Mass Action Law

General Remarks

This module is registered in the "advanced" part, because it uses the concept of the chemical potential also developed in the advanced part.

We use a rather general derivation, but do not go too deep into the details.

The mass action law is usually taught in high school chemistry, so we know what we want to find: We look at some chemical reaction, e.g.

2H₂ + O₂	Û	2H₂O

The mass action law, as we know it, than asserts that the concentrations of the particles (= molecules in this case) in equilibrium can be written as

[H₂]² · [O₂] [H₂O]²	=	K(T, p)

With K = reaction constant.

Or in words: The product of the concentration of the reaction partners with all concentrations always taken to the power of their stoichiometric factors, equals a constant K which has a numerical value that depends on the temperature and pressure. The constant K is called reaction constant.

This statement, however, includes already a generalization and a convention:

There can be any number of particles reacting or resulting from the reaction, and we always bring the results of the reaction, (in the example the H₂O), to the right side of the equation and assign a negative value to its stoichiometric factors - the reaction products thus end up in the denominator of the concentration products. We mostly use integers for the stoichiometric factors, but that is not de rigeur.

An alternative way of writing the reaction equations that shows the "minus" sign more clearly, is

2H₂ + O₂ – 2H₂O = 0

The mass action law is deceptively simple, it is however not so trivial to derive it from thermodynamics including a value for the reaction constant, and it is often quite tricky to use for real cases!

We will now give a standard derivation; an alternative way is given in another module.

Standard Derivation Using the Chemical Potential

First we define arbitrary reactions of any kind by the equation

n₁ · A₁ + n ₂ · A₂ + .... + n _f · A_f	=	n_g · A_g + n_g+1 · A_g+1 + .... + n_i · A_i

The A_x denote the particles (or reactions partners involved) - atoms, ions, molecules, vacancies, electrons, holes, .. - we want to be very general at this point. The corresponding stoichiometric factors are the n_x, and they are usually (but not always) integers. Bringing the products of the reaction to the left side of the equation which gives their stoichiometric factors a negative sign, leads to the simple version

i S 1	n_i · A_i	=	0

Chemical reactions as written down in standard notation always inherently assume that we have exactly the right amount of the chemicals (or, as we prefer to call it, particles) that are needed.

The reaction above, for our example, thus takes two mols of H₂ (= A₁) for every mol of O₂ (= A₂ ); or in our lingo, two H₂ particles (= molecules in this case) for one O₂ particle., yielding two H₂O (= A₃) particles.

We have n₁ = 2, n₂ = 1, n₃ = – 2.

Real life is different. You mix some number of H₂ particles with some number of O₂ particles, and after the reaction you have some number of all three particles involved (with one number probably being very low, or ideally zero, if the most scarce particle was completely used up in the reaction).

In deriving the mass action law, we have to allow for this by allowing arbitrary starting concentrations c_i⁰ of the particles involved including, if we wish, some concentration of the reaction products even before a reaction took place - nobody keeps us from filling some water into the container with H₂ and O₂ before we start the reaction.

We want to get a statement about the concentration of the particles in equilibrium for an arbitrary mix of concentrations at the start of the reaction in non-equilibrium; for ease of writing we denote the equilibrium concentration of the component i with c_i; the concentration at the start than is c_i⁰, and an arbitrary concentration is C_i.

The various c_i may be the number of mols, the absolute number of particles, or the concentration relative to some fixed value - it doesn't matter as long as the same definition is used throughout.

As pointed out above, it is important to realize, that the c_i⁰ can have any initial values whatsoever - you always can throw into a closed container whatever you want - but the dC_i; the changes in the concentrations, are tied to each other via the reaction equation.

If you produce one mol of H₂O from any initial quantity of H₂ and O₂; you will have reduced the H₂ concentration by 1 mol and the O₂ concentration by 0,5 mol - the dC_i thus are not independent.

The whole mixture of stuff - at whatever composition, i.e. for the whole range of the C_i - will have some free enthalpy G(C_i, p, T).

The important question is: For which concentration values of the various particles, do we have equilibrium and thus the minimum of G?

In other words: For what conditions is dG = 0?

Lets write it down. With G = G(C_i, p, T) we have for dG

dG	=	¶ G ¶C₁	· dC₁ +	¶ G ¶C₂	· dC₁ + ... +	¶G ¶C _i	· dC_i +	¶G ¶T	· dT +	¶G ¶p	· dp

The (¶G/¶C _i) by definition are the chemical potentials m_i of the particle sort x in the mixture, and the two last terms are simply = 0 if we look at it at constant pressure and temperature. For equilibrium.this leaves us with

dG	=	i S 1	m_i · dC_i = 0

Now comes a decisive step. We know that our dC_i are tied somehow, but how?

To see this, we "wiggle" the system a little and react some particles, changing the concentrations a little bit. As a measure of this change we introduce a "reaction coordinate" dx; a somewhat artificial, but useful quantity (without a unit).

The changes in the concentrations of the various particles of our system then must be proportional to dx and the proportionality constants are the stoichiometric indices n_i. Think about it! However you wiggle - if the concentration of O₂ changes some, the concentration of H₂ will change twice as much.

In other words, or better yet, in math, we have

dC_i	=	n_i · dx

Substituting that into the equation for dG from above, we obtain

d G	=	i S 1	m_i · n _i · dx	=	dx ·	i S 1	m_i · n_i = 0

Since dx is some arbitrary number, the sum term must be zero by itself and we have as equilibrium condition

i S 1	m_i · n_i	=	0

This looks (hopefully) familiar. It is the equilibrium condition we had before for particles not reacting with each other when we looked at the meaning of the chemical potential.

Now all we have to do is to take the "master equation" for the chemical potential so beloved by the more chemically minded, and plug it into the equilibrium condition for our reactions.

In order to stay within our particle scheme, we use k instead of R and the activity A_i of the component i instead of its concentration C_i. Feel free to read "activity" as " somewhat corrected concentration" if you are unfamiliar or uncomfortable with activities. We have

m_i	= m _i⁰ + kT · ln A_i

And, since we are treating equilibrium, the activity A_i now is the equilibrium activity a_i (= concentration c_i if everything would be "ideal") instead of the arbitrary concentration C_i because we are treating equilibrium now by definition.

Inserting this formula in the equilibrium condition from above (and omitting the index "i" at the sum symbol for ease of writing) yields

S (n_i · m_i⁰) + kT · S (n _i · ln a_i )	=	0

Going through the mathematical motions now is easy.

Expressing the sum of ln's as the ln of the products of the arguments, and rearranging a bit gives

ln	P	(a_i)ⁿⁱ	= –	1 kT	· S m_i⁰ · n_i = –	1 kT	· DG ⁰

Because Sn _i · m_i⁰ is just the sum over all standard reaction enthalpies involved, which we call DG⁰ .

The product on the right hand side is just a fancy way to write down one part of the mass action law, it would give exactly what we formulated for the case of 2H₂ + O₂ Û H₂ + O from above. Putting everything in the exponent finally yields the mass action law:

P (a_i)ⁿⁱ	= exp –	G⁰ kT	= K ^–1 =		(Reaction Constant) ^–1

It doesn't matter much, but it is standard to write K ^–1. In other words, put the products of the reaction in the nominator to get K.

There seems to be a bit of magic involved: We started with arbitrary amounts of components, let them react an arbitrary amount (we even defined a new quantity, the reaction coordinate x) - and none of this shows up in the final formula! There are certainly some questions.

What's left are only equilibrium concentrations (or activities) - what happened to the starting concentrations?

Can't we derive the mass action law then without introducing quantities that seem not to be needed?

Some short answers:

At some point, we essentially switched to changes (= derivatives) of prime quantities - and everything not changing is now gone. It is still there, however, if we do real calculations because then we need more information - the mass action law, after all, is just one equation for several unknown concentrations.

There probably is a more direct way to get the mass action law that does not involve the somehow superfluous reaction coordinate. However - I do not know it and I'm in good company. Several text books I consulted do not know a better way either. Still, try the link for some alternatives.

Lets go back to our original question and mix arbitrary amounts of whatever and than let the buggers react. What will we get, throwing in the reaction equation and possibly some reaction enthalpies?

The mass action law now gives us one relation between the equilibrium concentration, but not the absolute amounts. There are, after all, just as many unknowns for the equilibrium concentrations as you have components, and you need more than one equation to nail everything down.

Additionally, the way we have spelled out the mass action law here also has a number of pitfalls; if you want to really use it, you must know a bit more, in particular about conventions that must be strictly adhered to.

All that is essentially beyond the scope of this "Defect" lecture, but for the hell of it, a few more modules intertwining mass action law and chemical potentials were made; they are accessible via the following links.

Pitfalls and extensions of the mass action law

Some standard (chemical) examples of applying mass action law

Alternative derivations of the mass action law

Some defects in ionic crystal related applications of the mass action law

Appendix: Some Necessary Math

This appendix contains one of the necessary mathematical transformations used above to refresh your memory