intermediates

bellaha4F · Postby **bellaha4F** » Wed Mar 11, 2020 11:43 am

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

Jessica Booth 2F · Postby **Jessica Booth 2F** » Wed Mar 11, 2020 12:01 pm

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

LNgo 1G · Postby **LNgo 1G** » Wed Mar 11, 2020 12:44 pm

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

ABombino_2J · Postby **ABombino_2J** » Wed Mar 11, 2020 12:46 pm

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

Hailey Kim 4G · Postby **Hailey Kim 4G** » Wed Mar 11, 2020 1:21 pm

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 · Postby **Sean Cheah 1E** » Wed Mar 11, 2020 5:05 pm

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 · Postby **005384106** » Wed Mar 11, 2020 5:40 pm

Was O2 a catalyst or an intermediate in this example?

SimranSangha4I · Postby **SimranSangha4I** » Wed Mar 11, 2020 5:54 pm

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

Jessa Maheras 4F · Postby **Jessa Maheras 4F** » Wed Mar 11, 2020 7:19 pm

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 · Postby **Osvaldo SanchezF -1H** » Wed Mar 11, 2020 8:05 pm

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 · Postby **Timmy Nguyen Dis 1I** » Wed Mar 11, 2020 8:19 pm

you are correct!

William Chan 1D · Postby **William Chan 1D** » Wed Mar 11, 2020 10:07 pm

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 · Postby **Bryan Chen 1H** » Wed Mar 11, 2020 10:24 pm

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

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