# Arrhenius Equation for parallel reaction

• Saitama
In summary: I don't think it can be called a constant as such. It is a function of the temperature, albeit a linear one.
Saitama

## Homework Statement

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

## 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!

Pranav-Arora said:

## Homework Statement

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

## 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!

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

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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

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1 person
epenguin said:
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
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.

Chestermiller said:
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.

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?

Pranav-Arora said:
Is this wrong?

I think it is what they want.

epenguin said:
I think it is what they want.

Thanks for the help epenguin!

Pranav-Arora said:
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?

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

Chet

? 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?

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

Pranav-Arora said:

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.

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

epenguin said:

? 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?

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.

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

This is the desired result.

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.

epenguin said:
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.

Amen. I totally agree.

Chet

## 1. What is the Arrhenius Equation for parallel reaction?

The Arrhenius Equation for parallel reaction is a mathematical equation that relates the rate constant of a parallel reaction to temperature. It is expressed as k = Ae-Ea/RT, where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

## 2. How is the Arrhenius Equation used in chemistry?

The Arrhenius Equation is used in chemistry to determine how temperature affects the rate of a chemical reaction. It allows scientists to calculate the rate constant of a reaction at different temperatures and predict how the rate of the reaction will change with temperature.

## 3. What is the significance of the pre-exponential factor in the Arrhenius Equation?

The pre-exponential factor, represented by the symbol A, represents the frequency of collisions between reactant molecules. It is a measure of how often reactant molecules collide in the correct orientation and with enough energy to form products. A higher A value indicates a higher likelihood of successful collisions and a faster reaction rate.

## 4. How does the activation energy affect the rate of a reaction according to the Arrhenius Equation?

The activation energy, represented by the symbol Ea, is a measure of the minimum energy required for a reaction to occur. According to the Arrhenius Equation, a higher activation energy results in a lower rate constant, and therefore a slower reaction rate. This is because a higher activation energy means that more energy is needed for reactant molecules to reach the transition state and form products.

## 5. What are the limitations of the Arrhenius Equation for parallel reaction?

The Arrhenius Equation has several limitations, including the assumption that the reaction follows a simple two-step mechanism and that all reactant molecules have the same energy. It also does not take into account the effects of catalysts or changes in the reaction mechanism at high temperatures. In some cases, the equation may not accurately predict the rate constant for a reaction, and experimental data may be needed to validate the results.

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