Overall effective activation energy for parallel reactions

Question

Today, I was solving a question from Chemical Kinetics.

Consider the following parallel, first order reactions:

$$\begin{align} \ce{A -> P} \tag{1} \\ \ce{A -> Q} \tag{2} \end{align}$$

with rate constants $k_1$ and $k_2$, and activation energies $E_1$ and $E_2$ respectively.

I was asked to calculate the overall activation energy $(E)$ for the decay of A.

In my book a formula for $E$ is given to be :

$$E = \frac{E_1k_1+E_2k_2}{k_1+k_2}$$

This formula works well and provides correct answer, but I was trying to derive this expression using the Arrhenius equation. But I didn't succeed. I could deduce that,

Since:

$$ k_1=A_1 \mathrm e^{-E_1/RT}~; ~ k_2=A_2 \mathrm e^{-E_2/RT}$$

and

$$k_{\text{eff.}}= k_1+k_2 = A_{\text{eff.}}\mathrm e^{-E/RT}; $$

Using these both:

$$k_1+k_2= A_1 \mathrm e^{-E_1/RT} + A_2 \mathrm e^{-E_2/RT}$$

Now, I have no clue what to do.

Answer 1

The activation energy $E_\mathrm a$ should, in general, be viewed as an empirical quantity that characterises the sensitivity of the rate to temperature. From this perspective, it may be obtained via the equation

$$E_\mathrm a = RT^2 \left(\frac{\mathrm d \ln k}{\mathrm dT}\right) \tag{1}$$

This relation can be "derived" from the Arrhenius equation. (Quotation marks because the Arrhenius equation itself is only an empirical relation.)

$$\begin{align} k &= A\exp\left(-\frac{E_\mathrm a}{RT}\right) \\ \ln k &= \ln A - \frac{E_\mathrm a}{RT} \\ \frac{\mathrm d \ln k}{\mathrm dT} &= \frac{E_\mathrm a}{RT^2} \end{align}$$

If a reaction rate has Arrhenius-type behaviour, i.e. $k = A\exp(-E_\mathrm a/RT)$, then equation $(1)$ simply returns the usual interpretation of the activation energy. However, equation $(1)$ is useful in that it allows us to define an "activation energy" in the case where the rate constant does not have the mathematical form $A\exp(-E/RT)$. The parallel reactions that you have described fall exactly into this category. The sum of two exponential functions generally does not return an exponential function, so the overall rate constant $k = k_1 + k_2$ should not be expected to conform to an Arrhenius description.

If we substitute

$$k = k_1 + k_2 = A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)$$

into equation $(1)$, then we get:

$$\begin{align} E_\mathrm a &= RT^2 \cdot \frac{\mathrm d}{\mathrm dT}\left\{ \ln\left[A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)\right] \right\} \\[8pt] &= RT^2 \cdot \frac{A_1(E_1/RT^2)\exp(-E_1/RT) + A_2(E_2/RT^2)\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt] &= \frac{A_1E_1\exp(-E_1/RT) + A_2E_2\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt] &= \frac{E_1k_1 + E_2k_2}{k_1 + k_2} \end{align}$$

as desired.

Answer 2

The effective activation energy you quote is the weighted sum of the two individual ones to P and Q. The fraction of molecules going to P is $k_1/(k_1+k_2)$ and so $k_2/(k_1+k_2)$ go to Q. Thus the effective activation energy is the sum of these two fractional values, i.e. $$E_{eff}=\frac{k_1}{k_1+k_2}E_1 + \frac{k_2}{k_1+k_2}E_2 $$ which is your equation.

(Note that the rate constant for the decay of A and rise-time of P and Q is the same and is $k_1+k_2$; P and Q are only kinetically distinguished by the fraction of A that ends up as P and Q not by how quickly they are produced).