## intermediates

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

bellaha4F
Posts: 104
Joined: Wed Sep 18, 2019 12:20 am

### intermediates

intermediates should not be listed in the rate laws, right? but rather replaced with the products that the intermediates depend on?

Jessica Booth 2F
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

### Re: intermediates

Correct, they should be replaced using the equilibrium equation solved for the intermediate.

LNgo 1G
Posts: 100
Joined: Sat Aug 24, 2019 12:16 am

### Re: intermediates

Right, generally they are not present in the written form of the overall reaction.

ABombino_2J
Posts: 102
Joined: Thu Jul 11, 2019 12:15 am

### Re: intermediates

Yes. Because they are formed and used up they cancel each other out in the overall equation.

Hailey Kim 4G
Posts: 110
Joined: Sat Jul 20, 2019 12:16 am

### Re: intermediates

Reaction intermediates are formed in one step and then consumed in a later step of the reaction mechanism. Therefore, intermediates are not listed in rate laws.

Sean Cheah 1E
Posts: 105
Joined: Wed Sep 18, 2019 12:20 am

### Re: intermediates

Use the same concept behind the pre-equilibrium approximation to derive expressions for the intermediates in your rate laws. For example, for the reaction $2\textup{NO} +{\textup O_{2}} \rightarrow 2\textup{N}{\textup O_{2}}$ with elementary steps $2 \textup{NO} \rightleftharpoons \textup N_{2} \textup O_{2}$ (fast step; $\textup{rate} = k_{1}[\textup{N} \textup O]^{2}$) and $\textup N_{2} {\textup O_{2}} + \textup O_{2} \rightarrow 2 \textup N \textup O_{2}$ (slow step; $\textup{rate} = k_{2} [{\textup N_{2}} {\textup O_{2}}]$), treat the first step as if the intermediate $\textup N_{2} \textup O_{2}$ and the reactant NO were in equilibrium as represented by the double-sided arrow in the equation. The forward and reverse rates are equal at equilibrium so $k_{1}[\textup N \textup O]^{2} = k_{-1}[{\textup N_{2}} {\textup O_{2}}] \Rightarrow [{\textup N_{2}} {\textup O_{2}}] = \frac{k_{1}}{k_{-1}}[\textup N \textup O]^{2}$, where $k_{-1}$ is the rate constant for the reverse of the first elementary step. Plug that into the rate law for the slow step to get the overall rate law $\textup{rate} = \frac{k_{1}k_{2}}{k_{-1}}[\textup N \textup O]^{2} = k[\textup N \textup O]^{2}$, where k is the overall rate constant and is equal to $\frac{k_{1}k_{2}}{k_{-1}}$.

005384106
Posts: 101
Joined: Sat Aug 24, 2019 12:16 am

### Re: intermediates

Was O2 a catalyst or an intermediate in this example?

SimranSangha4I
Posts: 99
Joined: Sat Sep 14, 2019 12:17 am

### Re: intermediates

correct because that would mean its formed and then used up.

Jessa Maheras 4F
Posts: 121
Joined: Fri Aug 02, 2019 12:16 am

### Re: intermediates

Jessica Booth 2F wrote:Correct, they should be replaced using the equilibrium equation solved for the intermediate.

Can you explain how this works? This process is a bit confusing. Thank you!

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

### Re: intermediates

Intermediates are products that are formed in one reaction that are then used in another reaction to create the new products wanted. So therefore they are made and consumed so that means they should not be present in the final rate law.

Timmy Nguyen Dis 1I
Posts: 107
Joined: Sat Aug 17, 2019 12:17 am

### Re: intermediates

you are correct!

William Chan 1D
Posts: 102
Joined: Sat Sep 14, 2019 12:15 am

### Re: intermediates

Yes. Intermediates are molecules that are formed in one step of an overall reaction and consumed in one of the following steps. It will not show up in the overall reaction as it will cancel out once you add up the steps of the equation.

Bryan Chen 1H
Posts: 58
Joined: Mon Jun 17, 2019 7:24 am

### Re: intermediates

yes, since they are made in one step and used up in the next they are not included in the overall final one