physical chemistry - Why is the rate of a reaction proportional to the concentrations of reactants raised to their stoichiometric coefficients?

Consider a gaseous state elementary reaction $$\ce{aA(g) + bB(g)} \overset{k_\mathrm{f}}{\underset{k_{\mathrm{b}}}{\ce{<=>}}}\ce{ cC(g) + dD(g)}$$

I know that for this reaction, $$\Delta G = \Delta G^{\circ} + RT \ln \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) \tag1$$

Now, if $$K = \left( \frac{P_{\mathrm{C}}^c P_{\mathrm{D}}^d}{P_{\mathrm{A}}^a P_{\mathrm{B}}^b} \right) ,$$

then $\Delta G = \Delta G^{\circ} + RT \ln K$.

Now, how can we say that $$K = \frac{k_{\mathrm{b}}}{k_{\mathrm{f}}}$$

and $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{f}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$ $\,$where $r_{\mathrm{f}}$ and $r_{\mathrm{b}}$ are rates of forward and backward reaction as assumed in this answer.

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

Answer

I want that to know that how can one prove rigorously that $r_\mathrm{f} = k_{\mathrm{f}} \cdot P_{\mathrm{A}}^a \cdot P_{\mathrm{B}}^b \,$ and $\, r_\mathrm{b} = k_{\mathrm{b}} \cdot P_{\mathrm{C}}^c \cdot P_{\mathrm{D}}^d$.

I might try to give you some intuition to back up that, for a given (elementary) reaction $\ce{A ->[k] B}$, the reaction rate $r$ can be written as

$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} \propto P_\ce{A}\text{.}$$

(Observe that the second equality above is true due to $P_\ce{A} + P_\ce{B} = \text{constant}$.) First, for an ideal gas, $P_\ce{A} = \frac{n_\ce{A}}{V} RT$. This means that $P_\ce{A} \propto n$ for fixed temperature and volume.

Let's say the reaction happens as a random process. That is to say that, for every time interval $\Delta t$, we have a probability per unit time $p$ of having a single molecule $\ce{A}$ turning into $\ce{B}$. If we wait longer, proportionally more molecules will turn. We'll thus have, for initially $n_\ce{A}$ molecules of $\ce{A}$, after $\Delta t$ seconds,

$$\Delta n_\ce{B} = p n_\ce{A} \Delta t\text{.}$$

This means that, in the time interval $\Delta t$, the population of $\ce{B}$ goes from 0 to $\Delta n_\ce{B}$ (assuming no $\ce{B}$ initially). From the stoichiometry of the reaction, $\Delta n_\ce{B} = -\Delta n_\ce{A}$ (i.e., there's conservation of moles).

Thus,

$$\frac{\Delta n_\ce{B}}{\Delta t} = -\frac{\Delta n_\ce{A}}{\Delta t} = p n_\ce{A}\text{.}$$

After taking the limit of $\Delta t \rightarrow 0$ and multiplying by $\frac{1}{V}$ (i.e., dividing by volume), we get

$$\frac{d \frac{n_\ce{B}}{V}}{dt} = \frac{d [\ce{B}]}{dt} = p \frac{n_\ce{A}}{V} = p [\ce{A}]\text{,}$$

where $[\ce{A}]$ and $[\ce{B}]$ stand for the concentrations of $\ce{A}$ and $\ce{B}$, respectively. Since, again for an ideal gas, $[\ce{A}] = \frac{P_\ce{A}}{RT}$, we get

$$r = \frac{d P_\ce{B}}{dt} = -\frac{d P_\ce{A}}{dt} = p P_\ce{A}\text{,}$$

where the proportionality constant is the probability of reaction per unit time.