My first language is not English, so there might be some parts where it's hard to read. Sorry in advance.
I have some questions regarding the derivation of the equilibrium constant and the rate-determining step.
For a hypothetical reaction $$\ce{aA + bB <=> cC + dD}$$ the equilibrium constant ${K}$ would be $${K=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}$$
However, my textbook states that equilibrium is when the forward reaction rate ${R_{f}=k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$ and the reverse reaction rate ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$ becomes the same, so that $${R_{f}=R_{r}}$$ $${k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$$ $${\frac{k_{f}}{k_{r}}=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}=K}$$ which is a bit unsound because there is no guarantee that $$\ce{aA + bB -> cC + dD}$$ and $$\ce{cC + dD -> aA + bB}$$ are both elementary reactions, so ${R_{f}}$ and ${R_{r}}$ might not equal ${k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$ and ${k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$.
So I started searching, and found this:
Kinetic Derivation of the Equilibrium Constant
The usual method of finding the rate equation of a multistep reaction is that it equals the rate equation of the rate-determining step with the caveat that if it contains any reactant which is a product in a previous step, one must substitute the rate factor (e.g., ${\left[\ce{A}\right]^{a}}$, ${\left[\ce{B}\right]^{b}}$) of this reactant with the rate factors of the reactants of this prior step. The idea of this proof is to write this method in a form more amenable to algebraic manipulation, and then find use it to determine the rate equation for the forward reaction, ${R_{f}}$, and then for the reverse reaction, ${R_{r}}$. Once this has done, we can equate ${R_{f}}$ and ${R_{r}}$ and hence derive the equilibrium constant expression.
Consider a general multi-step reaction, but instead of written the usual way, written so that all products appear on the left-hand side of the equation with negative coefficients:
$${x_{1}\ce{X}_{1}+x_{2}\ce{X}_{2}+x_{3}\ce{X}_{3}+⋯+x_{n}\ce{X}_{n}\ce{<=>}0}$$
Imagine that there are ${s}$ steps making up this multi-step reaction, and each elementary step is written the same way as above. Then, for an elementary step ${j}$, define ${x_{ij}}$ so that it is the coefficient of ${X_{i}}$ in the ${j}$th step. Since, as stated above, the coefficients of the elementary steps must add up to the overall coefficients,
$${\displaystyle\sum_{j=1}^{s}x_{ij}=x_{i}}\tag{1}$$
Now multiply together the terms of each elementary step, to the power of their corresponding coefficients, up to but excluding the rate-determining step (step ${r}$).
$${\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}$$
The overall forward rate is proportional to this quantity multiplied by ${A}$, which is the reactants of the rate-determining step with their corresponding coefficients as exponents.
$${R_{f}=k_{f}A\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}$$
Now assume that we do the same as above, except in reverse. That is, we use the negative of each coefficient, and we start from elementary step ${s}$ (that is, the last step) and go until (but exclude) the rate-determining step:
$${\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}$$
The overall reverse rate is proportional to this quantity divided by ${B}$, which is the concentrations of the products of the rate-determining step with their corresponding coefficients as exponents.
$${R_{r}=k_{r}\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{B}}$$
Since ${R_{f}=R_{r}}$ in an equilibrium,
$${k_{f}A\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}=k_{r}\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{B}}$$
Rearranging this to get ${K}$,
$${\frac{k_{f}}{k_{r}}=\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{AB\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}}$$
Because there is a negative sign in the exponent in the numerator, we can move the numerator to the denominator if we switch the negative sign to a positive sign.
$${\frac{k_{f}}{k_{r}}=\frac{1}{\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}\right)\left(AB\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}x_{ij}}\right)}}$$
Recalling that ${A}$ is the product of the concentrations of the reactants of the rate-determining step with their coefficients as exponents, and that ${B}$ is the same for the products of the rate-determining step, we can write ${AB}$ as
$${AB=\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{ir}}}$$
Thus,
$${\frac{k_{f}}{k_{r}}=\frac{1}{\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{ir}}\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}x_{ij}}\right)}}$$
But then this can be written simply as,
$${\frac{k_{f}}{k_{r}}=\frac{1}{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{s}x_{ij}}}}$$
Recalling Equation 1, this is simply
$${\frac{k_{f}}{k_{r}}=\frac{1}{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{i}}}}$$
$${\frac{k_{f}}{k_{r}}=\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{-x_{i}}=K}$$
I did understand this, but new questions popped up.
First of all, this derivation of the equilibrium constant assumes that the rate-determining step for the forward reaction and reverse reaction are always identical (the ${r}$th step). Is this always true?
For example, in the reaction shown below, the 2nd step is the rate-determining step for both forward reaction and reverse reaction.
However, in this reaction below, the 2nd step is the rate-determining step for the forward reaction, while the 1st step is the rate-determining step for the reverse reaction.
I made the second graph myself, and I do not know if such a reaction is thermodynamically possible. But if reactions like this exist, then the derivation process written above would be flawed.
And here's my second question. Let's bring back the reaction $$\ce{aA + bB <=> cC + dD}$$ and suppose that this reaction is a 2-step reaction: $$\ce{aA <=> cC + eE}\tag{Step 1}$$ $$\ce{bB + eE <=> dD}\tag{Step 2}$$
If the 1st step is the rate-determining step, then the forward reaction rate would be ${R_{f}=k_{f}\left[\ce{A}\right]^{a}}$ and the reverse reaction rate would be... uh... ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$? I tried to derive ${K}$ from ${R_{f}}$ and ${R_{r}}$, but I got stuck in the middle:
The equilibrium constant of Step 2 of the reverse reaction is
$${K_{r2}=\frac{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}{\left[\ce{D}\right]^{d}}}$$
so
$${\left[\ce{E}\right]^{e}=\frac{K_{r2}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$
Equilibrium is when the forward reaction rate ${R_{f}=k_{f}\left[\ce{A}\right]^{a}}$ and the reverse reaction rate ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$ becomes the same, so
$${R_{f}=R_{r}}$$
$${k_{f}\left[\ce{A}\right]^{a}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}=k_{r}\left[\ce{C}\right]^{c}\frac{K_{r2}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$
Thus
$${K=\frac{k_{f}}{k_{r}}=K_{r2}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}\tag{???????}$$
What mistake did I make here? I have been trying to solve this for a few days but I can't seem to obtain a clear answer.
(Sorry for such a long question, and thanks in advance.)
EDIT
So about the second question... Thanks to @porphyrin, I've found my mistake. I shouldn't have assumed that the reverse reaction rate is ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$. It should be
$${R_{r}=k_{r}\left(\frac{\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}×\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}\right)=k_{r}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$
Knowing this and solving again, I was able to get
$${K=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}$$
EDIT 2
Thanks to @thewitness, I think I got the answer for the first question. So apparently, it is possible for the rate-determining step of the reverse reaction to be different from that of the forward reaction.
Though I got the explanations I needed, feel free to add an answer and give me more insight into these matters. Thanks.