 \,\cdot RT}\)

These equations apply only to the thermodynamic equilibrium constant K (and not to ^cK). And, the log K value – in contrast to K itself – behaves like an “energy”, which according to (2.3e) is also additive.

The temperature dependence of ΔG⁰ and K is discussed here.

Gibbs Energy. Beginners sometimes have trouble with the notation “ΔG⁰”. There are at least three questions. First question: why ΔG and not G itself?  Principally, a characteristic of energy is that there is no absolute energy at all; only energy differences are meaningful (which are measurable quantities). In our case, ΔG is the difference between the energies of the products and the educts. Infinitesimally small changes are abbreviated by dG.

Second question: why this Gibbs energy?  Of all energy types (internal energy U, enthalpy H, Helmholtz free energy A etc.) the Gibbs energy is tailored for chemical reactions in which the particle numbers change, i.e. dn_i ≠ 0. Under typical lab conditions (i.e. constant temperature and constant pressure), dG or ΔG are associated with the chemical potential μ_i of the species i:

(5.3)

dG  =  Σi μi dni

or

ΔG  =  Σi μi νi

The first equation is the Gibbs-Duhem equation; the second equation is obtained after integration of the first one.⁵ In the last equation, the particle change dn_i is replaced by the stoichiometric coefficients ν_i (which are well-known quantities taken from the reaction formula).

Third question: what is the meaning of the superscript “0” in ΔG⁰?  ΔG is the Gibbs energy change of any state, which may be a non-equilibrium state or an equilibrium state; ΔG⁰ refers exclusively to the equilibrium state. Both are related by:

(5.4)

ΔG  =  ΔG0 + RT ln Q

with

\(Q \ = \ \dfrac{\{C\}^c\{D\}^d}{\{A\}^a\{B\}^b}\)

Here, Q is the reaction quotient – a sort of “non-equilibrium constant”. In fact, Q only becomes the equilibrium constant K in 1.2 if ΔG = 0.

For the very special case of equilibrium, 5.3 also reduces to:

(5.5)

ΔG0  =  Σi μ0i νi

where μ⁰_i is the standard chemical potential of the species i (that can be looked up in tables).

Chemical Equilibrium from Two Perspectives

The concept of chemical equilibrium seems simple and obvious to us today. The way to understand this phenomenon, however, was long. Historically, there are two approaches to chemical equilibrium that came from opposite sides: from reaction kinetics and from classical thermodynamics.

From Reaction Kinetics. The kinetic approach relies on the temporal changes of educts and products. Experiments have shown that a forward reaction A ⟶ B is always accompanied by a backward reaction B ⟶ A. Over time, the forward and backward reaction rates converge. The chemical equilibrium is established when they become equal, k_f = k_b.⁶

Reactions do not stop but come to equilibrium (which then lasts forever when undisturbed).

From Thermodynamics. The second approach was a physico-mathematical approach. In 1873, Josiah Willard Gibbs introduced the idea of the chemical potential μ and its relationship to the free energy ΔG. It was the marriage of chemistry with classical thermodynamics.

Equilibrium is attained when the Gibbs energy is at its minimum value.

In fact, the first approach is easier to understand; it observes how educts and products evolve over time. In contrast, thermodynamics doesn’t know what “time” is. In thermo­dynamics, the physical quantity “time” is replaced by the abstract term “energy”.

The Effect of Temperature and Pressure

At this point a short but important statement: The K value is (strongly) temperature-dependent, but not dependent on pressure P.

In the thermodynamic jargon, it is expressed in this way:

The insensitivity of K to pressure results from the fact that the overall volume changes ΔV⁰ of liquids and solids are negligible in typical hydrochemical applications – in quite contrast to gases.

Remarks & References