Cascading chemical reaction

A -> 2B -> 6C -> 24D -> 24E. So guessed 1/24. :)

Since additional $A$ is added, we do not need to worry about the reaction rate equation for $A$ . Instead we consider the steady state equations for the intermediate products: $\begin{aligned} 0 \; = \; \tfrac{d}{dt}[B] & = \; 2[A] - 2[B]^2 - [B] + 3[C]^3 \\ 0 \; = \; \tfrac{d}{dt}[C] & = \; [B] - 3[C]^3 - [C] + 4[D]^4 \\ 0 \; = \; \tfrac{d}{dt}[D] & = \; [C] - 4[D]^4 - [D] \end{aligned}$ and, of course, $\tfrac{d}{dt}[E] = [D]$ . If $[D] \sim [A]^\alpha$ as $[A] \to \infty$ , then $[C] \sim [D]^4 \sim [A]^{4\alpha}$ , $[B] \sim [C]^3 \sim [A]^{12\alpha}$ and hence $[A] \sim [B]^2 \sim [A]^{24\alpha}$ as $[A] \to \infty$ , and hence $\alpha= \boxed{\tfrac{1}{24}}$ .

Increasing the concentration of substance A means that at some point the reversible reactions will reach equilibrium, so the rate of the forward reaction can be equated to the reverse reaction. Constant supply of the substance A is required as the last conversion is irreversible. As all the reaction rate constants are equal, the equations to solve are A=B^3, B=C^3, C=D^4 and E=D. This results in the overall reaction of A^(1/24)=E.

In a chemical equilibrium reaction ( $aA + bB \rightleftharpoons cC + dD$ ), $\frac{[C]^c[D]^d}{[A]^a[B]^b} = K_{eq} = \frac{k_\rightarrow}{k_\leftarrow}$ , for some $k$ 's (which chemists call the rates of the reaction).

Using this fact, we get a set of equations: $[A] = K_1 [B]^2$ , $[B] = K_2 [C]^3$ and $[C] = K_3 [D]^4$ . Hence:

$[A] = K_1 [B]^2 = K_1 {K_2}^2 [C]^6 = K_1 {K_2}^2 {K_2}^6 [D]^{24} \propto [D]^{24}$

Assuming the production rate of $E$ is proportional to $[D]$ (true for not too big concentrations of $[D]$ ), the production rate of $E$ is simply: $\frac{d}{dt}[E] \propto D \propto \boxed{[A]^{1/24}}$ .