[browniee]

A first order liquid phase irreversible reaction is being carried out in a batch reactor. A reaction at 40 Degrees C required 10 mins to achieve a conversion of 80%. The same conversion can be achieved in 5mins if the reaction was carried out at 60 Degrees C. Determine the activation energy for this reaction.

i formed 2 rate equations replacing the rate constant with the arrhenius equation and the only difference between the 2 equation is the temperature. One is 40 and another at 60.

I equate this 2 equation and find the activation energy (E). Is this the correct way of doing this problem?

Also, what if only 1 min is required to achieve 90% conversion of the reaction described above, how do i find the reaction temperature?

In batch reactor for 1st order reaction, im aware the equation to use is Ln (Ca) = -Kt + Ln (Cao)

[mjc123]

I equate this 2 equation and find the activation energy (E). Is this the correct way of doing this problem?

It depends what you mean by "equating the 2 equations". Why don't you show us exactly what you did and we can see if it's right?

[browniee]

The first equation is -> -r (a) = Ko e^ (-E/R(40) ) Ca ^ (n)

2nd equation is -> -r (a) = Ko e^ (-E/R(60) ) Ca ^ (n)

i equate this 2 together because what i understand is that the rate constant will be the same with the same fractional conversion. But with this method, i cant get to the final answer of the first part of my question.

[mjc123]

But one reaction is faster than the other. The rate constants can't be equal. If they were your equations would be inconsistent. (And by the way, T in the Arrhenius equation is in K, not °C.)

Look at your equation in your original post. It is the product Kt that is constant at constant conversion (for a first order reaction).

[browniee]

How do i find the activation energy then ? do i still need to equate something together?

[Corribus]

It is usually a decent approximation to assume the pre-exponential factor is independent of temperature over small temperature ranges. In that case, you basically have two equations and two unknowns, at which point this is more or less a straightforward algebra problem. You can use whatever method you like to solve it. For Arrhenius problems where the goal is to find the activation energy, the approach of choice is usually to divide one equation by the other, in which case the pre-exponential factor cancels out and you are left with only the activation energy to solve for.