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Main content
Current time:0:00Total duration:8:42

Reaction mechanism and rate law

AP.Chem:
TRA‑5 (EU)
,
TRA‑5.A (LO)
,
TRA‑5.A.1 (EK)
,
TRA‑5.A.2 (EK)
,
TRA‑5.A.3 (EK)
,
TRA‑5.B (LO)
,
TRA‑5.B.1 (EK)

Video transcript

a mechanism is the sequence of elementary steps by which a reaction proceeds so if we look at this reaction here we have nitrogen dioxide plus carbon monoxide giving us nitric oxide and carbon dioxide our two reactants don't go to our products in one step there's a sequence of elementary steps by which the reaction proceeds there's a mechanism for this reaction and one possible mechanism for this reaction shows the mechanism in two elementary steps in the first step step one we can see it's no.2 plus no.2 giving us n o plus no.3 and then in the second elementary step the no.3 that we just made right no.3 plus Co gives us no.2 plus co2 and any possible mechanism must have elementary steps that add up to give the overall reaction so if we add these two steps together we should get the overall reaction so let's put all of our reactants on one side so we have all of our reactants over here so that would be no.2 plus no.2 plus no.3 plus CE o and then we'll put all of our products on one side so all of this stuff so we have n o plus no.3 plus no.2 plus co2 now we look on both sides on our reactants and our products and let's see what we can cross out but we have an no.2 on the Left we have an no.2 on the right so we can cross that out we also have an no.3 on the left and an no.3 on the right so we cross those out and we see what we are left with we're left with we left with no.2 plus Co so we're left with nitrogen dioxide plus carbon monoxide and then for our products we have we have nitric oxide and O and then we also have carbon dioxide co2 so we added our two elementary steps of our mechanism together we added step 1 and step 2 together we got we got this for our overall equation right which is the same which is the same one that we talked about up here so this is a possible mechanism for this reaction also note the presence of no.3 so here we have this no.3 which we made in our first elementary step and then we can see in the second step the no.3 is consumed so we call this no.3 an intermediate so this is an intermediate it's made in 1 step of our mechanism and consumed in another step so the intermediate is not a reactant it's not a product it's something that we can detect sometimes and this helps us figure out mechanism so if you're able to to detect a certain intermediate that helps you figure out possible mechanisms for reactions a possible mechanism must also be consistent with the experimental rate law for the overall reaction our overall reaction is 1 no2 plus 1 co gives us 1 nm plus 1 co2 and the experimental rate law for this overall reaction is the rate of the reaction is equal to the rate constant K times the concentration of no.2 to the second power so our reaction is second order in nitrogen dioxides and if we look at our balanced equation here here's nitrogen dioxide notice it has a coefficient of 1 in our balanced equation so we can't just take the coefficient of 1 and turn that into the exponent because experimentally we've determined the exponent to be a 2 and not a 1 so you can only take the coefficient and turn into an exponent if you're talking about an elementary reaction also note that this reaction is zero-order in carbon monoxide alright so this you don't see carbon monoxide appearing in your rate law let's look at the mechanism and see if we can understand or explain the experimental rate law so the mechanism in step one of the mechanism we formed our intermediate which was no.3 and it turns out this is the slow step of the mechanism and the second step of the mechanism we took our intermediate no.3 and we formed we formed another product over here this is the fast step of the mechanism and the slow step of the mechanism is called the rate determining step so step one here of our mechanism is the rate determining step and the reason why this is the rate determining step let's let's make up some times here so we can understand this a little bit better let's say that this first step the formation of our intermediate takes one second so it takes one second to form our intermediate and then the second step of the mechanism is fast compared to the first so it should take it should take a much smaller amount of time and I'm just going to again make up a number let's say it takes 1 times 10 to the negative 8 seconds for the second step of our mechanism to occur so what's the total time for our reaction well we would just add those two times together 1 second plus 1 times 10 to the negative 8 seconds so we'd get approximately one second for the total time of our reaction so the effective rate of the reaction is determined by the time consumed during the slow step all right the slow step took 1 second and that's pretty much how long our overall reaction took so we can say that if we're trying to find the rate of our overall reaction right trying to find the rate of the overall reaction that should be equal to the rate of the rate determining step as we've just seen by this example over here so if we can figure out the rate law for the rate determining step that's a good approximation of the overall rate of our reaction so we can figure out the rate of the rate determining step because this is an elementary reaction step one is our rate determining step and this is an elementary reaction we talked about in an earlier video how to find the rate law for an elementary reaction right you would start by writing rate of the reaction is equal to the rate constant for step one the rate constant is K sub one so the rate constant is K sub one times the concentration of our reactant let's see we have no.2 so we put no.2 in brackets and since the coefficient here is a 1 remember for an elementary reaction you can take the coefficient and turn that into an exponent so this would be to the first power and then we have another no.2 over here so we multiply that by the concentration of no.2 and once again our coefficient is a 1 so since this is an elementary reaction we take our coefficient and we turn that into an exponent so now we have our rate law for our rate determining step and we could even go further we could say that the rate of our our rate determining step is equal to the rate constant k1 times the concentration of no.2 and this would be squared all right so concentration of no.2 to the first power times concentration of no.2 to the first power is concentration of no.2 squared and so we see that this has the same form right this this rate law that we just wrote has the same form as the experimental rate law and so our mechanism is consistent with the experimental rate law for the overall reaction so our overall reaction is second order in nitrogen dioxide and zero order in carbon monoxide notice that carbon monoxide doesn't appear until the second step right so here's where carbon carbon monoxide appears and that's not the rate determining step and so that's one way to think about why our reaction is zero-order in carbon monoxide also notice that this k1 here for the rate determining step K sub 1 should be equal to K the rate constant for our overall reaction