[hotvince]

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

The answer is a) the catalyzed rxn is 10,000,000 times faster at 25 degrees C and b)  the catalyzed rxn is 180,000 times faster at 125 degrees C.

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I just don't know what numbers to use. I think you use the Arrehenius Equation, but I don't know what numbers to put where.

[Borek]

You are right about using Arrhenius equation. Write it down and think, what is given and what you are looking for.

[hotvince]

well, Arrhenius Equation is k = Ae^-Ea/RT

R is 8.314 J/mol
T is 298 K
k is (95/55) ?
Ea = 50 x 10^3 joules

I tried that and I got an answer that was off by 3 decimal places.

[Yggdrasil]

The question is asking about the ratio of rate constants k.  If k is the rate constant of the uncatalyzed reaction and k' is the rate constant of the catalyzed reaction, the question is asking about the value of k'/k.

Note that when the question refers to the collision factor, it is referring to A.

[hotvince]

Oh. Thanks, I got the answer. 

I thought we had to solve for A instead of neglecting A.

[enahs]

Not to be picky, but you are not neglecting A, just A/A =1.