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Given the following three consecutive unimolecular reactions $$\ce{A ->[$k_1$] I_1 ->[$k_2$] I_2 ->[$k_3$] P}$$ write the rate of change for each.

I have gotten the following:$\newcommand{\diff}{\mathrm{d}}$ \begin{align} \frac{\diff [A]}{\diff t} &=-k_1[A]\\ \frac{\diff [I_1]}{\diff t} &=k_1[A]-k_2[I_1]\\ \frac{\diff [I_2]}{\diff t} &=-k_3[I_2]+k_2[I_1]\\ \frac{\diff [P]}{\diff t} &=k_3[I_2]\\ \end{align} Is this correct? If it is, then why cant we write \begin{align} \frac{\diff [A]}{\diff t} &=-k_1[A]\\ \frac{\diff [I_1]}{\diff t} &=k_1[A]-k_2[I_1]\\ \frac{\diff [I_2]}{\diff t} &=k_3[I_2]+k_2[I_1]+k_1[A]\\ \frac{\diff [P]}{\diff t} &=k_3[I_2]\\ \end{align}

andselisk
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MathCurious314
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    Your first set of rate laws looks correct to me. The only difference in the second set is the third line, the rate of I_2? If the first I_2 rate law is true, then why would you think the second I_2 rate law could be correct? I can't see where you're getting it. – electronpusher Dec 16 '16 at 03:03

2 Answers2

10

Your first set of four equations look correct. Next you need to integrate the first equation to get [A] vs. time i.e. $\ce{[A]}= \ce{A_0}\exp(-k_1t)$ where $\ce{[A_0]}$ is the value at $t=0$, and substitute [A] into the second equation. Integrate this equation (integrating factor method) and substitute it into the third and so on. It gets messy! You can obtain $\ce{[P]}$ easily as this is the sum of all other values less the amount you start with.

The general solution for a chain of n reactions in the special case that only the first species A is present at $t=0$ has the form

$$N_n = C_1\exp(-k_1t)~+ C_2\exp(-k_2t) ~+ ...+ C_n\exp(-k_nt) $$

$$C_1= \frac{k_1k_2...k_{n-1}}{(k_2-k_1)(k_3-k_1)...(k_n-k_1)}\ce{[A_0]} $$

$$C_2= \frac{k_1k_2...k_{n-1}}{(k_1-k_2)(k_3-k_2)...(k_n-k_2)}\ce{[A_0]}$$

$$ C_3 = .....$$ etc, for as many terms as necessary.

good luck!

porphyrin
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7

We have the following cascade of unimolecular chemical reactions

$$\ce{X_1} \xrightarrow{k_1} \ce{X}_2 \xrightarrow{k_2} \ce{X}_3 \xrightarrow{k_3} \ce{X}_4$$

where $k_1, k_2, k_3 > 0$ are the (distinct) rate constants. Assuming that every chemical reaction in this cascade has mass action kinetics, then we have the following system of linear ODEs

$$\begin{bmatrix} \dot x_1\\ \dot x_2\\ \dot x_3\\ \dot x_4\end{bmatrix} = \begin{bmatrix} - k_1 & 0 & 0 & 0\\ k_1 & - k_2 & 0 & 0\\ 0 & k_2 & - k_3 & 0\\ 0 & 0 & k_3 & 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}$$

where $x_i := [\ce{X}_i]$ is the (time-varying) concentration of the $i$-th species. Let the initial concentrations be $x_1 (0) =: x_{10} \neq 0$ and $x_2 (0) = x_3 (0) = x_4 (0) = 0$. Integrating the linear ODEs, we obtain

$$\begin{array}{rl} x_1 (t) &= x_{10} \, e^{- k_{1} t}\\ x_2 (t) &= x_{10} \left(\frac{k_{1}}{k_{2} - k_{1}}\right) e^{- k_{1} t} + x_{10} \left(\frac{k_{1}}{k_{1} - k_{2}}\right) e^{- k_{2} t}\\ x_3 (t) &= x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{2} - k_{1}\right) \left(k_{3} - k_{1}\right)}\right) e^{- k_{1} t} + x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{1} - k_{2}\right) \left(k_{3} - k_{2}\right)}\right) e^{- k_{2} t} + x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{1} - k_{3}\right) \left(k_{2} - k_{3}\right)}\right) e^{- k_{3} t}\\ x_4 (t) &= x_{10} \left( \frac{k_{2} k_{3}}{ \left(k_{2} - k_{1}\right) \left(k_{3} - k_{1}\right)}\right) \left( 1 - e^{- k_{1} t} \right) + x_{10} \left( \frac{k_{1} k_{3}}{ \left(k_{1} - k_{2}\right) \left(k_{3} - k_{2}\right)}\right) \left( 1 - e^{- k_{2} t} \right) + x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{1} - k_{3}\right) \left(k_{2} - k_{3}\right)}\right) \left( 1 - e^{- k_{3} t} \right)\end{array}$$

Python code

The following SymPy script

from sympy import *

t = Symbol('t')
x10 = Symbol('x10')
k1, k2, k3 = symbols('k1 k2 k3')

A = Matrix([[-k1,  0,  0, 0],
            [ k1,-k2,  0, 0],
            [  0, k2,-k3, 0],
            [  0,  0, k3, 0]])

V, D = A.diagonalize()

print "V = ", V
print "D = ", D
print "x (t) = ", simplify(V * diag(1, exp(-k1*t),exp(-k2*t),exp(-k3*t)) * V**-1 * Matrix([x10,0,0,0]))

produces the following output

V =  Matrix([[0, -(k1 - k2)*(k1 - k3)/(k2*k3), 0, 0], [0, k1*(k1 - k3)/(k2*k3), (k2 - k3)/k3, 0], [0, -k1/k3, -k2/k3, -1], [1, 1, 1, 1]])
D =  Matrix([[0,  0,  0,  0], 
             [0,-k1,  0,  0], 
             [0,  0,-k2,  0], 
             [0,  0,  0,-k3]])
x (t) =  Matrix([
[                                                                                                                                                                                                                                   x10*exp(-k1*t)],
[                                                                                                                                                                                        k1*x10*exp(-k2*t)/(k1 - k2) - k1*x10*exp(-k1*t)/(k1 - k2)],
[                                                                                          k1*k2*x10*((k1 - k2)*exp(t*(k1 + k2)) - (k1 - k3)*exp(t*(k1 + k3)) + (k2 - k3)*exp(t*(k2 + k3)))*exp(-t*(k1 + k2 + k3))/((k1 - k2)*(k1 - k3)*(k2 - k3))],
[x10*(-k1*k2*(k1 - k2)*exp(t*(k1 + k2)) + k1*k3*(k1 - k3)*exp(t*(k1 + k3)) - k2*k3*(k2 - k3)*exp(t*(k2 + k3)) + (k1*k2*(k1 - k2) - k3*(k1*(k1 - k3) - k2*(k2 - k3)))*exp(t*(k1 + k2 + k3)))*exp(-t*(k1 + k2 + k3))/((k1 - k2)*(k1 - k3)*(k2 - k3))]])
Rodrigo de Azevedo
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