I don't understand why the stoichiometric coefficients of reactants and products are expotentialized in law of mass action.

So I checked on Wikipedia. And this popped out:

In the 1879 paper[9] the assumption that reaction rate was proportional to the product of concentrations was justified microscopically in terms of collision theory, as had been developed for gas reactions. It was also proposed that the original theory of the equilibrium condition could be generalized to apply to any arbitrary chemical equilibrium.

How does collision theory explain this?


3 Answers 3


The law of mass action only rigorously applies to elementary reactions (in that otherwise the exponents do not match stoichiometric coefficients). These are reactions whose equations match which particles are colliding on the microscopic scale. Consider the reaction

$$\ce{$r_1$R1 + $r_2$R2 + \ldots + $r_n$R_n <=> $p_1$P1 + $p_2$P2 + \ldots + $p_m$P_m}\tag1.$$

Reaction $(1)$ is elementary if and only if it proceeds in a single step. At some time $t = t_c$ all reactant particles $\ce{R1, R2, \ldots, R_n}$ collide. Quantum mechanically this collision is not too well defined, so view it to be Newtonian in essence. Not to mention: the probability $P$ of the simultaneous collision goes to zero as $n \to \infty$. This is why in actuality most elementary steps involve at most three individual constituents.

We therefore sacrifice some generality and instead consider the easier to handle

$$\ce{R1(g) + R2(g) <=> P1(g) + P2(g)} \tag{1'}.$$


Collision theory postulates that particles only initiate a transformation when they have enough energy. This is called activation energy and denoted by $E_a$. Maxwell-Boltzmann distribution

$$\frac{\mathrm{d}N_v}{N} = 4\pi\left(\frac{m}{2\pi k_bT}\right)^{3/2}v^2\exp\left(-\frac{mv^2}{2k_bT}\right)\mathrm{d}v\tag2$$

allows us to estimate the fraction of particles that have the required amount of movement. Note we are technically assuming $\mathrm{total\ energy} = \mathrm{kinetic\ energy}$. Similarly relativistic effects are ignored, and no cap is placed on maximum velocity. With these simplifications, the fraction of interesting particles is

$$f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\tag3$$

which evaluates to

$$f = \exp\left(-\frac{E_a}{k_bT}\right)\tag{3'}.$$

Number of interesting collisions

Say $Z^*$ is the number of collisions that matter. We will look for a function $Z$ such that

$$Z^* = Z\cdot f.\tag4$$

The Maxwell-Boltzmann distribution $(2)$ gives for a single kind of particle (say two $\ce{R1}$'s) that there are

$$Z = 2N^2\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\tag5$$

collisions in $1$ second per $1\ \mathrm{cm^3}$ if $N$ is the number of particles per square centimeter. (This is the traditional starting point unit.) Hence the rate of such a reaction would be

$$v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\tag6\right).$$

Reaction rate $v_r$

If you like, you are now able to substitute $f$ from $(3')$ and $Z$ via $(5)$ into $(4)$. Then insert $(4)$ into $(6)$. After necessary manipulation, this yields $(7)$

$$v_r = \underbrace{4 \cdot 10^{-3} N_A\left(\sigma_\ce{R1}\right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}.$$

Coming back to our actual reaction $(1')$ involves more relative quantities. For example, $\sigma_\ce{R1}$ becomes $$\sigma_\text{aver} = 0.5\left(d_\ce{R1} + d_\ce{R2}\right).$$

So it is a bit more difficult. But the result is analogous to $(7)$.

$$v_r = \underbrace{2\sqrt{2} \cdot 10^{-3} N_A\left(\sigma_\text{aver}\right)^2\sqrt{\pi RT\left(\frac{1}{M_{\ce{R1}}} + \frac{1}{M_{\ce{R2}}}\right)}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot c_{\ce{R1}} \cdot c_{\ce{R2}}$$

More compactly, if $\ce{R1} = A$ and $\ce{R2} = B$

$$v_r = k[A][B]\tag8$$

which is what we set out to prove.

  • This derivation assumes that every active collision leads to a reaction. When theoretically computed $k$ were compared to experimental, an extra factor was introduced, $P$. This is called the steric factor, and in classical collision theory $P$ remains empirical in essence.

Collision theory is for elementary steps

As your quote suggests, collision theory is a first theoretical explanation for the proportionality to products of concentrations. It does not generally explain various exponentiation. It does not have to either. The higher (or non-integer or negative) exponents usually derive from the mechanism itself. In other words, the fact that common reactions are not elementary comes into play.

For instance, the transition

$$\ce{2Br- + H2O2 + 2H+ -> Br2 + H2O}$$

is experimentally found to follow

$$v_r = k\ce{[H2O2][H+][Br-]}\tag{a}.$$

Mathematically, we can verify that one possible mechanism is

$$\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium}$$ $$\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}$$ $$\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}$$


$$K_c = \frac{\ce{[H2O+-OH]}}{\ce{[H+]}\ce{[H2O2]}} \implies \ce{[H2O+-OH]} = K_c\ce{[H+]}\ce{[H2O2]}\tag{b}.$$

Applying the method of stationary concentration gives

$$v\left(\ce{HOBr}\right) = 0 \implies k_2\ce{[H2O+-OH]}\ce{[Br-]} = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]}\tag{c}.$$

The overall rate of the reaction is characterised by the rate of formation of bromine $\ce{Br2}$. So,

$$v_r = v\left(\ce{Br2}\right) = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]} \overset{(c)}{=} k_2\ce{[H2O+-OH]}\ce{[Br-]} \overset{(b)}{=} k_2K_c\ce{[H+][H2O2]}\ce{[Br-]}$$

or more briefly using $k_2K_c = k$

$$v_r = k\ce{[H2O2][H+]}\ce{[Br-]}\tag{d}.$$

I repeat: this only shows that it might be a valid mechanism, not that the reaction actually follows such a pathway. Therefore, while we can use collision theory to derive the law of mass action as a first approximation, the exponents for non-elementary reactions are determined via experiment.


Lets start off with the Wikipedia definition of the law of mass action:

"In chemistry, the law of mass action is the proposition that the rate of a chemical reaction is directly proportional to the product of the activities or concentrations of the reactants. It explains and predicts behaviors of solutions in dynamic equilibrium. Specifically, it implies that for a chemical reaction mixture that is in equilibrium, the ratio between the concentration of reactants and products is constant.
When two reactants, A and B, react together at a given temperature in a "substitution reaction," the affinity, or chemical force between them, is proportional to the active masses, [A] and [B], each raised to a particular power"

The "affinity", also describes the rate at which the reaction will happen. From this statement and the above definition we can also get the rate law for an irreversible chemical reaction as:

$$\ce{A + B -> C}$$

$$\ce{rate = k [A]^a[B]^b}$$

Here k is the rate constant for the particular reaction. For the simple reaction given above, the exponents are just one.

Collision theory ties into this in a very basic manner. Considering the simple reaction above, the frequency of collisions should be directly proportional to the concentrations of each reactant (as stated in the definition of the law of mass action above), thus the exponents should be one for the above reaction. Now consider this reaction:

$$\ce{2A -> B}$$
which we can write as:
$$\ce{A + A -> B}$$

Now the reaction rate (which is proportional to the "affinity" of the reactants for each other) is given by:

$$\ce{rate = k [A]^1[A]^1}$$ or
$$\ce{rate = k [A]^2}$$

I hope this description sufficiently ties together the law of mass action with collision theory, I realize in a fairly simplistic manner, in a way that also explains why the reaction coefficients represent the exponentials in the law of mass action. Please let me know of anything requiring clarification.

  • $\begingroup$ But why should Kc be proportional to the active masses? $\endgroup$ Feb 19, 2017 at 23:00
  • 1
    $\begingroup$ The collisions of the reactants is proportional to their concentration. A reaction cannot take place without a collision. Though not every collision of A and B reacts to make C, some fraction of them will. If you double the concentration of A, you will double the number of A-B collisions, increasing the collisions and thus the products by a factor of 2. Did I understand your question correctly? $\endgroup$
    – airhuff
    Feb 19, 2017 at 23:12
  • $\begingroup$ Great! Don't hesitate to ask if you have any other questions. $\endgroup$
    – airhuff
    Feb 19, 2017 at 23:35

I don't feel this is a full answer, but it may allow you to clarify your question.

The reactant molecules must collide in order for a reaction to take place. Therefore we must know the collision frequency in order to determine the reaction rate. This rate must be the product of some probability and the collision frequency.

Collision Theory gives a way to calculate this collision frequency. In doing so Maxwell-Boltzmann distributions are used to calculate the ratio of atoms that can react within a given concentration. Most of this 'stuff' gets swept up in the reaction rate constant. After multiplying all these probability ratios together you are left with the product of the concentrations of each reactant.

Since every reagent must be treated in this way, identical components end up with exponents. That is, stoichiometry goes to the exponent.

If you are still confused, or not confused enough: this approach assumes that this reacting system is deterministic and continuous. While many systems can be modeled by treating them this way, they are not deterministic or continuous. They are discreet, stochastic events.

  • $\begingroup$ But whether 2 moles of product is produced or any number of mole why should they be considered different? $\endgroup$ Feb 16, 2017 at 11:38
  • $\begingroup$ Need to think in a probabilistic way? $\endgroup$ Feb 18, 2017 at 1:18

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