## 15.49

$K = \frac{k_{forward}}{k_{reverse}}$

Emilie Hoffman 1E
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

### 15.49

For this question, they ask you to make rate laws for each elementary step of a two step reaction, the first slow and the second fast. The solution manual keeps the intermediate in the rate law for the second step. Is this done because you can't assume anything is in excess because there isn't a buildup of any material involved in the second step?

Here's the full question:
The following mechanism has been proposed for the gasphase reaction between HBr and NO2:
Step 1 HBr + NO2 --> HOBr + NO (slow)
Step 2 HBr + HOBr --> H2O + Br2 (fast)
(a) Write the overall reaction.
(b) Write the rate law for each step and indicate its molecularity.
(c) What is the reaction intermediate?

Angela G 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.49

I think they're just asking for the rate laws of each individual step, not the overall reaction. For the overall reaction you wouldn't keep the intermediate in the rate law for the slow step and would instead substitute it using the pre-equilibrium approach.

Katie Lam 1B
Posts: 52
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.49

The rate law for each individual step is just the rate constant times the reactant concentrations for the reactants (including intermediates) in that particular equation.

andrewr2H
Posts: 29
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.49

The intermediate is substituted out in the total reaction rate law.

Justin Folk 3I
Posts: 43
Joined: Wed Sep 21, 2016 2:56 pm

### Re: 15.49

Yes rate law is only the slow step