## Questions about 14.85

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

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

### Questions about 14.85

In question 14.85, why does the textbook put brackets and a symbol around the activated complex? What does this mean & do we have to follow this convention in our answers?

Also, for part b, I understand that the reaction is termolecular from the rate law, but why doesn't the solutions manual include Ar in the elementary reaction? I realize that Ar is both a reactant and a product so therefore cancels out, but if it's in the rate law shouldn't it be included in the elementary reaction for the sake of clarity? (And would we be marked wrong if we included it in the elementary reaction on a test?)

Thanks!!

Anuk Burli 2C
Posts: 29
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Questions about 14.85

I'm not sure about part a but for part b I believe that inert noble gases don't affect the rate of the reaction so they wouldn't be included. I don't think you ever add noble gases because they're inert similar to how they didnt affect equilibrium

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Questions about 14.85

In the question they say that for this certain reaction, Ar has a role (it removes energy as the product forms) and they do include it in the rate expression. Just not the elementary reaction for some reason

Regina Chi 2K
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Questions about 14.85

I don't remember the question, but could Ar possibly be a catalyst?

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Questions about 14.85

Yes, you should put brackets and the double plus sign when writing out your activated complex. The double plus sign is used to indicate that there is a transition state.

The Ar is technically in the reaction, but it appears in both the reactants and products. It gets canceled out which is why you don't see it in the chemical equation.

I think the Ar could be a catalyst but the problem doesn't specify.

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