The slowest step will be the step where we pass over a transition state with the highest energy, e.g.
[latex]\text{C}_{60}\text{O}_3\rightarrow \text{O}_2+\text{C}_{60}\text{O}[/latex]The rate of this reaction can be monitored by measuring the absorbance of the solution. @MathewMahindaratne Actually, my point of commenting is to include the information of K being large as well, otherwise it gives the false of impression that the different energies decide the kinetics of the reaction.
Every reaction has a "slowest step", therefore, every reaction has an rds. We need to write this rate law in terms of reactants only. c) What is the rate law for each step (in terms of [O 3], include the reverse of step 1) ?
We want to hear from you.\[ \underbrace{A +A \xrightarrow{k_1} A_2}_{\text{slow}}\]\[ \underbrace{ A_2 \xrightarrow{k_2} C + A}_{\text{fast}}\]the rate determining step approximation suggests that the rate (expressed in terms of the appearance of product \(C\)) should be determined by the slow initial step, and so the rate law will bematching the order of the rate law to the molecularity of the slow step.
[latex]2\;\text{NO}(\text{g})+\text{O}_2(\text{g})\rightarrow 2\;\text{NO}_2(\text{g})[/latex]The overall equation suggests that two NO molecules collide with an oxygen molecule, forming NO[latex](1)\quad\quad 2\;\text{NO}(\text{g})\rightarrow \text{N}_2\text{O}_2(\text{g})\quad\quad \text{rate}=\text{k}_1[\text{NO}]^2[/latex][latex](2)\quad\quad \text{N}_2\text{O}_2(\text{g})+\text{O}_2(\text{g})\rightarrow 2\;\text{NO}_2(\text{g})\quad\quad \text{rate}=\text{k}_2[\text{N}_2\text{O}_2][\text{O}_2][/latex]Note that the two steps here add to the overall reaction equation, as the intermediate NThe rate of a multi-step reaction is determined by the slowest elementary step, which is known as the rate-determining step.Describe the relationship between the rate determining step and the rate law for chemical reactionsChemists often write chemical equations for reactions as a single step that shows only the net result of a reaction.
the activation energy) for the hydride shift would not be higher than that of the carbocation formation. So, for example, if the reaction (12.4.1) A + B → C Both cases can be addressed by using what is known as the steady state approximation.Recall our mechanism for the reaction of nitric oxide and oxygen:[latex]2\;\text{NO}(\text{g})\rightleftharpoons \text{N}_2\text{O}_2(\text{g})\quad\quad\quad\quad \text{Step 1}[/latex][latex]\text{N}_2\text{O}_2(\text{g})+\text{O}_2(\text{g})\xrightarrow{\text{k}_2} 2\;\text{NO}_2(\text{g})\quad\quad \text{Step 2}[/latex]Before, we assumed that the first step was fast, and that the second step was slow, thereby making it rate-determining.
For example, consider the following reaction:[latex]\text{H}_2(\text{g}) + 2\;\text{ICl}(\text{g})\rightarrow \text{I}_2(\text{g})+2\;\text{HCl}(\text{g})[/latex]The proposed reaction mechanism is given as follows:[latex](1)\;\text{slow reaction:}\quad \text{H}_2+\text{ICl}\rightarrow \text{HI}+\text{HCl}\quad\quad \text{rate}_1=\text{k}_1[\text{H}_2][\text{ICl}][/latex][latex](2)\;\text{fast reaction:}\quad \text{HI}+\text{ICl}\rightarrow \text{I}_2+\text{HCl}\quad\quad \text{rate}_2=\text{k}_2[\text{HI}][\text{ICl}][/latex]Since the first step is the rate-determining step, the overall reaction rate for this reaction is given by this step: [latex]\text{rate}=\text{k}[\text{H}_2][\text{ICl}][/latex]. This will be explored later in more detail.Elementary reactions are classified in terms of their Termolecularity is not common due to the rarity of three molecules colliding at the same time, and in just the right way for reaction to occur.The molecularity of the elementary step, and the reactants involved, will determine what the rate law will be for that particular step in the mechanism.