## Lavelle's review slides

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

JosephineF
Posts: 49
Joined: Wed Sep 18, 2019 12:17 am

### Lavelle's review slides

On the last slide of Lavelle's review it asks to choose which students' mechanism is valid based on the observed rate law. I'm not sure where the K1,k1, and k2 are coming from and how they are used to determine the rate limiting step. If someone doesn't mind walking me through that whole problem, it will be much appreciated!

asannajust_1J
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am
Been upvoted: 1 time

### Re: Lavelle's review slides

For student A, start by writing each rate law as if it were the rate-limiting step. If the second step is the slow step, an intermediate is found in your rate law. the k in the rate law from stoichiometric coefficients corresponds with the rate constant for step two. Then, since intermediates are not allowed in the rate law, you can use pre-equilibrium to assume that step 1 is in equilibrium. Thus set up the equilibrium expressions with capital K1 and solve for the intermediate. plug this new expression in as the intermediate for the rate law for step 2 which will introduce k2and K1 into the same expression. k2 is the rate constant for step 2 and K1 is the equilibrium constant for step 1.

Repeat the same process for student B. I hope this helps!

Amy Luu 2G
Posts: 105
Joined: Wed Sep 18, 2019 12:19 am

### Re: Lavelle's review slides

I am confused on step 2 of student B. The rate limiting reaction rate in step 2 is k2[Cl2(g)][O2(g)]2. Since O2(g) is an intermediate, we use K1 to replace it. However, I am confused since [O2(g)]=K1[ClO2]2/[Cl2(g)]. How come when [O2(g)] in the rate law is replaced with that value, it is not squared? I thought it would be squared since the rate is k2[Cl2(g)][O2(g)]2?

Amy Luu 2G
Posts: 105
Joined: Wed Sep 18, 2019 12:19 am

### Re: Lavelle's review slides

Nvm! i just realized that Cl2 is the intermediate and not O2