[moonlitbubbles]

The activation energy of an uncatalyzed reaction is 94 kj/mol . The addition of a catalyst lowers the activation energy to 52 kj/mol.  Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 24 C?

Will someone please show me (in steps) how to solve this, and whether or not it is with the Arrhenius Equation.  Thanks!
 

[Borek]

Direct application of Arrhenius equation. Just plug in numbers.

[moonlitbubbles]

That is the problem.  I used the Arrhenius equation and got 7.01 x 10^-10.  But the answer is 2.4 x 10^7.  I don't understand how they got that answer.  That is why I really need this equation worked out step by step.  I am not that great at chemistry, or math.  But I have done this over and over, and I keep getting the same numbers.  I have the activation energy = 52kj/mol (52,000 J/mol-K), R = 8.314 T= 294K.  I multiplied R x T.  Divided -52,000 J by R x T.  Then took e^x of that answer.  What am I doing wrong???

[Yggdrasil]

You're supposed to compare the rate of the uncatalyzed reaction (E_a = 94 kJ/mol) and the catalyzed reaction (E_a = 52 kJ/mol).  So, you're halfway to the answer.