# Arrhenius Equation for parallel reaction

1. Jul 8, 2013

### Saitama

1. The problem statement, all variables and given/known data
For the two parallel reactions $A \stackrel{k_1}{\rightarrow} B$ and $A \stackrel{k_2}{\rightarrow} C$, show that the activation energy $E'$ for the disappearance of $A$ is given in terms of activation energies $E_1$ and $E_2$ for the two paths by
$$E'=\frac{k_1E_1+k_2E_2}{k_1+k_2}$$

2. Relevant equations

3. The attempt at a solution
I don't know what should be the way to approach this problem. I can find the concentration of A as a function of time which is
$$A=A_0e^{-(k_1+k_2)t}$$
(I guess I can call $k_1+k_2$ as the equivalent rate constant. Is it correct to say so?)
Applying the Arrhenius equation to both the reactions
$$k_1=A_1e^{-E_1/(RT)}$$
$$k_2=A_2e^{-E_2/(RT)}$$
I am clueless about what to do.

Any help is appreciated. Thanks!

2. Jul 8, 2013

### Staff: Mentor

Try the following. I haven't tried it myself, so I don't know whether it will work.
Rewrite the two rate constants as

$$k_1=k_{10}e^{-E_{1}/(RT)}$$
$$k_2=k_{20}e^{-E_{2}/(RT)}$$
Substitute these into your equation for A. Also set $$A=A_0e^{-kt}$$

and $$k=k_0e^{-\frac{E'}{RT}}$$
Then take the natural logs of both sides, and cancel out identical terms. I hope that this works for you.

Chet

Last edited: Jul 8, 2013
3. Jul 8, 2013

### epenguin

It's looking a bit odd to me. Although as usual it's late night.
My first and then second reaction is to think there is no way the sum of two exponentials can be identical to a single exponential so that the overall A disappearance can be represented by an apparent activeation energy E' unless there is something special about the parameters, so have you told us everything that was in the problem asked?

Third (after-) thought: but if we don't look at temperature dependence laws as a whole, but just allow the infinitesimal local temperature-dependence to define an apparent activation energy, i.e. use

d ln (k1 + k2)/dT = E'/RT2

to define this apparent activation energy E', probably that is what they're after. That is a temperature-dependent apparent activation energy

Last edited: Jul 8, 2013
4. Jul 8, 2013

### Staff: Mentor

Yes. This is a correct assessment. There is no exact solution to this problem. As epenguin suggests, one needs to expand in a Taylor series about some specific temperature (and retain the linear term) to get the result they are asking for. The dead giveaway is that E' (as the problem determines it) is a function of temperature.

5. Jul 9, 2013

### Saitama

Taylor series? I solved this problem and I did not use that anywhere. The equation epenguin wrote gives the answer. Here are the steps:
$$\frac{d\ln k_1}{dT}=\frac{E_1}{RT^2} \Rightarrow \frac{d k_1}{dT}=\frac{k_1E_1}{RT^2}$$
Similarly
$$\frac{d k_2}{dT}=\frac{k_2E_2}{RT^2}$$
Also,
$$\frac{d(\ln(k_1+k_2))}{dT}=\frac{E'}{RT^2} \Rightarrow \frac{dk_1+dk_2}{dT}=\frac{(k_1+k_2)E'}{RT^2}$$
It can be easily seen that
$$(k_1+k_2)E'=k_1E_1+k_2E_2$$
Is this wrong?

6. Jul 9, 2013

### epenguin

I think it is what they want.

7. Jul 9, 2013

### Saitama

Thanks for the help epenguin!

8. Jul 9, 2013

### Staff: Mentor

No, it is correct. I was just thinking of the same thing in a more formalized way. Sorry if I caused any confusion.

Chet

9. Jul 9, 2013

### epenguin

? There is no requirement or difficulty of justifying talking about the E' like an honorary constant for which

$$\frac{d\ln (k_{1} + k_{2})}{dT} = \frac{E'}{RT^2}$$

when the RHS should strictly be

$$\frac{1}{RT}(\frac{E'}{T} − \frac{dE′}{dT})$$

so we have to be able to say that last term in the bracket is negligible?

10. Jul 9, 2013

### Saitama

11. Jul 9, 2013

### epenguin

There is however a temperature dependence of your E': it is there, implicit, hidden in your first formula

$$E'=\frac{k_1E_1+k_2E_2}{k_1+k_2}$$

since the k's are temperature dependent (while E1 and E2 are constants).

I hope we get some other opinion/contribution on this.

I hope you understand where that last equation of mine comes from.

Don't worry, I think you did get what you were being asked to get. I think some more clarification would be good though.

Last edited: Jul 9, 2013
12. Jul 9, 2013

### Saitama

Yes. I already knew about the equation but never had the opportunity to use it so it didn't hit me at the first look on the question. :)

I guess we have to assume that there is not much difference between $T_1$ and $T_2$ so the activation energy stays almost constant but then again, it isn't mentioned in the problem statement.

13. Jul 10, 2013

### Staff: Mentor

This is where the Taylor series expansion I was referring to comes into play.

Let's say we want to consider the changes in the reaction rate with respect to temperature, relative to a specific temperature T0.

Then $$\ln(k_1(T)+k_2(T))=\ln(k(T))$$
$$\ln(k_1(T_0)+k_2(T_0))=\ln(k(T_0))$$

If we subtract these two equations, we obtain
$$\ln(k_1(T)+k_2(T))-\ln(k_1(T_0)+k_2(T_0))=-\frac{E'}{R}\left(\frac{1}{T}-\frac{1}{T_0}\right)$$

Now, let x = 1/T and x0 = 1/T0

Then, $$-(x-x_0)\frac{E'}{R}=\ln(k_1(x)+k_2(x))-\ln(k_1(x_0)+k_2(x_0))$$

Now, expanding the rhs in a Taylor series about x0, we get

$$-(x-x_0)\frac{E'}{R}=-\left(\frac{x-x_0}{R}\frac{E_1k_1(x_0)+E_2k_2(x_0)}{k_1(x_0)+k_2(x_0)}+...\right)$$

This is the desired result.

14. Jul 10, 2013

### epenguin

After the next term maybe - otherwise it's just the initial result.

I don't think we should too much give students the impression that kinetics is much about mathematical subtleties. I had the uncomfortable feeling all along that this was an "academic exercise" not necessarily in the best sense. However the student might ask his teachers about these last doubts of #9.

More important IMO than this slightly artificial apparent activation energy would be to know what the graph looks like for A, B, and C concentrations against time as you change temperature.

And also what ln k against 1/T for these three cases looks like. The only slightly remarkable case I think is A. But that might arise in practice in cases where only [A] can be conveniently measured.

15. Jul 10, 2013

### Staff: Mentor

Amen. I totally agree.

Chet