## 15.67 Homework Activation Energies

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

jenshan
Posts: 23
Joined: Wed Nov 18, 2015 3:00 am

### 15.67 Homework Activation Energies

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?

Christina Pham 3H
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

For a, set rate catalyst over the rate without a catalyst. Use the equation $k=Ae^{\frac{-Ea}{rt}}$. They key is setting the top Ea as 0.6X and the bottom Ea as X. This is because if you divide 75 by 125, it equals 0.6. The A's cancel out, and then you should after simplifying get $e^{^{\frac{0.4x}{rt}}}$ Substitute 125 into X, and 298 for T, Note, make sure r is $8.314*10^{-3}$, and you should get your answer.
For Part B, instead of 298, substitute 350 for T.
I suggest looking at the solutions manual to clarify your question even further.

jenshan
Posts: 23
Joined: Wed Nov 18, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

Thank you so much!

I don't have a solutions manual which is why I asked the question

Camila Bautista 3I
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

Why do you substitute 125 into X?
How did you simplify the equation to get e^(.4x/RT)?

Christina Pham 3H
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

You're welcome! :)

Sorry for the late reply. But in response to your question, you substitute in 125 kj/mol since 0.4 is the factor of Ea that changed in regards to the original Ea which is 125. I explained it above, but I can try to explain it again. Basically you are setting rate catalyst over rate w/o catalyst (so $\frac{A*e^{\frac{-Ea}{RT}}}{A*e^{\frac{-Ea}{RT}}}$.
For Ea of catalyst, you divide 75 by 125 which equals 0.6X. You substitute 0.6X into Ea catalyst, and X into Ea w/o catalyst. The A's cancel out, and you should end up with $e^{\frac{0.4X}{RT}}$

Alexander Vong 3I
Posts: 56
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

To add, I find it more intuitive to set r, or R, to 8.314 J/K mol and keep x as 125 x 103 J/mol

Of course, the approach described above works too

Carrie Huang 2F
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

Instead of setting Ea,cat=0.60Ea,uncat, one can simply set Ea,cat=75 and Ea,uncat=125, correct?

Christina Pham 3H
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

I tried it and it works! Basically just make sure when substituting T, both T for cat and uncat are the same. (Part a. 298 K and Part b. 350 K)

Christina Pham 3H
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

### Re: 15.67 Homework Activation Energies

Edit: "Instead of setting Ea,cat=0.60Ea,uncat, one can simply set Ea,cat=75 and Ea,uncat=125, correct?" No. It must have been a fluke. There's a similar problem on a practice final and you must find the answer by solving it the way the sol'n manual did.