## Homework 7E.3

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

MaryBanh_2K
Posts: 101
Joined: Wed Sep 18, 2019 12:21 am

### Homework 7E.3

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kJ?mol−1 to 75 kJ?mol−1. (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?

Why is activation energy of the catalyzed reaction equal to (75/125) x activation energy of the uncatalyzed reaction?

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

### Re: Homework 7E.3

The activation energy of the catalyzed reaction is 75 kJ/mol. The activation energy of the uncatalyzed reaction is 125 kJ/mol. Therefore, the activation energy of the catalyzed reaction is 75/125 times the activation energy of the uncatalyzed reaction, which is 125 kJ/mol. Therefore, (75/125) times (125) is still 75, which is the number given in the problem. The purpose of rewriting the activation energy of the catalyzed reaction in this way is to make the ratio of the rate constants easier to simplify when cancelling out terms in the numerator and denominator. You don't have to do this; if you plug in 75 kJ/mol and 125 kJ/mol individually into the expressions for k and solve for the ratio of the 2 constants you will get the same answer.