# What is the constant of integration in a series reaction?

Consider the series reaction

$$\ce{A ->[k_1] B ->[k_2] C}$$

Then

$$\frac{\mathrm d}{\mathrm d t} [\ce{B}] = k_1 [\ce{A}] - k_2 [\ce{B}]$$

$$[\ce{A}]=[\ce{A}]_{0} \exp(-k_1 t)$$

Thus far, I have used the method integrating factors, with $u=\exp(k_2t)$, to arrive at the equation:

$$[\ce{B}]= \frac{k_1}{k_2-k_1} \left( \exp(-k_1 t) + C \exp(-k_2 t) \right) [\ce{A}]_{0}$$

What would the constant of integration be in this situation?

• Did you take in mind that you use a definite integral? You should get a dependence of [B] on [B]o. – EJC Sep 24 '14 at 19:10
• As @Marko already stated, the problem seems to be similar to this one. It might be resolved by using definite integrals which should eliminate the unknown constant from you equation. – Philipp Sep 24 '14 at 21:04

There might or there might not be a problem with your final equation. Either you have made a mistake while using the integrating factor to solve your differential equation or you made some further substitutions so that your constant $C$ contains something more than just the "standard stuff" I would have expected from the method. Thus it's best I go through the derivation from the beginning so that it'll become clear what I mean.

The method of solving differential equation using an integrating factor is explained well here. You have used the correct integrating factor $u(t)$. For the differential equation at hand, i.e.

\begin{align} \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= k_1[\ce{A}] - k_2[\ce{B}] \\ \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} + k_2[\ce{B}] &= k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \ , \end{align}

it is indeed

\begin{align} u(t) = \exp\biggl( \int_{0}^{t} k_2 \, \mathrm{d} t \biggr) = \mathrm{e}^{k_2 t} \ . \end{align}

With the integrating factor you can get $[\ce{B}]$ according to

\begin{align} u(t) [\ce{B}] &= \int_{0}^{t} u(t) k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \, \mathrm{d} t + \tilde{C} \\ [\ce{B}] \mathrm{e}^{k_2 t} &= \int_{0}^{t} k_1 [\ce{A}]_{0} \mathrm{e}^{(k_2 - k_1) t} \, \mathrm{d} t + \tilde{C} \\ &= k_1 [\ce{A}]_{0} \biggl[\frac{1}{k_2 - k_1} \mathrm{e}^{(k_2 - k_1) t} \biggr]_{0}^{t} + \tilde{C} \\ &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{(k_2 - k_1) t} - 1 \bigr) + \tilde{C} \end{align}

Rearranging for $[\ce{B}]$ gives

\begin{align} [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \underbrace{\frac{ \mathrm{e}^{(k_2 - k_1) t} - 1 }{\mathrm{e}^{k_2 t}}}_{= \, \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t}} + \frac{\tilde{C}}{\mathrm{e}^{k_2 t}} \\ &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) + \tilde{C} \mathrm{e}^{-k_2 t} \ . \end{align}

This equation differs from yours a little if your integration constant $C$ would be the same as my $\tilde{C}$. But if you used some substitutions so that your integration constant is $C = \frac{k_2 - k_1}{k_1 [\ce{A}]_{0}} \tilde{C} - 1$ then your solution and mine would be equivalent. If you indeed made a mistake then I hope my derivation cleared things up a bit and if you had intended the equation to look that way by introducing the substitution then at least you know what integration constant I mean when I am going to explain the method for determining its value in the following.

The integration constant can now be determined by enforcing boundary conditions. In this particular case this means that at the start of your reaction ($t=0$) you require that $[\ce{B}]$ is equal to the initial concentration of your reactant $\ce{B}$, i.e. $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$. So, plugging $t=0$ and $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$ into the equation for $[\ce{B}]$ you get

\begin{align} [\ce{B}]_0 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{0} - \mathrm{e}^{0} \bigr) + \tilde{C} \mathrm{e}^{0} \\ [\ce{B}]_0 &= \tilde{C} \ . \end{align}

In the case of your reaction $\ce{A -> B -> C}$ the initial concentration of $[\ce{B}]$ will usually be $0$ as $\ce{B}$ is only an intermediate product, so $\tilde{C} = [\ce{B}]_0 = 0$ and the final equation is

\begin{align} [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) \ . \end{align}

• If I may be a little bit pedantic, your answer is not quite complete, as it does not cover the case where $k_1 = k_2$, which leads to "resonance", i.e., $t \exp(-k_1 t)$ instead of $\exp(-k_2 t)$. I lack the training in chemistry to tell whether this kind of "resonance" is interesting or not. – Rodrigo de Azevedo Oct 15 '17 at 12:33

Assuming that the cascading chemical reaction $\ce{A} \xrightarrow{\kappa_1} \ce{B} \xrightarrow{\kappa_2} \ce{C}$ has mass action kinetics, we have the following pair of linear ODEs

$$\begin{array}{rl} \dot a &= - \kappa_1 \, a\\ \dot b &= \kappa_1 \, a - \kappa_2 \, b\end{array}$$

where $\kappa_1, \kappa_2 > 0$ are the rate constants, $a := [\ce{A}]$ and $b := [\ce{B}]$. For the time being, we neglect $[\ce{C}]$.

Integrating the first ODE, we obtain

$$a (t) = a_0 \exp( -\kappa_1 t )$$

Hence, the second ODE can be written as follows

$$\dot b + \kappa_2 \, b = \kappa_1 \, a_0 \exp( -\kappa_1 t )$$

Eventually, we arrive at the solutions

$$b (t) = \begin{cases} a_0 \left(\dfrac{\kappa_1}{\kappa_2 - \kappa_1}\right) \exp( -\kappa_1 t ) + \left(b_0 - a_0 \left(\dfrac{\kappa_1}{\kappa_2 - \kappa_1}\right)\right) \exp( -\kappa_2 t ) & \text{if } \kappa_1 \neq \kappa_2\\\\ (b_0 + \kappa_0 \, a_0 \, t) \exp( -\kappa_0 t ) & \text{if } \kappa_1 = \kappa_2 =: \kappa_0\end{cases}$$