Reaction temperature to maximize selectivity of product

In summary: Rayburn said:In summary, increasing the temperature will help the desirable product formation, but it does so more rapidly than the undesirable product formation rate.
  • #1
gfd43tg
Gold Member
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Hello,

I am confused a little bit on why for a reaction with a given activation energy, one should run at high temperatures if the activation energy of the desirable product, D, is greater than the undesirable product, U.

To illustrate this, I have an example with two reactions
##A + B \rightarrow D##
##A + B \rightarrow U##

##r_{D} = 10 \hspace{0.05 in} exp(- \frac {8000}{T})C_{A}C_{B}##
##r_{U} = 100 \hspace{0.05 in} exp(- \frac {1000}{T})C_{A}^{1/2}C_{B}^{-3/2}##

So I know to maximize the selectivity here, I will want to use a high concentration of A and low concentration of B, so I will use a semibatch reactor with a slow feed of B. However, as far as temperature is concerned, I am a little bit stumped. Why is it that increasing the temperature will help the desirable product formation? exp(-8000/T) is going to be much smaller than exp(-1000/T).

I run it in MATLAB and I see that the desirable is a flat line, so the difference between the two is actually increasing as temperature increases, so the undesirable reaction is increasing much faster than the desirable reaction rate as temperature increases. That means ##r_{U}## is growing a lot quicker with temperature increase than ##r_{D}##, which seems to be the opposite of what I want to accomplish.

Code:
T = linspace(300,1000,1e4);
fD = @(T) exp(-8000./T);
fU = @(T) exp(-1000./T);
plot(T,fD(T),T,fU(T))
legend('D','U')
 

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  • #2
Maylis said:
exp(-8000/T) is going to be much smaller than exp(-1000/T).
True, but the ratio of the two depends on temperature. Just plug in two different values and calculate the ratio.

Your plot does not allow to see that as D is too low. Maybe use a logarithmic plot or directly plot the ratio.
 
  • #3
Maylis said:
Hello,

I am confused a little bit on why for a reaction with a given activation energy, one should run at high temperatures if the activation energy of the desirable product, D, is greater than the undesirable product, U.

To illustrate this, I have an example with two reactions
##A + B \rightarrow D##
##A + B \rightarrow U##

##r_{D} = 10 \hspace{0.05 in} exp(- \frac {8000}{T})C_{A}C_{B}##
##r_{U} = 100 \hspace{0.05 in} exp(- \frac {1000}{T})C_{A}^{1/2}C_{B}^{-3/2}##

So I know to maximize the selectivity here, I will want to use a high concentration of A and low concentration of B, so I will use a semibatch reactor with a slow feed of B. However, as far as temperature is concerned, I am a little bit stumped. Why is it that increasing the temperature will help the desirable product formation? exp(-8000/T) is going to be much smaller than exp(-1000/T).

Yes. But it increases with temperature (percentage-wise) much more rapidly than exp (-1000/T)
I run it in MATLAB and I see that the desirable is a flat line, so the difference between the two is actually increasing as temperature increases, so the undesirable reaction is increasing much faster than the desirable reaction rate as temperature increases. That means ##r_{U}## is growing a lot quicker with temperature increase than ##r_{D}##, which seems to be the opposite of what I want to accomplish.
Are you sure of your calculations? This should not happen: The ratio of D to U should increase with temperature for a given total product of U + D. That's in the mathematics.

Chet
 

1. What is the ideal reaction temperature to maximize selectivity of product?

The ideal reaction temperature to maximize selectivity of product depends on the specific reaction and the desired product. Generally, a lower reaction temperature will result in higher selectivity, but this can vary depending on the reactants and catalyst used.

2. How does reaction temperature affect product selectivity?

Reaction temperature can affect product selectivity by influencing the rate of reaction and the stability of intermediates. Higher temperatures can lead to faster reactions and the formation of byproducts, while lower temperatures can promote the formation of the desired product.

3. Can changing the reaction temperature improve product selectivity?

Yes, changing the reaction temperature can often improve product selectivity. By optimizing the temperature, researchers can control the kinetics and thermodynamics of the reaction to favor the desired product over unwanted byproducts.

4. What factors should be considered when determining the reaction temperature for maximum selectivity?

Several factors should be considered when determining the reaction temperature for maximum selectivity, including the desired product, reactant concentrations, reaction kinetics, and catalyst activity. It is important to carefully balance these factors to achieve the desired outcome.

5. Is there a specific range of reaction temperatures that is generally recommended for maximizing product selectivity?

There is not a specific range of reaction temperatures that is universally recommended for maximizing product selectivity. Each reaction is unique and may require different temperature conditions for optimal selectivity. It is important to carefully evaluate the reaction and adjust the temperature as needed to achieve the desired results.

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