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

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– [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

order of your reaction.

Audrey StatenPost authorthanks for posting this video.

Vera JansenPost authorThank you so much! Btw you sound like Tyler Oakley

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

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

Tiffany HewittPost authorTHANK YOU!!!!!!

McDonalds Lunch-ablePost authorThis was REALLY good at explaining the topic clearly and understandably. Thanks!!

debora vidalPost authoramazing video. Thank you!!

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

Master ExplosivePost authorVery helpful, thanks.

Daniel KrichavetsPost authori love this guy

Aminah EmeranPost authorthis way so helpful!

Michael SehPost authorthis guy teaches me so well

PandaLover JessPost authorThankieeeeew so much this was a big help!!

Honest LiePost authorI have a Chem final in 3 days and omg this made everything so clear

Thank you ππ½

GAMERFLAMERPost authorWHERE'S SAL?!?!?!!!!

Caroline LanaPost authorHELP! 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?

eragon2121Post authorIs the teacher for organic chemistry Mr. Anderson from Bozeman Science?

TicoPost authorI needed Zero order reaction explanation. Thanks, bro.

Monkey D. LuffyPost authorThank you T^T

OwenGTAPost authorFuck taking chem 2 over the summer.

JemuzuDatsWhoPost authorthank you!

rabya banoryPost authorthanku man it was very helpful

Me, a rat. BTS, the Pied PiperPost authorI have A levels chemistry finals tomorrow and you just saved me.

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

Frank JPost authorSuch a good job explaining this.

Taylor ClayPost authorthank you! this was a big help

Khanh TranPost authorThere's a spelling error in the title. Goddamn it Khanacademy.

VanessaPost authorThank you so much

Steve GerrishPost authorTHANK YOU!!!!

Anthony Chinedu IfedigboPost authorI got a negative and a number in ratio form as my answer eg. -1.765. Any help?

Navoda De SilvaPost authorYou are awesome! Thank you so much!

GarenPost authorAnyone 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 MajeedPost authorbless your soul

Dania HasanPost authorthis was a great way to explain this, thanks a lot ! π

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

Komal BhardwajPost authorthanks

NellioStylePost authorbest explication ever Thanks you

Official KelbyPost authorMaking MCAT studying 10x manageable. Thank you!

HEHEHE I AM A SUPAHSTAR CHIMERAPost authorMakes sense.

c h a m c h a mPost authorThank you very much! You saved me from my report tomorrow!

Vanessa EscaleraPost authorsaved my life

XetoDesignsPost authordo pseudo first order reactions

JessicaPost authorThank you so much , this was very very helpful ππ©ππΎππΎ

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Anita RanaPost authorThanks

retoo luvyuhxPost authorYOU ARE THE BEST

Sharia AlamPost authorThe sound is TOO low!

Gerson Cocjin Jr.Post authorwhat if concentration [A] stays the same and the rate increased by 2 what would happen in that case

R. Abdul WahidPost authorexcellent explanation <3 thank you so much!

Samuel MichaelPost authorWhy there is a difference in the order between different reactants?

Why there is a zero order or a first order .. etc

james pooPost authorkinetics

Ray GeorgePost authorabsolute legend! xx

Alvin StjohnPost authorBetter teacher than my fucking useless waste of a cunt university professor

Pooria KarimiPost authorthis guy is SUPERB he reminds me of high school forgoten topics

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

Gift ChonkaPost authorIs rate always m/s or it can change?

RTG SuperPost authorThanks very helpful

Nishat NawajPost authorO my allah..!!..thank you for the essential tutorial