The Law of mass action has 2 parts - the equilibrium rate part & the reaction rate part. The "reaction rate part" states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to it's stoichiometric coefficient. The rate law states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to some power which may or may not be equal to stoichiometric coefficient of the reactant.

Why are the two laws different when in fact we are talking about the powers to which active masses of reactants should be raised in both the cases ? And which law is correct?

  • $\begingroup$ As far as my knowledge is concerned reactant can be one or more. In case when there is one reactant first part is applied and when there are two or more reactants second part is applied. $\endgroup$
    – AksaK
    Jul 4, 2017 at 16:04
  • $\begingroup$ This question has been answered in a separate question:chemistry.stackexchange.com/questions/68195/… $\endgroup$
    – Withnail
    May 1, 2019 at 12:28

2 Answers 2


First of all, you need to understand the basic difference between the two.

Consider the reaction $$\ce{A + B -> C}$$ Law of mass action states that in a rate equation the powers that the active masses of components raised are always stoichiometric factors.

For a reaction they are always fixed. However, the rate of reaction differs with the physical conditions. This is applicable to both equilibrium reactions and moderate reactions that are unidirectional.

This makes law of mass action kind of theoretical in nature. In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.

Both the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.

  • $\begingroup$ In which cases do the law of mass action give the exactly correct relations? $\endgroup$ Aug 6, 2017 at 18:10

Your statement of the Law of Mass Action is wrong. From Wikipedia:"The law is a statement about equilibrium and gives an expression for the equilibrium constant, a quantity characterizing chemical equilibrium." You apparently confuse the dynamic equilibrium where concentrations of reactants and products do not change with the rate of chemical reactions not at equilibrium (generally at constant temperature, and always with changes in concentrations). Keq is the equilibrium rate constant, the Reaction Quotient is Q. Q changes as the reaction proceeds and reaches Keq as dynamic equilibrium is attained.
It is NOT true that the rate of a (forward) chemical reaction is proportional to the concentration of the active masses of the reactants raised to some empirical power. Some reactions proceed at constant rates. These are known as zeroth order reactions - catalyzed reactions can be examples of these. (While you can argue that if R → P and if d[R]/dt = -k then d[R]/dt = -k[R]° (since concentration taken to the zeroth power is 1) that the proportionality holds, but this is trivial and not useful.)
In the historical development of the Law of Mass Action, there have been many instances where the powers for the Reaction Quotient at equilibrium have matched (more or less) the empirically determined powers of the rate expression. This is quite different than CONFUSING a situation at equilibrium with the situation which must, by definition, be non-equilibrium (since concentrations are changing). Which is correct?
Logically invoking a dynamic equilibrium requires concentrations to be unchanging. Say R → P. With [R] and [P] constant, then taking the ratio of [R]α/[P]ß for some arbitrary constants α & ß is also going to produce a constant. Choosing the stoichiometric coefficients for α & ß doesn't change that.
For the rate of a reaction, the claim is that the rate of (forward) reaction is proportional to [R1]α[R2]ß...[Rn]n. This often isn't even an acceptable approximation. (although over a small enough time interval, any continuous (non-pathological) data can be approximated by a straight line). The more difficult question to answer here is does the rate of reaction for simple single-step reactions follow this rule? I'd say in dilute gas phase reactions (without significant effects from products) (and of course at constant temperature and pressure) it generally will. The problem with this, is determination of which are and which are not "simple single step" reactions isn't trivial. You could start there and confirm the proportionality. You'd have to confirm it, which makes it mostly unnecessary to assume it to begin with, but you have to start somewhere.

  • $\begingroup$ The argument that you could choose any powers to raise the concentrations to (when at equilibrium as the concentrations don’t change, so neither will the quotient of those concentrations) is not a justification for always choosing to raise to the power of the coefficients. Raising to other powers will still give a constant, but it won’t generally have the same value as the actual equilibrium constant. $\endgroup$
    – Withnail
    Apr 30, 2019 at 21:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.