Relation between chemical kinetics and chemical equilibrium In my chemistry book, the law of chemical equilibrium is derived from the law of mass action:

For a reversible chemical reaction $$\ce{aA +bB\rightleftharpoons cC + dD}$$ where $a$, $b$, $c$ and $d$ are stoichiometric coefficients in the balanced equation, we have, from the law of mass action, the rate of the forward reaction $$r_\mathrm{f} = k_1 \ce{[A]^a[B]^b}\label{1}\tag{1}$$ where $[\ce{A}]$ represents the concentration of $\ce{A}$ and so on. The rate of the backward reaction is $$r_\mathrm{b} = k_2\ce{[C]^c[D]^d} \label{2}\tag{2}$$
At equilibrium, $r_\mathrm{f} = r_\mathrm{b}$, so, we have $$\frac{k_1}{k_2} = \frac{\ce{[C]^c[D]^d}}{\ce{[A]^a[B]^b}}$$ Replacing $\displaystyle \frac{k_1}{k_2}$ by $K_\mathrm{c}$, we have
$$K_\mathrm{c} = \ce{\frac{[C]^c[D]^d}{[A]^a[B]^b}}\tag{3}\label{3}$$
which is the law of chemical equlibrium. $K_\mathrm{c}$ is supposed to be constant, for any concentration of the reactants, at a particular temperature.

However, in chemical kinetics, they tell us that the rate equation, derived from the law of mass action is not always correct. That is, the exponents of the concentration terms in the rate equations may not be equal to their stoichiometric coefficients. If this is true then, the above two rate equations ($\eqref{1}$ and $\eqref{2}$) may not be correct for a particular reversible reaction.
Then, how can we apply the law of chemical equilibrium on every reaction, without experimentally verifying the rate equations? If the rate equations derived from the law of mass action are wrong, then the value of $K_\mathrm{c}$ obtained by the final equation will not be constant.
The derivation that you cite only holds true if both forward and backward reactions are elementary steps. In this case, the rate of the forward reaction is indeed given by

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

and so on, and hence the derivation is logically sound. If they are not elementary steps, then even though the conclusion is correct, the method of arriving there is not, for the reasons you have already pointed out. See also: hBy2Py's answer and Ben Norris's answer where the assumption of an elementary step is explicitly mentioned.



In general the equilibrium constant does not need to be derived in this manner; a more conventional method of arriving at it is by using (pressure here is taken to equal an arbitrary but specified standard pressure $p^\circ$)

$$\mu_i = \mu_i^\circ + RT\ln a_i$$

where $\mu_i$ is the chemical potential of species $i$, $\mu_i^\circ$ is the standard chemical potential of species $i$, and $a_i$ is the chemical activity of species $i$. We now have, for a reaction

$$\ce{a A + b B -&gt; c C + d D},$$

the following equation:

$$\begin{align}
\Delta_\mathrm r G^\circ &amp;= c\mu_\ce{C}^\circ + d\mu_\ce{D}^\circ - a\mu_\ce{A}^\circ - b\mu_\ce{B}^\circ \\
&amp;= c(\mu_\ce{C} - RT\ln a_\ce{C}) + d(\mu_\ce{D} - RT\ln a_\ce{D}) - a\mu_\ce{A} - RT\ln a_\ce{A} - b(\mu_\ce{B} - RT\ln a_\ce{B}) \\
&amp;= c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B} - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \\
&amp;= \Delta_\mathrm r G - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]
\end{align}$$

However, at equilibrium, $\Delta_\mathrm r G = 0$ and hence

$$\Delta_\mathrm r G^\circ = - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]$$

The term in square brackets can then be identified as the equilibrium constant, $K$. As you can see there is no need for any recourse to kinetic rate laws.

Why is $K$ a constant? Well, the standard chemical potentials $\mu_i^\circ$ are thermodynamic constants, just like standard enthalpies of formation, standard entropies, etc. So $\Delta_\mathrm r G^\circ$ - a sum of constants - is itself a constant, and it follows that $K = \exp(-\Delta_\mathrm r G^\circ/RT)$ must be a constant.