# What are the rate laws for each elementary step?

Overall reaction

$$\ce{2NO + H2 -> N2O + H2O}$$

Mechanism

\begin{align} \ce{NO + NO &-> NO2}\\ \ce{NO2 + H2 &-> NO + H2O}\\ \ce{N + NO &-> N2O}\\ \end{align}

I am aware that this is not a valid mechanism for the overall reaction however I am still supposed to find the rate law for each elementary step. I am having trouble determining what this rate law would be in step 3 of the mechanism.

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. Oct 19, 2015 at 4:50

For elementary reactions, the rate law can be directly read from the chemical equation. In your last equation this is a bimolecular reaction, i.e. \begin{align} \ce{N + NO &-> N2O} & r &= k\cdot c(\ce{N}) \cdot c(\ce{NO}) \end{align}

To be more precise here, one molecule of $\ce{NO}$ reacts with on atom $\ce{N}$, and in a wider sense, that is two (bi) molecules reacting. The order of the reaction is two.

That being said, there are a couple of things you need to pay close attention to. An elementary reaction is always a reaction between the number of molecules given as reactants, with exactly one transition state leading to the products. If this number or reactants is greater than three it is highly unlikely to happen this way, since a concerted transition state is statistically very unlikely. An elementary reaction therefore also has to be balanced, so your first reaction needs to be: $$\ce{2NO -> NO2 + N}$$ In this case we could see $[\ce{ON\bond{~}O\bond{~}N}]^\ddagger$ as a transition state.

I very highly doubt that your second reaction as an elementary reaction, from my point of view, there are too many bond that would need to be broken. Instead, the following reactions might be possible and/or necessary decompositions: \begin{align} \ce{NO2 + H2 &-> NO + OHH}\\ \ce{OHH &-> OH + H}\\ \ce{NO2 + H &-> NO + OH}\\ \ce{2OH &-> H2O + O}\\ \ce{NO + O &-> NO2}\\\hline \ce{NO2 + H2 &-> NO + H2O} \end{align} And there are more possibilities.

In all but one the elementary reactions are bimolecular - guess which one would be unimolecular.

As for the mechanism itself, I think these are reactions that occur, just not all of them, maybe not even the majority of them.

Remember that when a reactant is used in more than one step you have to use the rule of

rate of production = rate of consumption

This leads to $k1[NO]^2 = k2[NO_2][H_2]$

See where you can go from there remember in the last reaction its $k3[N][NO]$

Hope that helps.

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. It appears that you are commenting on the mechanism, which is only part of the question for context. The actual question is: What is the rate law in the third step? Oct 19, 2015 at 5:25
• I don't see why your saying that is wrong. He posted a reaction mechanism, and the rate law for the third step assuming it's the rate determining step is what I commented. He even posted that neglect that it is an invalid mechanism and you posted why it isn't a valid mechanism. I think you misread his question.
– ATP
Oct 20, 2015 at 5:59
• "I am having trouble determining what this rate law would be in step 3 of the mechanism." You did not answer how to determine the rate law for that step. Oct 20, 2015 at 6:04
• I showed him that if one of the species is used in more than one step then the rate law has to be made to account for that. I gave him everything he would need to determine it....
– ATP
Oct 20, 2015 at 6:09
• The question is not about the rate law of the mechanism. It is only about the rate law of the last step. Oct 20, 2015 at 6:10