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## MCAT

### Course: MCAT > Unit 9

Lesson 18: Kinetics- Kinetics questions
- Introduction to reaction rates
- Rate law and reaction order
- Worked example: Determining a rate law using initial rates data
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reaction (with calculus)
- Collision theory
- The Arrhenius equation
- Forms of the Arrhenius equation
- Using the Arrhenius equation
- Elementary reactions
- Reaction mechanism and rate law
- Catalysts
- Kinetic and thermodynamic enolates

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# Reaction mechanism and rate law

A reaction mechanism is the sequence of elementary steps by which a chemical reaction occurs. Many reaction mechanisms contain one step that is much slower than the others; this step is known as the rate-determining step. If the rate-determining step is the first step in a mechanism, the rate law for the overall reaction can be derived directly from the stoichiometry of the step's balanced equation. Created by Jay.

## Want to join the conversation?

- How does one go about determining the elementary steps? how do you know which molecules will form together, before you get to the original equation? is there a certain technique, or is it just trial and error with different molecules?(24 votes)
- Usually, the steps are either given or determined experimentally. One way to go about writing the steps is if you are given the original equation and any intermediates, as you know that they intermediates must be produced and then used. So, to answer your question, it is trial and error in experiments, as far as I am aware. Hope this helps.(13 votes)

- so you use the slow step in the reaction rate equation?(5 votes)
- Yes. Imagine the slowest step as a bottleneck of a ketchup bottle. It hardly matters if the ketchup bottle is full but if the neck of the bottle is too constricted, ketchup will come out slowly.Similarly.the slowest step will determine the rate of formation of products.

Hope this helps. :D(23 votes)

- Doesn't this approximation require adjustments if you were dealing with elementary steps that took much longer than just 1 second? For example, if you had a slow reaction that occurred after 30 minutes and the fast one after 15 minutes, leaving the fast reaction out seems like it won't make for a good approximation. Or do most elementary steps in these mechanisms occur within seconds?(6 votes)
- Think about it this way instead. The slow step is the weakest link in the chain of reaction steps. A fast step will already be complete by the time the slow step finishes one cycle, so eventually every step faster than the slow step is slowed down to the slow step's rate, because the products from the slow step are needed to proceed to a faster step. That means that the overall reaction must proceed at the slow step's rate.(11 votes)

- How would you create a rate law for a mechanism in which the slow step is NOT the initial step?(7 votes)
- In this case, I suppose the overall rate would near the 2nd step rate(the slower one) because of the same reason. The time it takes for step one would be negligble compared to the time it takes for step 2, making our overall rate near the second step's rate.(4 votes)

- In this reaction, can we also consider NO2 the catalyst?(3 votes)
- I see why you are asking this since one NO2 is regenerated, but reactants are not considered to be catalysts. Catalysts are something other than a reactant that are added to a reaction to lower the activation energy and thereby speed up the reaction.(4 votes)

- What if there are 2 fast steps and 1 slow? How do you get the rate from there?(4 votes)
- Why is the rate law R=k[NO2]^2(3 votes)
- what is difference betn parallel, series and normal reactions?(1 vote)
- In a "normal" reaction, reactants give products: A → B.

In a "series" reaction, reactants give a product that then reacts to give another product: A → B → C.

In a "competitive parallel" reaction, a reactant gives two different products: A → B and A → C.(4 votes)

- At2:50,Can we also consider nitrogen dioxide as intermediate?(2 votes)
- No because the nitrogen dioxide is one of the reactant causing the chemical reaction to take place.(1 vote)

- what if the question asks to deduce the initial rate of formation of another product that's formed in the reaction?(2 votes)

## 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 NO2 plus NO2 giving us NO plus NO3 and then in the second elementary step the NO3 that we just made, right, NO3 plus CO gives us NO2 plus CO2. 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 NO2 plus NO2 plus NO3 plus CO and then we'll put all of
our products on one side so all of this stuff so we have NO plus NO3 plus NO2 plus CO2. Now we look on both sides on our reactants and our products and let's see what we can cross out. Well we have an NO2 on the left and we have an NO2 on the right so we can cross that out. We also have an NO3 on the left and an NO3 on the right
so we cross those out and we see what we are left with. We're left with NO2 plus
CO, so we're left with nitrogen dioxide plus carbon monoxide and then for our products we have nictric oxide, NO, and then we also have carbon dioxide, CO2. So we added our two elementary steps of our mechanism together, we added step one and step two together, and we got this for our
overall equation, alright, 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 NO3. So here we have this NO3 which we made in our first elementary
step and then we can see in the second step the NO3 is consumed so we call this NO3 an intermediate. So this is an intermediate. It's made in one 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 mechanisms. So if you're able 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 one NO2 plus one CO gives us one NO plus one 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 NO2 to the second power. So our reaction is second order in nitrogen dioxide and if we look at our
balanced equation here, here's nitrogen dioxide, notice it has a coefficient of one in
our balanced equation. So we can't just take
the coefficient of one and turn that into the exponent because experimentally we've determined the exponent to be a two and not a one. So you can only take the coefficient and turn it into an exponent if you're talking about
an elementary reaction. Also note that this reaction is zero order in carbon monoxide. So 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 NO3 and it turns out this is the
slow step of the mechanism and the second step of the mechanism we took our intermediate NO3 and 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 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 a much
smaller amount of time and I'm just gonna again make up a number. Let's say it takes one times ten to the
negative eight 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, one second plus one times ten
to the negative eight seconds so we 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. Alright, the slow step took one 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, 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 and we talked about in an earlier video how to find the rate law
for an elementary reaction. Alright, you would start by writing the 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 NO2 so
we put NO2 in brackets and since the coefficient here is a one 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 NO2 over here so we multiply that by
the concentration of NO2 and once again our coefficient is a one 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 rate determining step is equal to the rate constant
k1 times the concentration of NO2 and this would be squared, alright? So concentration of NO2
to the first power times concentration of NO2 to the first power is concentration of NO2 squared. And so we see that this has the same form, 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 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 one should be equal to k, the rate constant for
our overall reaction.