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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 stoichiometric coefficients. The left-right arrow tells us that the reaction proceeds in both directions: forwards and backwards. It is quite common to replace the left-right arrow with an equals 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 simple and 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 is 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 is more convenient to work with the decadic logarithm of K, K, rather than with K itself. Equation (1.2) then becomes:¹

(2.1)

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

Instead of K, the pK value is also 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 values	(e.g. wateq4f2, …)

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

Today there are dozens of software tools in hydrochemistry. The vast majority of them are LMA-based and of high numerical accuracy. Using the same input water and the same thermodynamic data the calculations will lead to the same results. In practice, however, each software has its own thdyn database, so the same input can lead to different results (i.e. speciation, saturation indices, etc.).

Extensive model calculations (and comparisons) have shown that the cause of different results almost always lies in the differences between the thdyn data used. The question, “Which software was used?” is incomplete. One should always ask additionally: “Which database was used?”

Inside the Database. The thdyn database³ comprises at least two items for each species:

• the reaction formula (stoichiometry)
• 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 is a good idea to always take and write both together (in one line):

(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)

\(0\ =\ \sum\limits_i \, \nu_i A_i\)

where A_i (= A, B, C, D) denotes the i^th chemical constituent and the small Greek letter \(\nu_i\) symbolizes the corresponding stoichiometric coefficient with the convention: \(\nu_i>0\) for products and \(\nu_i<0\) 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 \(\{i\} = \gamma_i\, [i]\), 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 \(\gamma_i = 1\), 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)

\(p\,^cK \,=\, pK - (\lg\gamma_a + \lg\gamma_b - \lg\gamma_c - \lg\gamma_d)\)

log K as an “Energy”

There is a fundamental relationship between the equilibrium constant K and the Gibbs energy change:⁴

(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:⁵

(5.2)

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

These equations only apply 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.

• Question 1: Why ΔG and not G itself?

A fundamental property of energy is that there is no absolute energy; only differences in energy 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.

• Question 2: Why this Gibbs energy?

Of all energy types (internal energy U, enthalpy H, Helmholtz free energy A), 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 \(\nu_i\)

The first equation is the Gibbs-Duhem relation; the second equation is obtained after integration of the first one.⁶ In the last equation, the particle change dn_i is replaced by the stoichiometric coefficients \(\nu_i\) (which are well-known quantities taken from the reaction formula).

• Question 3: 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 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 the special case ΔG = 0, Q becomes the equilibrium constant K in 1.2 and 5.3 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. The velocity of the forward (backward) reaction depends on the concentration of educts (products):⁷

(6.1a)		vf = kf · [educts]
(6.1b)		vb = kb · [products]

where the proportionality constants are the rates: k_f and k_b. During the reaction, the amount of educts shrinks and the v_f decreases; vice versa, the amount of products rises and v_b increases. Over time, the forward and backward velocities converge.⁸ The chemical equilibrium is established when they become equal: v_f = v_b.

Since v_f and v_b are always nonzero, the reactions do not stop, but come to equilibrium (which then lasts forever if it is not disturbed).

From Thermodynamics. The second approach was a physico-math 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”.

Effect of Temperature T and Pressure P

At this point a short but important statement: The K value is (strongly) T-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 change ΔV⁰ of liquids and solids is negligible in typical hydrochemical applications — in quite contrast to gases.

Remarks & References