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### Course: MCAT > Unit 10

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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# Worked example: Determining a rate law using initial rates data

The rate law for a chemical reaction can be determined using the method of initial rates, which involves measuring the initial reaction rate at several different initial reactant concentrations. In this video, we'll use initial rates data to determine the rate law, overall order, and rate constant for the reaction between nitrogen dioxide and hydrogen gas. Created by Jay.

## Want to join the conversation?

- When we talk about initial rate of a reaction, is that a INSTANTANEOUS RATE of a product or sum of all the products or sum of all reactant ? I have an practice question in my AP Chemistry book by Pearson and they dont have answer key.

Given a reaction C2H5Br + OH- ---> C2H5OH + Br- , has rate law has rate= k[C2H5Br][OH] . When [C2H5Br}= 0.0477 and [OH-]=0.100 M , the rate of disappearance of ethyl bromide is 1.7 x 10^-7 M/s. What is the value of k, rate constant?

If it is a sum of all reactants, I got k= 7.12 x 10^-5(12 votes)- An instantaneous rate is the slope of a tangent to the graph at that point.

An average rate is the slope of a line joining two points on a graph.

If the two points are very close together, then the instantaneous rate is almost the same as the average rate.

So the initial rate is the average rate during the very early stage of the reaction and is almost exactly the same as the instantaneous rate at t = 0.

If rate = k[C₂H₅Br][OH⁻], then

k = rate/([C₂H₅Br][OH⁻]) = 1.7 × 10⁻⁷mol·L⁻¹s⁻¹/(0.0477 mol·L⁻¹ × 0.100 mol·L⁻¹) =

3.6 x 10⁻⁵ L·mol⁻¹s⁻¹(17 votes)

- You've mentioned in every video, the unit of concentration of any reactant is (M) that is (Mol) and the unit of rate of reaction to be (M/s). But what we've been taught is that the unit of concentration of any reactant is (mol.dm^-3) and unit of rate of reaction is (mol.dm^-3.s^-1) . Can you please explain that?

Thank you.(4 votes)- M is the symbol for molarity, not moles.

1 M = 1 mol / L = 1 mol / dm³(17 votes)

- how can you raise a concentration of a certain substance without changing the concentration of the other substances?(7 votes)
- One of the reagents concentrations is doubled while the other is kept constant in order to first determine the order of reaction for that particular reagent. This is done because in the equation for the rate law, the rate equals the concentrations of the reagents raised to a particular power. The order of reaction with respect to a particular reagent gives us the power it is raised to(3 votes)

- In our book, they want us to tell the order of reaction by just looking at the equation, without concentration given! How would you decide the order in that case?

Thanks.(3 votes)- Late, but maybe someone will still find this useful.

Sometimes you can tell the reaction order from stoichiometry, but ONLY if the reaction is*elementary*- that means that it happens in one step, with no intermediates. I'm assuming that's what your textbook meant. In that case, the reaction order with respect to a reactant is equal to its stoichiometric coefficient, and the overall reaction order is the sum of those. For example:

A + B -> C

The stoichoimetric coefficient of A is 1, so the reaction is 1st order in A. It's also 1 for B, so the reaction is 1st order in B as well. Then you add their orders together and find that the reaction is 2nd order overall.

In practice, reaction order is only determined experimentally, and it's actually what helps us determine the reaction mechanism. For example, if reaction orders line up with stoichiometric coefficients in the equation, you know it's a one-step mechanism. If it doesn't, you can tell there are more steps, and you have to devise a mechanism that agrees with the experimentally determined rate. It only really works the other way around in, yeah, textbook problems ("assuming xyz is an elementary reaction, determine reaction order").(9 votes)

- I get k constant as 25 not 250 - could you check?(2 votes)
- Make sure the number of zeros are correct. I'm getting 250 every time. You should be doing 1.25x10^-5 / ((.005^2) x (.002))(7 votes)

- is it possible to find the reaction order ,if concentration of both reactant is changing .

is there anyway to cancel out the effect of other reactant ??(2 votes)- Yes. To the first part, the changing concentrations have nothing to do with the order, and in fact, the way in which they change
**is**in fact the order. So, if they were not changing, then we cannot determine the order.

As to the second part, yes. In fact, cancelling out the effect of all but one reactant is the standard method of finding the order. The method used is called flooding. In this, all but one are taken in very high concentrations (excess) such that even if the entire reaction were to take place and our reactant (in small amount) were to be entirely consumed, the net change in the concentrations of the remaining would be negligible, and hence can be taken as zero (constant concentration). Hence all their concentration along with the actual rate constant now form a new constant which can be computed while measuring the rate with respect to one reactant only. By doing this for every reactant, the original constant can be determined. To note, although we are taking the concentrations to be constant, it is just a mathematical step (the change being closed to zero) and not in reality, since their concentrations need to change for the reaction to occur (or at least, they need to be used up).(4 votes)

- Why did he first use the difference between experiment 1 and 2 for NO and didn't use the difference between let's say exp. 2 and 3?

And why is it the opposite for H2?(2 votes)- When getting the concentration change of NO, he wanted to make sure the H2 was constant, so it wasn't affecting his calculation. Same for H2. Actually, he answers better than I can: go to 1 min. 10 sec. of the video.

Hope this helps.(2 votes)

- at1:20so we have to use the experiment/trial that have a constant concentration when we want to determine the order of reaction?(2 votes)
- You need to run a series of experiments where you vary the concentration of one species each time and see how that changes the rate.(2 votes)

- What if i was solving for y (order) of a specific concentration and found that 2^y=1.41? I know that y has to be an integer so what would i round 1.41 to in order to find y?(2 votes)
- "y" doesn't need to be an integer - it could be anything, even a negative number.(2 votes)

- What if one of the reactants is a solid? How would you measure the concentration of the solid?(1 vote)
- You can't measure the concentration of a solid. But what would be important if one of the reactants was a solid is the surface area of the solid. The finer the solid is ground (and hence the larger the surface area), the faster the reaction will take place.(4 votes)

## Video transcript

- [Voiceover] Now that we
understand how to write rate laws, let's apply this to a reaction. Here we have the reaction of
nitric oxide, which is NO, and hydrogen to give us nitrogen and water at 1280 degrees C. In part A, our goals is
to determine the rate law. From the last video, we
know that the rate of the reaction is equal to K,
which is the rate constant, times the concentration of nitric oxide. Nitric oxide is one of our reactants. So the rate of the reaction
is proportional to the concentration of nitric
oxide to some power X. We don't know what X is yet. We also know the rate of
the reaction is proportional to the concentration
of our other reactant, which is hydrogen, so
we put hydrogen in here. But we don't know what the
power is so we put a Y for now. Alright, we can figure
out what X and Y are by looking at the data in our experiments. So let's say we wanted to
first figure out what X is. To figure out what X is
we need to know how the concentration of nitric oxide affects the rate of our reaction. We're going to look at
experiments one and two here. The reason why we chose
those two experiments is because the concentration of hydrogen is constant in those two experiments. The concentration of hydrogen is point zero zero two molar in both. If we look at what we
did to the concentration of nitric oxide, we went
from a concentration of point zero zero five to a concentration of point zero one zero. We increased the concentration of nitric oxide by a factor of two. We doubled the concentration. What happened to the
initial rate of reaction? Well the rate went from
one point two five times 10 to the negative five to five
times 10 to the negative five. The rate increased by a factor of four. We increased the rate by a factor of four. If you have trouble doing
that math in your head, you could just use a
calculator and say five times 10 to the negative five
and if you divide that by one point two five times
10 to the negative five, this would be four over one, or four. This rate is four times this rate up here. Now we know enough to figure
out the order for nitric oxide. Remember from the previous
video, what we did is we said two to the X is equal to four. Over here, two to the X is equal to four. Obviously X is equal to two,
two squared is equal to four. So we can go ahead and put
that in for our rate law. Now we know our rate is equal
to K times the concentration of nitric oxide this would
be to the second power. So the reaction is second
order in nitric oxide. Next, let's figure out the
order with respect to hydrogen. So this time we want to
choose two experiments where the concentration of
nitric oxide is constant. That would be experiment
two and three where we can see the concentration of
nitric oxide has not changed. It's point zero one molar for
both of those experiments. But the concentration
of hydrogen has changed. It goes from point zero zero
two to point zero zero four. So we've increased the
concentration of hydrogen by a factor of 2 and what happened to the rate of reaction? Well it went from five times
10 to the negative five to one times 10 to the negative four so we've doubled the rate. The rate has increased by a factor of two. Sometimes the exponents bother students. How is this doubling the rate? Well, once again, if you
can't do that in your head, you could take out your
calculator and take one times 10 to the negative
four and divide that by five times 10 to the
negative five and you'll see that's twice that so the rate
goes up by a factor of two. Now we have two to what
power is equal to two? So two to the Y is equal to two. Obviously Y is equal to one. Two to the first power is equal to two. So know we know that our reaction is first order in hydrogen. We can go ahead and put that in here. We can put in hydrogen and we know that it's first order in hydrogen. We've now determined our rate law. In part B they want us to find the overall order of the
reaction and that's pretty easy to do because we've already determined the rate law in part A. We know that the reaction is second order in nitric oxide and
first order in hydrogen. To find the overall order, all we have to do is add our exponents. Two plus one is equal to three so the overall order of
the reaction is three. Let's compare our exponents
to the coefficients in our balanced equation
for a minute here. It's very tempting for
students to say oh, we have a two here for our
coefficient for nitric oxide, is that why we have a two down here for the exponent in the rate law? But if you look at hydrogen,
hydrogen has a coefficient of two and we determined that the exponent was a one
down here in the rate law. You can't just take your
coefficients and your balanced chemical equation
and put them in for your exponents in your rate law. You need to look at your
experimental data to determine what your exponents are in your rate law. Later we'll get more into mechanisms and we'll talk about
that a little bit more. Alright, let's move on to part C. In part C they want us
to find, or calculate, the rate constant K. We could calculate the
rate constant K by using the rate law that we determined
in part A and by choosing one of the experiments and plugging in the numbers into the rate
law so it doesn't matter which experiment you choose. You could choose one, two or three. I'm just going to choose
one here, so experiment one. We're going to plug all of
our information into the rate law that we just determined. For example, in our rate law we have the rate of reaction over here. Well, for experiment one,
the initial rate of reaction was one point two five times
10 to the negative five. And it was molar per second
so we're going to plug this in to our rate law. So let's go down here
and plug that value in, one point two five times
10 to the negative five and this was molar per second. Alright, so that takes care
of the rate of the reaction. Next, we have that equal
to the rate constant K, so we're trying to solve for K, times the concentration
of nitric oxide squared. Let's go back up here and
find the concentration of nitric oxide in the first experiment. The concentration is point
zero zero five molar. We're going to plug in point
zero zero five molar in here. We have point zero zero five molar. Next, we're going to multiply
that by the concentration of hydrogen to the first power. We go back up to experiment
one and we find the concentration of hydrogen which is point zero zero two
molar so we plug that in. We have zero point zero zero two molar. Next, all we have to do is solve for K. Let's go ahead and do that so let's get out the calculator here. We could say point zero
zero five squared gives us two point five times 10
to the negative five, we need to multiply that
by point zero zero two. On the right side we'd have five times 10 to the negative eight. So we have five times 10
to the negative eight. The thing about your units,
this would be molar squared times molar over here
and all of this times our rate constant K is equal to one point two five times 10 to the
negative five molar per second. Let's go ahead and find
the number first and then we'll worry about our units here. To find what K is, we just
need to take one point two five times 10 to the
negative five and if we divide that by five times
10 to the negative eight then we get that K is equal to 250. We can go ahead and put that in here. K is equal to 250, what
would the units be? Well, we have molar on the left,
we have molar on the right, so we could cancel one
of those molars out. On the left we have one over
seconds and on the right we have molar squared so
we divide both sides by molar squared and we
get, for our units for K, this would be one over
molar squared times seconds. We've found the rate
constant for our reaction. And notice this was for
a specific temperature. Our reaction was at 1280
degrees C so this is the rate constant at 1280 degrees C. Finally, let's do part D. What is the rate of the reaction when the concentration of nitric
oxide is point zero one two molar and the concentration of hydrogen is point zero zero six molar. Well, we can use our rate law. Our rate law is equal
to what we found in A, our rate law is equal to
K times the concentration of nitric oxide squared
times the concentration of hydrogen to the first power. Our goal is to find the rate
of the rate of reaction. We're solving for R here
and we know what K is now. K is 250 one over molar
squared times seconds. The concentration of nitric
oxide is point zero one two, so we have point zero one two
molar and then we square that. All I did was take this
and plugged it into here and now we're going to
take the concentration of hydrogen, which is
point zero zero six molar and plug that into here. Let's go ahead and do
that, so that would be times point zero zero six molar, let me go ahead and
put in the molar there, so point zero zero six
molar to the first power. And we solve for our rate. The rate is equal to,
let's do the numbers first. We have point zero one two squared. We're going to multiply
that, so times point zero zero six and then we also
need to multiply that by our rate constant K so times 250. This gives us our answer of two point one six times 10 to the negative four. Let's round that to two
point two so we have two point two times 10
to the negative four. In terms of our units, if
we think about what happens to the units here, we would
have molarity squared, right here molarity
squared molarity squared so we end up with molar
per seconds which we know is our units for the rate of
reaction, so molar per seconds. We found the rate of our reaction.