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Catalysis and its effect on rate

Submitted by Sunil on Tue, 02/26/2008 - 19:09

The activation energy of an uncatalyzed reaction is 95 KJ/mol. The addition of a catalyst lowers the activation energy to 55 KJ/mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) 25*C (b) 125*C

We had trouble with this, even my friends couldn't figure it out. Please help.


do you know the arrhenius equation? are you using it to try to solve the problem


Yea I am using that, I use arrhenius equation to find the rate constant at 95 KJ and then to find rate constant at 55 KJ. I eliminated A, because the frequency factor is the same so its constant. Then I divide the first one by the second one, but it seems like its not the right answer.


Interesting problem in that I've never used Arrhenius's equation in this form.

                   k  = A e -Ea/RT    where k is the rate constant and A is the frequency factor.  

If you calculate the rate constant for the catalyzed and uncatalyzed reactions you could then compare them at the different conditions (since the value of A is constant and you are only interested in the relationship between the rate constants and not the actual value of the rate constants assign A=1 for all calculations).


Well the original problem uses the term rate of the reaction, I am assuming that we have to use the rate constants. Otherwise there must be a different equation to use. Also, isnt excluding the A the same thing as assingning it a value of 1?

Correct Answer: (a) approximately 10,000,000 (b) approximately 180,000


Yes, excluding A is the same as assigning it an arbitrary value of 1.

Here's what I got for a at 25:
                                     k  = A e-Ea/RT
uncatalyzed                    k  = 1 x e -95x103 J/mol/8.314 J/mol K /298K) = 2.2 x 10-17

catalyzed                        k = 1 x e -55x103 J/mol/8.314 J/mol k / 298K)  = 2.2 x 10-10

so the ratio of catalyzed to uncatalyzed would be
                                        2.2 x 10-10 / 2.2 x 10-17  = 1 x 107 = 10,000,000


Oh my gosh, I did the same exact thing before and came up with really weird numbers on my calculater, I really need to practice more on how to plug in numbers on the Calculator.

Thanks mate.


Hi-

I'm having a similar problem as the one listed above.  Here it is:

The activation energy of an uncatalyzed reaction is 85kJ/mol . The addition of a catalyst lowers the activation energy to 53kJ/mol .

Here's the associated question-

Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 134?

Our answer needs to be two significant figures and the answer I get is 12,000, which is incorrect when I submit it on mastering chemistry for my class.

 

Please help!!

 

Daniel

 


I'm assuming that "at 134" means at 134C (407K).  When I did the math, I came up with a factor of 12,845.  When you round this to two significant digits you would get 13,000.