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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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# Second-order reaction (with calculus)

Deriving the integrated rate law for second order reactions using calculus. How you can graph second order rate data to see a linear relationship.

## Want to join the conversation?

- What's the highest order a reaction can be theoretically?(11 votes)
- I'm not sure that there is a theoretical limit to it, but there is definitely a real limit. Reactions of order 3 or above are very rare. In reactions it is necessary for the reacting components to meet in a particular orientation in one instant, and the probability of this occurring decreases as the number of molecules increases. Let me know if that helped :)(14 votes)

- How would you figure out which integrated rate equation (first, second, zero) to use based on the units of the rate constant?(5 votes)
- There is a pattern.

If n is the order of the reaction, the units of k are (mol·L⁻¹)^(1-n)s⁻¹.

If n = 0, the units are (mol·L⁻¹)^(1-0)= (mol·L⁻¹)¹s⁻¹ = mol·L⁻¹s⁻¹

If n = 1, the units are (mol·L⁻¹)^(1-1)= (mol·L⁻¹)⁰s⁻¹ = s⁻¹

If n = 2, the units are (mol·L⁻¹)^(1-2)= (mol·L⁻¹)⁻¹s⁻¹ = L·mol⁻¹s⁻¹

If n = 3, the units are (mol·L⁻¹)^(1-3)= (mol·L⁻¹)⁻²s⁻¹ = L²·mol⁻²s⁻¹

To go the other way, you reverse the process.

If the units of k are (mol·L⁻¹)^n, The order of the reaction is 1 - n.

If the units of k are mol·L⁻¹s⁻¹, n = 1. The order is 1 - 1 = 0.

If the units of k are s⁻¹, n = 0. The order is 1 - 0 = 1.

If the units of k_are L·mol⁻¹s⁻¹, n = -1. The order is 1 – (-1) = 2.

If the units of k are L²mol⁻²s⁻¹, n = -2. The order is 1 – (-2) = 3.

Then you use the integrated rate law that corresponds to the order of the reaction.(23 votes)

- What if the reaction is A+B->C, and is second order? (R=k[A][B]) Then how do you derive the integrated rate law?(5 votes)
- The derivation is too complicated to reproduce in this comment box, but this link explains how to derive the integrated rate law for an overall second order reaction with two reactants:

https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions.(4 votes)

- What is the law that he used to integrate [A]^-2*d[A] can anyone tell me that?that is what formula he used?(3 votes)
- I thought 2nd order is 2A ---> products. Im confused now about the difference between 1st and 2nd order.(1 vote)
- The order respect to a reactant does not necessarily correspond to the coefficients of the reaction, this because a reaction can occur in multiple elementary reactions.

For example the reaction 2NO + Br2 ----> 2NOBr has the experimental rate law

Rate = k {NO} {Br2}

This can be explained considering that the reaction occurs in 2 steps:

NO + Br2 ----> NOBr2 ( The first reaction is the slow step, therefore it dictates the rate)

NOBr2 + NO -----> 2NOBr

-----------------------------------------

2NO + Br2 ----> 2NOBr ( The sum of the 2 reactions above gives the original reaction)(5 votes)

- Why did you not move the negative to the other side when integrating like you did in the beginning of the first order derivation video?(3 votes)
- My guess is because the initial concentration is in the denominator and so a larger value indicates a smaller denominator..hence... the concentration is getting lower as time continues(1 vote)

- why a stright line for second order reaction ?

how could u indicate it ?(1 vote)- Any equation of the form y=mx plus b is essentially a straight line. (where m is the 'slope', that is how fast the y value changes with respect to the x value and b is the y intercept, i.e. the value of y when x is zero) why? figure it out because its so much fun doing it yourself...(3 votes)

- When do you know to use the integrated rate law?(1 vote)
- whenever the question states concentration of a substance with respect to time, it is an integrated rate law.(3 votes)

- At1:47wouldn't we take the derivative of both sides? In order for the equation to not change?(1 vote)
- Is it possible to have a negative K value in a second order reaction? If so how would a negative k be graphed if second order should have a positive slope?(1 vote)
- What would a negative value for
**k**mean?

In other words, what would decreasing the value of`1/[A]`

mean about the reaction? And Is this result physically possible?(1 vote)

## Video transcript

- Let's say that our reaction here is second order in A. So A turns into our products. And when time is equal to zero, we're starting with our
initial concentration of A. And after some time, t would have the concentration of A at that time, t. So if we're expressing
the rate of our reaction, we know we can do that in
a couple different ways. So we can say that the rate is equal to the negative change in
the concentration of A, over the change in time. So we've done this in earlier videos. We can also say that our rate is equal to uh, think about the rate law. The rate is equal to the rate constant k times the concentration of A. And since I said this
is second order in A, this is a second order reaction, this is the concentration
of A to the second power. Now we can set these two rates equal. We can set this equal to
this, so let's do that. We have the negative change in the concentration of A,
over the change in time, and that's equal to the rate constant k times the concentration of A squared. Next, we need to think about our calculus. So instead of expressing
this as an average rate, change in A over change in time, we can think about the instantaneous rate. So the rate of change of
the concentration of A with respect to time. So we can write negative
d[A], d concentration of A, over dt, so the rate of change in the concentration of A with respect to time. And this is equal to k A squared. And we have our differential equation. All right, and we can solve
our differential equation and get a function, and the first thing that we do to solve a differential equation is
to separate our variables. We need to put the As on one side, and we need to put the
t on the other side, the dt on the other side. So we divide both sides by a squared, so now on the left we have negative d[A] over concentration of A squared, and we multiply both sides by dt. So we get kdt on the right side. Next, we're ready to integrate. So we're ready to integrate, and let's rewrite what
we have on the left here. So this would be the negative. I'm going to say this is
the concentration of A to the negative two power, d[A], just to make the
integral a little bit easier to follow. And on the right here, we still have kdt. All right, so we're ready to integrate. So we're going to integrate the left here, and we're going to integrate the right. K is a constant, so we can
pull it out of our integral. And let's go back up
here and remind ourselves what we would be integrating from. So for time, all right
let's look right here. We're going from time is equal to zero, to time is equal to t. And for our concentration,
we'd be going from our initial concentration
to our concentration at time of t. So let's go ahead and put those in, so we have, we're
integrating from zero to t, so from zero to t. And we're integrating from
our initial concentration, our initial concentration
to our concentration at some time, t. All right, so on the left side, what do we have here? So what's the integral. This is like X to the negative two DX. So it's like the integral
of X to the negative two DX here. So that would give us
the concentration of A to the negative first, over negative one. So obviously you have to had some calculus if you're watching the
video at this point. And then we still have our negative sign. So let me go ahead and put this in here. This negative right sign
here is still there. And we will be evaluating this, this would be equal to, let me just rewrite this here. So this is the same as,
this will be the same as one over the concentration of A. So you have two negative signs, and so now you have a positive. So this would be one over
the concentration of A, since A to the negative
first is one over the concentration of A. We're evaluating this from
our initial concentration to our concentration at time t. And on the right side, the integral dt would just be t. So we have kt, evaluated from zero to t. Next we use the fundamental
theorem of calculus. So we go ahead and plug in what we have. So we plug in this first. So we have one over the
concentration of [A]t minus one over the initial
concentration of A, and then on the right side, this of course would just be kt. So we found our integrated rate law for a second order reaction. So this is our integrated rate law. Some people call this the
integrated rate equation. It doesn't really matter which
one you're thinking about. But this, this is very useful, because we can rearrange this. We can move the one over
initial concentration of A to the right side, so we get one over the concentration of A is equal to kt plus one over the
initial concentration of A. And now this looks very familiar. This is the equation for a straight line. We can see that this
would be Y is equal to MX plus B. So if you're graphing time on your X axis, and on your Y axis you have one over the concentration of A, you're gonna get a straight line. And the slope of that straight line, the slope which is M, the slope should be K, your rate constant. And the Y intercept here which is B, should be one over the
initial concentration of A. So let's really quickly
sketch out this graph here. So we have our axes. And on our X axis, we're gonna put time. So on our X axis we put time, and on the Y axis we put one over the concentration of A. So we have one over
the concentration of A. And if this is a second order reaction, your graph should be a straight line. So I'm gonna go ahead and
draw a straight line in here, or attempt to draw a straight line. So you get the idea. And the slope of this line, the slope of our line is equal to k. It's equal to the rate constant. That's what we saw up here, the slope is equal to the rate constant. And the Y intercept, so
this point right here, this should be one over the
initial concentration of A. So this point right here is one over the initial concentration of A. So that's the idea of
the integrated rate law, or the integrated rate equation for a second order reaction.