physical chemistry - Law of mass action & Collision Theory

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?

Answer

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

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

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

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

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 $(7)$

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.

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

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

Indeed,

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

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.