Hw 15.67

[Juanalv326]

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kJmol1 to 75 kJmol1
. (a) By what factor does the rate of the reaction increase at 298 K, all other factors being equal?
How would this problem be done? I know we would have to use the arrhenius equation, however it is still confusing as i get the incorrect answer.

[Peter Dis1G]

Divide the catalyzed over uncatalyzed since it says all other factors equal.

[Magdalena Palavecino 1A]

In this equation you cancel out A, but if you didn't what is A and what do you plug into the equation for it?

[Kayla Ikemiya 1E]

Where did this catalyzed/uncatalyzed set up come from?

[Joyce Lee 1C]

Use the equation k = Ae^(-Ea/RT). Because you're just trying to find the factor by which the catalyzed and uncatalyzed reactions differ, divide the k of the catalyzed reaction by the k of the uncatalyzed to get the term:
cat refers to catalyzed; uncat refers to uncatalyzed
kcat/kuncat = Ae^(-Ea,cat/RT)/ Ae^(-Ea,uncat/RT) = e^(-Ea,cat/RT)/ e^(-Ea,uncat/RT)
To make the problem a bit easier, you can rewrite Ea,cat as a function of Ea,uncat:
Ea,cat = 75 kJ/mol and Ea,uncat = 125 kJ/mol
Ea,cat = 75/125 kJ/mol*Ea,uncat = 0.60 kJ/mol*Ea,uncat
Plug this new value in into the equation for kcat/kuncat and keep the Ea,uncat in the denominator the same:
kcat/kuncat = e^(-0.60 kJ/mol*Ea,uncat/RT)/ e^(-Ea,uncat/RT) = e^(0.40 kJ/mol*Ea,uncat/RT) = 6 x 10^8
They A's cancel out. Because the values of A are not given, you can't plug in any numbers.

[Jana Sun 1I]

I understand your overall process, but I don't get why we need to do this step:

"To make the problem a bit easier, you can rewrite Ea,cat as a function of Ea,uncat:
Ea,cat = 75 kJ/mol and Ea,uncat = 125 kJ/mol
Ea,cat = 75/125 kJ/mol*Ea,uncat = 0.60 kJ/mol*Ea,uncat"

Why can't we just keep Ea,cat = 75 kJ/mol and Ea,uncat = 125 kJ/mol as is and just plug them directly into our equation?