 # Rate law and reaction order | Knetics | Chemistry | Khan Academy

– [Voiceover] Let’s take
a reaction where A plus B gives us our products. And the lower case a and the lower case b represent the coefficients
for our balanced equation. It makes sense if we increase
the concentration of A and B, right, A and B would be
closer together in space and more likely to react, therefore increasing the
rate of our reaction. And this is true for most reactions. If you increase the
concentration of your reactants, you increase the rate of your reaction. We can check this by
doing some experiments. So let’s say we wanna figure out what the effect of the
concentration of A has on our rate of our reaction. So we’re gonna hold the
concentration of B constant, so we hold the concentration of B constant in our experiments. We change the concentration of A, and we see what effect that has
on the rate of our reaction. We’re going to use the
initial rate of the reaction. And that’s because as
our reaction proceeds, the concentration of
products will increase. And since reactions are reversible, if we have some products present, right, that can affect the rate of our reaction. And that’s not our goal. Our goal is to figure out
what the concentration, what effect the concentration
of our reactants has on our rate. And so we use the initial rate, where we have only reactants present, and no products. So in our first experiment, let’s say the concentration
of A is one molar, and the rate of our reaction, the initial rate of our reaction is .01 molar per second. And our second experiment, we increase the concentration
of A to two molar. We hold the concentration of B constant, and we observe the rate of our reaction to increase to .02 molar per second. So we’ve increased the concentration of A by a factor of two. And what happened to our rate? Our rate went from .01 to .02. So the rate increased by two as well. All right, let’s compare
our first experiment with our third experiment now. We’re going from a
concentration of A of one, to a concentration of A of three. So we’ve increased the concentration of A by a factor of three. And what happened to the rate? The rate went from .01 to .03. So the rate increased
by a factor of three. All right, to figure out the relationship, if you think to yourself,
two to what power X is equal to two? Obviously that would be two to the first. Two to the first is equal to two. All right, we could have done it for our other comparison as well. Three to what power X is equal to three? Obviously three to the
first is equal to three. So the rate, the rate of our
reaction is proportional to, and that’s what this
funny symbol means here, the rate of our reaction is proportional to the concentration of
A to the first power. All right, let’s do the same thing for the concentration of B. So we do some experiments where we change the concentration of B, and we see what effect that
has on our initial rate. So for all of these, we’re gonna hold the
concentration of A constant, therefore, whatever we
do to B is reflected in the rate of our reaction. So in our first experiment, the concentration of B is one molar and the rate is .01 molar per second. And then we change the
concentration of B to two molar. Right, we double the concentration of B while holding the
concentration of A constant. And we observe the initial
rate of our reaction to be .04 molar per second. So we’ve increased the
concentration of B, not A, and let me change that (laughs). We’ve increased the concentration of B by a factor of two. We’ve gone from one molar to two molar. And what happened to the rate? The rate went from .01 to .04. So we’ve increased the
rate by a factor of four. Let’s compare our first experiment with our third experiment now. We’re going from a concentration of B of one molar to three molar. So we’ve increased the concentration of B by a factor of three. And what happens to the rate? The rate goes from .01 to .09. So we’ve increased the
rate by a factor of nine. So now we think to
ourself, two to what power, I’ll make it Y, two to what power is equal to four? Obviously Y would be equal to two. Two to the second power is equal to four. Or three to what power Y is equal to nine? Obviously, three to the
second power is equal to nine. So we’ve determined that
the rate of our reaction is proportional to the concentration of B to the second power. All right, now we can put those together. We can put these together to write what’s called a rate law. Ok, So we know that the
rate of our reaction is proportional to the concentration of A to the first power, and we know that our rate is proportional to the concentration of
B to the second power. And then we put in, we
put in what’s called a rate constant here, K. And this represents our rate law. So let’s go through these one by one here. So, capital R is the rate
of our reaction, right? This is the rate of our reaction. All right? K is what’s called the rate constant. So this is the rate constant. And there’s a difference
between the rate of our reaction and the rate constant. If we change the concentration
of our reactants, we change the rate of our reaction. But if we change the
concentration of our reactants, we don’t change the rate constants, right? And this is constant. It does depend on the temperature, though, so we’ll talk about that in later videos. Here we have that the reaction is concentration of A to the first power. We say the reaction is first order in A. So we say that our
reaction is first order, first order in A. And we found, we found that
it’s second order in B. Right, so we had a two here. So this is second order, second order in B. And we can also talk
about the overall order of our reaction. So if we’re first order in A, right, we’re first order in A, and second order in B, the overall order, the overall order would be one plus two,
which is equal to three. So the overall order of
our reaction is three. All right, let’s go back up
here to the general reaction that we started with, all right, so let’s go back, right back up to here. We have, we have this. And let’s write a general rate law. So if this is your reaction,
your general rate law would be R is equal to your rate constant, times the concentration
of A to some power, I’ll make it X, times
the concentration of B to some power which I will make Y. And the reason why I’m showing you this, is to show you that you can’t
just take your coefficients, right, you can’t take your coefficients and stick them into here. Right? So it doesn’t work that way. You’d have to know the
mechanism of your reaction. So these orders have to be
determined experimentally. So you have to look at your
experimental data here. And the orders affect the
units for your rate constant. For example, let’s go back down to here. And let’s figure out the
units for the rate constant for this example. So the rate of our reaction, the rate of our reaction was
in molar per seconds, right? This is molar per second. We’re trying to find the units for K. The units for concentration are molar. All right, so this would be molar, and this would be to the first power. And this would be molar
to the second power. So we’d have molar to the second power. All right, so solving for K, right, you could just go ahead and cancel out one of these molars right
here, and solve for K. So you would get, this
would be one over seconds now on the left. So one over seconds, right,
and divide by molar squared. So one over seconds times molar squared. Or you could write this one over molar squared times seconds. Those would be your units for K for this reaction, right? With an overall order of three. But it can change. Right? It can change depending on the order. Now let’s look at this reaction. We have only one reactant, A, turning into our products. And if we look at the two experiments, in our first experiment,
the concentration of A is one molar, and the
initial rate of reaction is .01 molar per second. If we double the concentration
of A to two molar, the rate stays the same. It’s still point zero
one molar per second. So even though the concentration of A is going from one molar to two molar, right, that’s doubling the concentration, or increasing the concentration
of A by a factor of two, the rate stays the same. So you could say, the rate,
it’s the rate times one. ‘Cause it’s the same rate. So two, all right, so two to what power X, two to what power X is equal to one? Obviously X would have
to be equal to zero. Two to the zero power is equal to one. So any number to the zero
power is equal to one. So this reaction is zero order, it’s zero order in A. Now if we wanted to write our rate law, we would write the rate of the reaction is equal to the rate constant K times the concentration of A. We have only one reactant here. And since this is zero order in A, we could just write the
rate of the reaction is equal to the rate constant K. And so if you wanted to know the units for the rate constant K, well, the rate is in molar per second. And so those would also
be your units for K. K would be in molar per second. So here’s an example of how
your units for K change, depending on the overall

## 58 thoughts on “Rate law and reaction order | Knetics | Chemistry | Khan Academy”

• ### Audrey Staten Post author

thanks for posting this video.

• ### Vera Jansen Post author

Thank you so much! Btw you sound like Tyler Oakley

• ### KunaiKrazy Post author

Why can't HE be my chemistry professor? I actually get this.

• ### Pratik Patel Post author

the best explaination ever on chemical kinetics …Bingo U hit the target of my mind Thanx

• ### Tiffany Hewitt Post author

THANK YOU!!!!!!

• ### McDonalds Lunch-able Post author

This was REALLY good at explaining the topic clearly and understandably. Thanks!!

• ### debora vidal Post author

amazing video. Thank you!!

• I was freaking out because my lecture notes didn't have anything on "third order rate laws". Thanks so much!

• ### Master Explosive Post author

• ### Daniel Krichavets Post author

i love this guy

• ### Aminah Emeran Post author

• ### Michael Seh Post author

this guy teaches me so well

• ### PandaLover Jess Post author

Thankieeeeew so much this was a big help!!

• ### Honest Lie Post author

I have a Chem final in 3 days and omg this made everything so clear
Thank you 🙌🏽

• ### GAMERFLAMER Post author

WHERE'S SAL?!?!?!!!!

• ### Caroline Lana Post author

HELP! what if the one im trying to work out only has 1 table of initial rates? do i do what you did but compare A and B to that one table of initial rates?

• ### eragon2121 Post author

Is the teacher for organic chemistry Mr. Anderson from Bozeman Science?

• ### Tico Post author

I needed Zero order reaction explanation. Thanks, bro.

• ### Monkey D. Luffy Post author

Thank you T^T

• ### OwenGTA Post author

Fuck taking chem 2 over the summer.

• ### JemuzuDatsWho Post author

thank you!

• ### rabya banory Post author

thanku man it was very helpful

• ### Me, a rat. BTS, the Pied Piper Post author

I have A levels chemistry finals tomorrow and you just saved me. THANK YOU

• ### lollopisemis Post author

I'm not even given any concentrations but they still want me to calculate the reaction order… I have no idea what to do -.-

• ### Frank J Post author

Such a good job explaining this.

• ### Taylor Clay Post author

thank you! this was a big help

• ### Khanh Tran Post author

There's a spelling error in the title. Goddamn it Khanacademy.

• ### Vanessa Post author

Thank you so much

• ### Steve Gerrish Post author

THANK YOU!!!!

• ### Anthony Chinedu Ifedigbo Post author

I got a negative and a number in ratio form as my answer eg. -1.765. Any help?

• ### Navoda De Silva Post author

You are awesome! Thank you so much!

• ### Garen Post author

Anyone know when to use this
average rate of reaction = (-1/a)(deltaA/deltat)
First order and second order
ln(k2/k1) = -Ea/R(1/t2-1/t1)

• ### Abdul Majeed Abdul Majeed Post author

• ### Dania Hasan Post author

this was a great way to explain this, thanks a lot ! 🙂

• ### Ben Roberts Post author

At the end, shouldn't the units for k be mol/L*s ?

• ### Komal Bhardwaj Post author

thanks

• ### NellioStyle Post author

best explication ever Thanks you

• ### Official Kelby Post author

Making MCAT studying 10x manageable. Thank you!

• ### HEHEHE I AM A SUPAHSTAR CHIMERA Post author

Makes sense.

• ### c h a m c h a m Post author

Thank you very much! You saved me from my report tomorrow!

• ### Vanessa Escalera Post author

saved my life

• ### XetoDesigns Post author

do pseudo first order reactions

• ### Jessica Post author

Thank you so much , this was very very helpful 💕😩🙏🏾👏🏾

• ### dua syed Post author

I love u

• ### Anita Rana Post author

Thanks

• ### retoo luvyuhx Post author

YOU ARE THE BEST

• ### Sharia Alam Post author

The sound is TOO low!

• ### Gerson Cocjin Jr. Post author

what if concentration [A] stays the same and the rate increased by 2 what would happen in that case

• ### R. Abdul Wahid Post author

excellent explanation <3 thank you so much!

• ### Samuel Michael Post author

Why there is a difference in the order between different reactants?
Why there is a zero order or a first order .. etc

• ### james poo Post author

kinetics

• ### Ray George Post author

absolute legend! xx

• ### Alvin Stjohn Post author

Better teacher than my fucking useless waste of a cunt university professor

• ### Pooria Karimi Post author

this guy is SUPERB he reminds me of high school forgoten topics

• ### Michael Kawamura Post author

Two years later and I'm surprised no one has realized that the title is misspelled. It's "kinetics", not knetics.

• Is rate always m/s or it can change?

• ### RTG Super Post author

• 