Solving Equations for Consecutive Reactions (Not first order)

Question

$$\frac{dA}{dt} = -k_1 A^{1.5} ,\quad A(0)=A_0;$$ $$\frac{dB}{dt} = k_1 A^{1.5} - k_2 B^{1.5} ;$$

Can I get an analytical expression when the system is not first-order? Either for $1.5^{th}$ order or for a more generic $n^{th}$ order?

Answer 1

You can write $$\frac{dA}{A^{3/2}}=-k_1dt$$ and integrate. An we get by integrating $$\frac{1}{\sqrt{A(t)}}-\frac{1}{2}k_1t-C=0$$ Solve this equation for $$A(t)$$ and plug this into the second equation. It is $$A(t)=\frac{1}{(\frac{1}{2}k_1t+C)^2}$$ and also Maple can find an answer.

Answer 2

This actually still is a first-order DE. The order of a DE has absolutely nothing to do with anything happening to the functions, and everything to do with the orders of the derivatives that appear in the DE's. In your case, the first DE is separable: \begin{align*} A^{-1.5}\,dA&=-k_1\,dt \\ \frac{A^{-0.5}}{-0.5}&=-k_1t+C \\ \frac{1}{\sqrt{A}}&=0.5k_1t+C,\quad\text{re-writing }C\\ A&=\frac{1}{(0.5k_1t+C)^2}. \end{align*} The second equation is not separable. You'd get \begin{align*} \frac{dB}{dt} &= k_1 A^{1.5} - k_2 B^{1.5} \\ \frac{dB}{dt} &= \frac{k_1}{(0.5k_1t+C)^3} - k_2 B^{1.5}. \end{align*} This is a much more difficult DE; Mathematica's answer is: $$\text{Solve}\left[-2 \text{RootSum}\left[-\text{$\#$1}^3+\text{$\#$1}^2 \left(-\frac{i k_1}{k_2}\right)^{2/3}-1\&,\frac{\ln \left(\frac{1}{2} \sqrt{B(t) \left(\frac{-8 C^3 k_2-12 C^2 k_1 k_2 t-6 C k_1^2 k_2 t^2-k_1^3 k_2 t^3}{k_1}\right)^{2/3}}-\text{$\#$1}\right)}{-3 \text{$\#$1}+2 \left(-\frac{i k_1}{k_2}\right)^{2/3}}\&\right]=C_2-\frac{2 k_2 (2 C+k_1 t) \ln (2 C+k_1 t)}{k_1 \sqrt[3]{-\frac{k_2 (2 C+k_1 t)^3}{k_1}}},B(t)\right] $$ As you can see, $B(t)$ appears in this expression, and so Mathematica is not able to give an explicit answer. You're probably better off doing the best you can numerically.

Answer 3

I know this is math.stackexchange, but seeing your username, I thought I'd add a more 'chemical engineering' viewpoint to this. Approximations are essential in engineering, but they also require some mathematical insight to justify what we're doing.

As the other answers have shown, an explicit solution for $B(t)$ does probably not exist. Also, worrying about nth order reactions is somewhat nonsensical, as there are virtually no reactions of order $\geq3$. The reaction network you're interested is

$$A \rightarrow B \rightarrow \mathrm{Product}$$

with rate constants $k_1$ and $k_2$. We can't usually get an explicit solution for anything other than first and second order consecutive reactions (don't take my word on this, but it's mostly a good guess), so theres an important set of simplifications you need to be able to apply.

Say $A$ is present in large excess ($[A]>>[B]$). If there is so much A, the concentration of A will not markedly decrease over time. It is then fair to assume that $$\frac{\mathrm{d}A}{\mathrm{d}t} \approx 0 \rightarrow [A] \approx A_0$$

You did not specify that it is an equilibrium reaction, so we'd have to be careful with that approximation.

The other, more important one is the one of the rate determining step. If one rate constant is much smaller than the other, this reaction will be the bottleneck of the network, and the overall kinetics will be dominated by the bottleneck.

If $k_1>>k_2$, the bottleneck is the reaction $B \rightarrow C$; the concentration of $B$ will then mainly be governed by the formation of $B$ via $A \rightarrow B$, so

$$r = \frac{\mathrm{d}B}{\mathrm{d}t} \approx k_1 [A]^{1.5}$$ and $$ \frac{\mathrm{d}[Product]}{\mathrm{d}t} \approx k_2 [B]^{1.5} = k_2 A(t)^{1.5}$$ which has an explicit solution.

If the first reaction $A \rightarrow B$ is the bottleneck, so $k_1<<k_2$, $B$ is formed very slowly and reacts on quickly, so the formation of product is governed by the progress of the first reaction

$$\frac{\mathrm{d}[Product]}{\mathrm{d}t} \approx \frac{\mathrm{d}A}{\mathrm{d}t}$$

which again simplifies the equations a lot.

I'd like to say sorry to every pure mathematician that felt uncomfortable reading this.