The Rate Law

036 - The Rate Law
Paul Andersen explains how the rate law can be used to determined the speed of a reaction over time. Zeroth-order, first-order and second-order reactions are described as well as the overall rate law of a reaction. The rate of a reaction can be determined experimentally.
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All of the images are licensed under creative commons and public domain licensing:
"File:Ammonia-3D-vdW.png." Wikipedia, the Free Encyclopedia. Accessed October 19, 2013. en.wikipedia.org/wiki/File:Amm....
File:Crystal Violet in Aqueous Solution.jpg, n.d. en.wikipedia.org/wiki/File:Cry....
"File:Nitric-oxide-3D-vdW.png." Wikipedia, the Free Encyclopedia. Accessed October 19, 2013. en.wikipedia.org/wiki/File:Nit....
"File:Nitrogen Dioxide at Different Temperatures.jpg." Wikipedia, the Free Encyclopedia. Accessed October 19, 2013. en.wikipedia.org/wiki/File:Nit....
"File:Nitrogen-dioxide-3D-vdW.png." Wikipedia, the Free Encyclopedia. Accessed October 19, 2013. en.wikipedia.org/wiki/File:Nit....
"File:Oxygen Molecule.png." Wikipedia, the Free Encyclopedia. Accessed October 19, 2013. en.wikipedia.org/wiki/File:Oxy....
File:Spektrofotometri.jpg, n.d. en.wikipedia.org/wiki/File:Spe....
Izmaelt. Slovenčina: Príprava Oxidu Dusičitého z Medi a Kyseliny Dusičnej. Fotené Na Slovenskej Technickej Univerzite., October 27, 2010. Own work. commons.wikimedia.org/wiki/Fil....

[@deleteeeeeeed]

Bozeman out there saving lives 🙌

[@Shellshocked97]

I'm a second semester uni student and you are seriously the only thing that allows me to do well sometimes. Thank you x1000 for all of your videos!

[@demonlayer4414]

so you graduated now ??

[@dayayelluripati6091]

Mr. Anderson,i would like to thak you for making efforts in explaining very complex chemistry , in simple words .thank you ,you are a life saver

[@connerpace4972]

who else is here at 5 in the morning before a test?

[@melissarainchild]

ten on a sunday morning, my weekends are just this: study study study...and coffee... :)

[@diszylam1018]

Conner Pace I got the AP exam tomorrow:/

[@dereksmith1301]

I’m taking the AP Chem test in 3 hours

[@timmyglasgow4979]

Bruh I have it in 30 minutes

[@starwarfan8342]

@@timmyglasgow4979 How did that go?

[@kristina48046]

I couldn't get it from Atkins' book for days, but you explained it so clearly!! Thank you for your work. You are a talanted teacher.

[@IbbIxMonster]

is the answer: rate = k[A][B] so a reaction order of 2?

[@jass2312]

what's the ans?

[@Yankee4ever2]

@@jass2312 it is a second order reaction cuz you add the exponents of A & B and you get 2 cuz each has an exponent of 1

[@user-gb2mf7dy5j]

@@Yankee4ever2 wait can u explain how u got the rate for B

[@Yankee4ever2]

@@user-gb2mf7dy5j to get B: keep A the same and find 2 experiements where B doubles and A stays the same (exp 1 & 2): look at the rate, the rate doesnt change, therefore its a 0th order for B hope that makes sense

[@R.Cruz9]

⁠​⁠@@user-gb2mf7dy5jif you include the exponents it would look like: rate = k•[A]^1•[B]^1 Therefore when you add up the exponents it is = 2. (When you raise a number to the 1 power, it does not change. Which is why they don’t include the ^1 in either A or B) But when you ask did we know that both A and B are first order (^1)? I’m basing my answer off the fact when you look at the chart, when you double the concentration of A you get a Rxn rate of 6, and when you double the concentration of B you get the same answer as you do with A. That informs us that they are equal and likely first order. Had A or B been a second order, the Reaction rate would’ve gone up to 9.

[@Fatescervix]

7 years of college and I finally get this concept thanks to this video. Keep up the good work, you're a life (and test score) saver!

[@nahfid2003]

Bro help a fellow bro out. Please tell me some sources for physical chem

[@RonShenkar]

The only rate law video that finally breached through my thick skull. Thank you Mr Andersen.

[@Bozemanscience1]

Hey

[@juanguio5932]

Hey, thanks for sharing your knowledge to help others learn too

[@user-tk8cz3ry8l]

Sup

[@calvinfiebich4979]

thank you! helped me review for the ap chem test tomorrow

[@steff4413]

yo

[@milliemogudala9982]

This is SO timeless, thank you SO much for helping out students even after 10 years!!

[@frankied.2828]

You explain things so well and with quality. Thank you for the solid explanations

[@braydonmorris3518]

Wishing you were my college chemistry teacher the way you teach things so clearly and concisely

[@rosenguyen4181]

My man explained clearly in 3 minutes what my professor hasn't been able to clarify in three days. Wow.

[@ryanwaddell6156]

I believe the answer for the 6:50 question Is the rate order is 2. I believe this is correct because the M/s (Molarity/second) concentrations are multiplied by 2. I understand the difference between first-order and second-order as, first order the M/s concentrations are multiplied and the second-order M/s concentrations are squared, cubed, or to a greater power.

[@Aryan-qz2fq]

I agree

[@shereecarter304]

studying for the MCAT and this super helpful! Thank you!

[@mitchleung8500]

Thanks a lot. You teach well and concisely. Students will be benefited from you. Keep up!

[@nnekaude]

You need a Nobel prize.. You are out here saving lives!

[@Shammer1]

That because some of us are fools studying right before test or just have teacher who can't teach

[@thechair124567]

chemistry is hard for people though. So we use this to understand more.

[@eddielookingbeautifulbravo8]

Nobel Prize for Peace

[@yaquanliang4994]

Thank you so much Mr. Anderson, this have made my life soooo much easier!!!

[@colinmccutcheon5698]

Thank you. You explained it better than my textbook did.

[@dausonlarrabee5875]

When you talk about second order reactants at 5:13 and 6:25, I think there is a mistake on the tables. Doubling the concentration of a second order reactant will multiply the rate by FOUR between the trials, not square it. At 5:13, if you work out the rate order for A algebraically (rate1/rate2 = ([reactant]/[reactant])^order of reactant) the math is as follows: 2x10^-3/ 4x10^-3 = (0.1M /0.2M)^x which simplifies to 1/2 = (1/2)^x. X is 1 in this case which would indicate a first order reactant, not a second order one. For it to be second order, the rate for experiment two would have to be 8x10^-3 M/s. Similarly, the rate for experiment 1 should be 32x10^-3 M/s if you doubled the concentration of A again. There is a similar issue with the example problem at 6:25. The algebraic set up to find the order of A would be 3x10^-3/ 9x10^-3 = (0.1M/ 0.2M)^x which simplifies to 1/3 = (1/2)^x. X would have to equal 2 for it to be a second order reactant, but it does not equal 2. For the table to indicate a second order reactant I believe experiment 3 would have to have a rate of 12x10^-3 M/s. Here is my work: 3x10^-3/ 12x10^-3 = (0.1M/ 0.2M)^x which simplifies to 1/4 = (1/2)^x which simplifies to x=2. I would love to hear what someone think about this. I am a chemistry student teacher who is relearning kinetics in order to eventually teach it myself so naturally I want to get my facts straight. I very well could be wrong here but in my quest to master kinetics I have found that this video's explanation of second order reactants doesn't jive with other sources I've found such as this resource on page 11: apchemistrynmsi.wikispaces.com/file/view/12+Kinetics.pdf

[@siebertcollegiate]

I just went over this with my students, who watched this video for homework. You are right. I think that Paul was thinking that the rate would be squared (e.g., 3^2 = 9), which is where he got the 9 in the table instead of 12. But the rate should increase by a factor of four, like you are saying.

[@user-xd9pw4dr9v]

Dauson Larrabee great job wow I think you are right

[@user-xd9pw4dr9v]

Dauson Larrabee and thank you

[@karenrandazzo]

Agree, there is a mistake in the video.

[@sunnysherman7915]

Hello, Mr. Anderson. Thanks for the tips.

[@maggiefei1995]

The overall reaction order is 2, no?

[@ahmedelsaid173]

1 and 1, right?

[@Stanly-oo8zq]

@@ahmedelsaid173 yes I think so too

[@ryanwaddell6156]

I believe the difference between zeroth-order, first-order, and second-order are the rate/time graphs that each create. The first-order is a constant rate over time (straight horizontal line). The second-order is a straight line with a decreasing slope. The second-order is a curved decrease of rate over time.

[@saalimsayyed279]

Nice explanation and easy tricks to solve problems. Thank you :)

[@user-ng7fj6sf7i]

I have question. I heard that first order reaction substances in transition state in slowest step in reaction, only one mole of that first order substance is involved (in slowest step of transition state). I just do not understand why. because I thought order is just the indicator that says about the number between the rate and concentration of reactant??? I am so confused

[@chriscola22]

phenomenal. please continue making videos you are an excellent teacher

[@johnyapple8447]

This was incredibly helpful. Thank you so much!

[@valentinaavilespaizy7055]

Would the correct answer for the table at 7:00min be first order reaction (rate=k[A])?

[@user-yu1jp5sz8d]

I really like how you explain everything, but I still have a little confuse on the overall reaction order. Can you explain the part were you identified if it is a zeroth-order, first-order and second-order? I would appreciated very much.

[@PaolaSanchezBME]

I have two questions: 1. Considering the information on reaction orders discussed in the video, how would you determine the overall reaction order for a reaction involving multiple reactants? 2. After discussing the rate law and reaction orders in the video, how would you explain the concept of rate constants (K) to someone new to chemistry?

[@CORPSE77]

Hello Mr. Andersen, at 6:21 : isn't the rate proportional to [A], so if the rate is increased by a factor of 3 (between the two trials), then [A] doesn't make a sense that it's doubling, right?. I am trying to match it with what I have in my textbook but I don't think i understand

[@Bellabeautylove]

this was so helpful, my exam is tomorrow and I feel a lot better already! THANK YOU

[@Jasminemakeupxoxo]

when he is talking about the a+b=products , did he say that the rate=k(a) m x k(b) m or n? thanks

[@_lazzie_]

Very good and simple explanation of the topic highlighting how we can identify each order. If I may, can you elaborate on any limitations or exceptions where graphing might not be the best method to determine these orders?

[@narthanavarsa7015]

thankyou for this nice explanation and example which you took has made me to realize the concept well

[@ladymaefire]

I wanted to know more about spectrophotometric analysis...is the absorbance then equal to the concentration?

[@mrcs2572]

Great video, I have a question and I was curious to learn more about spectrophotometric analysis. Is the concentration equal to the absorbance at that point?

[@WinstonEstelritz]

Would a reaction whose overall rate is second-order behave similarly to a simple second-order reaction?

[@nyahmolina]

What methods can I employ to confirm the accuracy of the data I've collected for analyzing the rate of a reaction in an experiment?

[@Manuelb778]

At the min 3:28, the decomposition of ammonia was discussed in terms of reaction order. If the concentration of ammonia does not affect the rate; does this imply that temperature and catalyst presence are the only factors that can change the rate of a zeroth-order reaction?

[@daileenrivera8169]

I understood that the decomposition of ammonia is a zeroth-order reaction because the rate remains constant regardless of concentration changes. But how do we determine the reaction order for more complex reactions with multiple reactants?

[@Catalystheory]

Can you help me with this question :What is the half-life of a reaction with an initial concentration of 0.300 M and a rate constant of 0.00385 1/M*s

[@karinarivera8717]

How does the fluidity of the cell membrane change in response to changes in temperature?

[@detonyeadickson]

Bozeman’s the real MVP. I actually feel like I might pass the AP exam tomorrow 😊

[@isabellabussch5796]

did u pass lmfao

[@blazethunder6909]

When one has more than two reactions orders, how does one establish who goes first in the rate order, how does one add up both rates and how does one establish the graph of two reactions?

[@ElierSantiago823]

Considering the information about reaction orders discussed in the video, how would you determine the reaction order for a reaction involving multiple reactants?

[@yazen1988]

I think the answer at 8:40 min should've been R=K[A] Since they both double which means they are proportional and that says it should be the first order NOT the 2nd

[@kianfoster]

7:26 How do you calculate which order it is if after making a spreadsheet and filling the data, none of the graphs are a straight line?

[@valeriaayala3365]

How does the rate of reaction change over time for a zeroth-order reaction compared to a first and second-order reaction, as described in the integrated rate laws for each reaction order?

[@casanova9250]

How does the half-life of a reaction change with concentration for zeroth-order, first-order, and second-order reactions?

[@hernandezmil3230]

After plotting the values, identify the reaction as zero order. In the graph of [A] vs. time I could see the straightest line. What do you think, am I correct?

[@captainmexico4585]

Amazing job!!! Went into Chem lecture confident, left confused as fuck!! But after watching this simplified version the basics help make total sense of whats to come. Thank you bro!!!

[@haniehfahimpour7652]

I think the straight slope for the 2nd order reactions must be increasing with time since it inverse of the concentration

[@paolafranco8972]

How does changing the concentration of reactants in a chemical reaction impact the determination of the rate law order, and what role do experimental methods play in determining the order of reaction?

[@yarielacastro3200]

what are the key steps involved in determining the rate constant and overall reaction order from experimental data?

[@Bozemanscience1]

After I finish the AP Chemistry series I will start on an AP Physics series.

[@xiiixiiih.16]

🍎Oh Soo fun! Like sugar before bed.🤣

[@greychalispenaayala8278]

Excellent video, I have a question. How do changes in temperature and concentration of reactants influence the rate of a reaction? And with respect to the spectrophotometric analysis, is the absorbance equal to what the concentration has been? thank you

[@astridramos8872]

How do we determine that graphing the inverse of concentration over time results in a straight line connected to k without requiring advanced calculus, as discussed from 5:40 to 5:45 min? Is there a specific concept or method that support this?

[@amandarodriguez3951]

Thank you! I have a question: We know that in a zeroth-order reaction, we can change the concentration, and the rate stays the same, but what happens if we change the temperature?

[@yamiletteperez1903]

Is the slope positive or negative in the first order reaction?

[@RobertoGormon]

Greetings! I was wondering if there are any exceptions where the concentration does not have a direct relation to the rate, specifically, that I should watch out for?

[@shadaflorida4248]

Absolutely amazing! thank you so much, this was so helpful :)

[@ShantiRamirez]

I have a question, Would a reaction whose overall rate is second-order behave similarly to a simple second-order reaction?

[@anna-xf3jb]

Would individual reactants affect the rate law of a chemical reactant?

[@edwardsantiagotorres1004]

Two questions: 1. In the context of reaction orders for a reaction involving multiple reactants, how does one determine the order with respect to each individual reactant and subsequently calculate the overall reaction order? 2. Explaining rate constants (K) to someone unfamiliar with chemistry: How can you elucidate the significance of rate constants in the context of the rate law and their role in quantifying reaction rates?

[@javiercolon7014]

How is the rate law for a zero-order reaction written? What are the units of the rate constant for a zero-order reaction?

[@adosar7261]

Does rate of law gives us the rate at a specific time?

[@JavierMorales-mk4pl]

Does every straight line on a slope automatically determine what type of order it is? Either zeroth, first or second

[@zachmyers8424]

Since f is concave up shouldn't f' be concave up on the first order reaction?

[@andreachristina512]

Will the first order reaction rate always be negative/descending?

[@rokr9999]

Thank u for Your Gr8 work Mr Anderson ! I love your videos..these r really helpful to even a guy like me who's sittin here in India..! i hope someday i'll be able to meet u and express my gratitude to u in person..!

[@kianfoster]

6:02 If both concentrations of A and B are changing and it changes the rate, it there a way or formula to precisely figure the rate of A and B?

[@stephaniequintana9371]

I have doubt about a specific section, would a reaction whose overall rate is second-order behave similarly to a simple second-order reaction?

[@E5Diaz]

If one were to be utilizing a method of collecting the data of a reaction, like spectroscopy for example, could the data collected of the emission or absorption be thus put in place of the concentration in the table and graph from there?

[@rickmenros3892]

Does the atmospheric pressure influence the rate of reaction? If so, how do the changes in atmospheric pressure influence the different orders of reaction?

[@mirjanavrdoljak899]

Helou, I need a help: THE DRAWING SHOWS THE PROGRESS OF REACTION 2A-> A2 ibb.co/12cgBL3 Based on the drawing, conclude at what time the composition of the reaction mixture will correspond to drawing D. I do not understand the procedure of this task, so I ask for help. Thank you

[@moefrd]

For the question at 6:50, i got rate=k[A][B] (I am assuming I don't need to put the 1's since it is already implying first order for both). Correct me if I am wrong please.

[@moefrd]

Also, that video was very helpful.

[@matthiasrambally1899]

+MoeBro25 yes that is correct

[@OVOXOIsMyMotto]

It its correct but overall its a 2nd order reaction

[@Oliwolf78]

Incorrect. It should be rate = k[B] because A is zero order. When you doubled [A] the rate stayed the same, meaning it would be to the 0 power.

[@mrweather5961]

I think the rate of each experiment is considered overall for that experiment. In experiment 1 they both started out with 100 M and that resulted in a rate of 3 *10^-3. In the second experiment the change was made to reactant B where it M was increased to 200 (reactant A remained at 100 M), then the "overall" rate of that experiment is 6*10^-3( first order because it went from 3 to 6). And in experiment 3, a change was made to reactant A but not B (I'm assuming that we are making our basis of changes around the standard 100 M of a reactant, so be is at 100). Similarly, the increase in M of reactant A in experiment 3 causes the "overall" rate of to be 6*10^-3 (first order because it went from 3 to 6 [compare experiment 1 and 3]). According to this video, the "overall" rate of reaction for this REACTION (not saying experiment here, we concluded this based on the EXPERIMENT though) would be to add the two order (exponential properties here?) of the reactants together. 1 + 1 = 2 or 2nd order.

[@nathanderhake839]

4:38 shouldn’t it the rate over time be exponential too since rate is proportional to the concentration of A. Wouldn’t the graph of the rate over concentration of A make the straight like instead?

[@mwansahtyller9002]

This video is really helpful 🙌🙌 Thank you so much 🌟🙌

[@lorcresiakonopasek9793]

The over all reaction? So if they are acting in regards above all ?

[@alondrarosario635]

Can the rate of law expend any sum of reactants?

[@camilaavictoriaa4883]

How can I be sure that the data I got to analyze the rate of law in an experiment is right?

[@nicecoolgurl1]

What is the answer of the overall reaction?

[@MCuniuppr87]

Could the second-order also have a positive slope?

[@Idk-bs6in]

This helped me understand my lab, we did the crystal violent thing I have no idea what we did I did my lab report with general things like when x=0 and 1 it'll be decreasing, and only when x=2 it'll increase, for the second trial I got a bit of a curve and I put that I there must have been inaccuracies when I did the lab, not entirely sure what it meant but I do understand it a bit better after the video I'll re watch this when editing my lab since it's not due for another 2 days

[@mwansahtyller9002]

Is it always that the concentration reduces in that same order like from 0.400M to 0.200M to 0.100M because I'm thinking if it reduces in a different manner let's say ( from 0.400M to 0.350M to 0.300M) it might have some effects on everything

[@nicoleaponte6565]

Can the rate of law consume any amount of reactants?

[@munjir7618]

Is there a limit on orders that exist?

[@ruthcabajes299]

That was very helpful! Thank you!

[@gayatritandel8549]

how to find n or power of concentrations

[@alexsheremett3097]

just amazing explanation thank you

[@kevmc7711]

Outstanding. Bravo.

[@NoahWilson-ms2vr]

@user-up8se8yf7s Yes, a reaction can have a zero order with respect to one reactant while having a non-zero order with respect to another reactant. This situation arises in complex reaction mechanisms where the reaction rate is determined by the slowest step, which may involve different reactants.

[@user-up8se8yf7s]

Can a reaction have a zero order with respect to one reactant and a non-zero order with respect to another?

[@naomimankrado8510]

This video was extremely helpful however when looking at the table of data shown at 6:08 you said that for compound A we double the concentration but because the rate is proceeding as a square the overall order must be 2. but isn't 3 multiplied by 10 to the -3 squared , 9 multiplied by 10 to the -6 instead of 9 multiplied by 10 to the -3?

[@tigerfish4628]

So helpful! Thank you!

[@lisatouyon1456]

This was helpful. Thanks a lot

[@ahmedalkarkhi]

Isn't the reaction to the "zeroth" power suppose to be a straight line because (k) is a constant for that specific reaction and the first order is the one that makes a diagonal line ?

[@ahmedalkarkhi]

never mind its the [] to the time graph not a rate to time graph my bad xd

[@ahmedmunif5862]

What is the rate is

[@justinekwiecinski]

You're awesome, thank you!