## 7E.3A

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

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 7E.3A

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?

Can someone explain how to do this problem? How would you use the equation k = Ae^(-Ea/RT) to find the factor?

Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am

### Re: 7E.3A

I found an equation in book which helped.
Ea(with catalyst)-Ea(w/o catalyst)/RT= a number. That number you then add a negative sign to and raise it to the e^power. Basically, its putting k2/k1 with the Arrhenius equations both on the top and bottom, respectively.
so for part a i got 75000-125000/(8.314x298)=-20.18, I then raised it to the negative e power to get e^-(-20.18) to get 5.81x10^8 at 298K. This corresponds to k2=k1(5.18x10^8), which is the factor by which the rate increases. Hopefully, this was a coherent answer.

Return to “Arrhenius Equation, Activation Energies, Catalysts”

### Who is online

Users browsing this forum: No registered users and 1 guest