# Law of Mass Action

- Page ID
- 83031

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} } \label {1}.\]

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

\[\ce{R1(g) + R2(g) <=> P1(g) + P2(g)} \label {1'}.\]

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.

## Energies

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

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

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

\[f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\label{3}\]

which evaluates to

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

## Number of *interesting* collisions

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

\[Z^* = Z\cdot f. \label {4}\]

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

\[ Z = 2N^2 \left( \sigma_{\ce{R1}} \right)^2 \sqrt{\dfrac{\pi RT}{M_{\ce{R1}}}} \label{5}\]

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

\[v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\right). \label{6}\]

## Reaction rate \(v_r\)

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

\[v_r = \underbrace{4 \times 10^{-3} N_A \left(\sigma_{\ce{R1}} \right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right) }_k \cdot \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}. \label{7}\]

Coming back to our actual reaction (Equation \(\ref{1'}\)) 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 Equation \(\ref{7}\).

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

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

\[v_r = k[A][B]\label{8}\]

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

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

Mathematically, we can verify that *one possible* mechanism is

\[\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium} \]

\[\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}\]

\[\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}\]

Indeed,

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

Applying the method of stationary concentration gives

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

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 \(k_2K_c = k\)

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

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.

#### Contributors

- Linear Christmas (StackExchange)