4
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – electronpusher Dec 16 '16 at 3:03
7
$\begingroup$

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!

$\endgroup$
5
$\begingroup$

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))]]) 
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.