Homework Help: Physical Chemistry

Posted by Mark on Thursday, April 1, 2010 at 8:30pm.

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 192 kJ·mol-1 to 72 kJ·mol-1.
(a) By what factor does the rate of the reaction increase at 281 K, all other factors being equal?

(b) By what factor would the rate change if the reaction were carried out at 343 K instead?

What equation do i use to solve this???

Physical Chemistry - DrBob222, Thursday, April 1, 2010 at 9:12pm
Use the Arrhenius equation. You can solve for k1/k2 for each activation energy, then compare them.
Physical Chemistry - Brad, Monday, April 5, 2010 at 5:04pm
For part A, I'm plugging into the Arrhenius equation, ignoring A, and then dividing the 2 rates for each activation energy. I'm getting 1.027 for my particular set of numbers (Ea = 120 & 56) @ 293K. This isn't correct though.
Physical Chemistry - Z, Monday, April 5, 2010 at 5:54pm
If any clarification is needed on the question, it wants to know the difference between the rates with an Ea of 192 at T=281K and an Ea of 72 at T=281K.

Use the rearranged Arrhenius like DrBob suggested: k=Ae^(-Ea/RT)

Take k2/k1 and you get your answer. If you want to check your answers, my numbers were 178 kJ·mol-1 to 54 kJ·mol-1 with the reaction carried out at 333K. My rate changed by a factor of 2.82774e19

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