The Rate Law 

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!

Bozeman out there saving lives 🙌

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

Hey

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

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.

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!

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

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

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

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

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

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

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

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

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

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

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

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.

This was incredibly helpful. Thank you so much!

Hello, Mr. Anderson. Thanks for the tips.

phenomenal. please continue making videos you are an excellent teacher

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

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

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

The overall reaction order is 2, no?

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

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

That was very helpful! Thank you!

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

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..!

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!!!

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

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.

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

Outstanding. Bravo.

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.

just amazing explanation thank you

This was helpful. Thanks a lot

So helpful! Thank you!

Thank you that was really helpful :D

Thanks! this is very helpful :D

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?

You're awesome, thank you!

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?

Is the a physics series ? Your videos are always awesome and very informative! Keep up the great work :-D

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.

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

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

you da man mr A

this is great! thank you

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

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

Great vid, thanks

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?

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?

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

Thank god you did this , exam in two weeks

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

thanks alot Mr Andrewson :***

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

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?

this guy just saved my midterm !!!!!!!

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

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?

Steve Kerr if he didn't hoop.

God bless you!

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

Thanks a lot much respect

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

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

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?

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

Thank you.

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?

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?

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?

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?

You are awesome!

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?

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

thank you!!

Thank youuu!!

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

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?

you're a genius

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

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?

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?

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

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

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?

You are awesome. Thanks.

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?

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 ?

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

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

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

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

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