## HW problem 15.67

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

Jaira_Murphy_2D
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### HW problem 15.67

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kJ/mol to 75 kJ/mol.

(a) By what factor does the rate of the reaction increase at 298 K, all other factors being equal?

(b) By what factor would the rate change if the reaction were carried out at 350. K instead?

I've seen this problem a lot on here but I cannot get the same answer as the textbook. I've tried to solve it every way that's mentioned and I'm still not getting the correct answer. Can someone please explain each step. (P.S. I do not have the solutions manual which is why I'm stuck)

Chem_Mod
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### Re: HW problem 15.67

In order to answer this question you have to set a ratio for the Arrhenius Equations and solve for the ratio of the rate constants. You then realize that by whatever factor the rate constants changed is the factor by which the rates changed because there is a linear relationship between the two given by the differential rate law ( rate = k[A]l ).