14.2 Rate Laws

Struggling with Rate Laws? Chad shows you how to determine a reaction's rate law from experimental data whether it be zero, first, or second order.
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Happy Studying!

[@TheEpiphany101]

This man deserves a Noble Prize, I swear. Genius

[@ChadsPrep]

I love your use of hyperbole! Glad you're finding my videos helpful!

[@user-ry2ur6vj6h]

They didn’t laugh at your jokes enough smh, I would’ve been like “AHHAHAHA” You’re a great professor, I’ve been stuck on these forever but I finally understand it for my test next week. Thank youuuu Back to binge watching, all your vids

[@ChadsPrep]

I appreciate that, Lea - good luck on your test and enjoy all our videos!

[@Ongoingpidgeon6]

You are an great communicator. Keep up the good work. You made this really confusing topic from my class and textbook easy to understand.

[@ChadsPrep]

Thanks! That is precisely what I hope to accomplish!

[@Nithishthomas1991]

The battery and starter analogy was spot on for explaining this. Thank you.

[@ChadsPrep]

You're welcome and Thank You.

[@AlleEsor28]

I watched these videos 3 years ago for AP chemistry and I'm back again... this time for general chemistry in college. Thank you Chad! Your content is always quality and reliable :)

[@ChadsPrep]

Welcome back, Alle Esor - glad the channel has served you - Happy Studying!

[@keniagiselle]

So happy I found your videos! You're such a good teacher!

[@ChadsPrep]

Thanks Kenia! Happy Studying!

[@bryanpulliam1500]

Love your videos! They've been super helpful during chemistry 2 in college.

[@ChadsPrep]

Glad you like them, Bryan - good luck in chem2!

[@EliasH-video]

Far and away the best vid for rate law I've seen

[@ChadsPrep]

Awesome Elias and thank you!

[@farangizsautova5477]

Professor how good teacher u r!It is my fortune that I was recommended to watch ur videos,they r incredible and priceless!Thankssssssssss😁😇

[@ChadsPrep]

You are welcome, Farangiz - thank you for saying so.

[@justmuhd03]

This helps me and my friends alot. Thank you for your great lecture as you've made it easy for us to understand. definitely will recommend your channel to my coursemates !

[@ChadsPrep]

Glad to hear it and Thank You - Happy Studying!

[@LittleMissDanyte]

thank you so much for this. I was gonna cry cos im studying for my finals and NONE OF THE VIDEOS made sense until i came across this one. thank you so so so much

[@ChadsPrep]

Glad you finally got the help you needed, Diala - good luck in your studies.

[@heretocomment2337]

Hello Chad, I have some questions I was hoping you can answer. When comparing trials/experiments, you said (around 11:40) that you need to compare the one where only one of the reactants have changed, so that if the reaction rate changes, you know what was responsible for the change. That makes total sense. What doesn't make sense is in your table around 12:00, you compared trial 2 and 3 for NO. Then at 13:56 you compared trials 1 and 2 for Cl2. My questions are: 1) Why can't you compare the first trial and the third for NO and the Cl2? Why specifically 2 and 3 for NO and 1 and 2 for Cl2? 2) In trial 2, the concentration of Cl2 changed to 0.2 while NO stayed the same at 0.1. You can only change one concentration at a time in each trial, and the first row at the top is the numbers for the original before you changed anything. So why did you change the concentration of both NO and Cl2 in the 3rd trial? Both are now 0.2 moles/L. Since you already changed the concentration of Cl2 in trial 2, for trial 3, shouldn't the concentration now be 0.2 for NO and back to 0.1 for Cl2? Thanks in advance.

[@spicytrailmix3531]

Your videos are the first I have found online that I truly understand, better than Khan academy and everything. You just make it make sense. Thank you so much :)

[@ChadsPrep]

Thanks for saying so, spicytrailmix - Happy Studying!

[@alejandracruz1245]

Extremely helpful! Thank you!

[@ChadsPrep]

Thank you for the feedback, Alejandra!

[@gelogemina409]

Hey, I really want to know why the concentrations of reacting substances must be multiplied and even raised to some powers. Please enlighten me about this.

[@meiyaoli2258]

Wooow love the examples you give!!!

[@ChadsPrep]

Thanks again! A good example or analogy is worth its weight in gold!

[@HZMusic613]

Chad, you are a legend! Thank you very much.

[@ChadsPrep]

You are very welcome!

[@iAmDubem]

You explained this SO well

[@ChadsPrep]

Thanks for saying so, Dubem.

[@deborahaigbedion6323]

should we know anything about 4th order reaction ?

[@alphamale5496]

You really make this easy. (: Thanks a lot.

[@ChadsPrep]

You're welcome Lucwilerme! 😊 Glad your finding these helpful!

[@lulucharmaine6510]

What a great way of teaching!!!!! I wish I was in his class

[@ChadsPrep]

Thanks and welcome to the channel, Lulu!

[@lulucharmaine6510]

@@ChadsPrep Its my pleasure to be here!

[@ChadsPrep]

@@lulucharmaine6510 Ours as well!

[@eileenhernandez156]

How you calculate this? For a particular reaction, the hydroxide ion is one of the reactants. Doubling the concentration of the hydroxide ion, while holding all other concentrations constant, changes the rate of reaction from 0.002 to 0.008. What is the order with respect to the hydroxide ion?

[@avrohomhousman5958]

watch this video. it will explain how to do that type of calculation. kzbin.info/www/bejne/jInJk62tdttlesU

[@user-uh9uv2yw9h]

I suggest actually learning this material lol that question is too easy. You'll need to know this for college.

[@Jencap276]

What a smart teacher wowww

[@ChadsPrep]

Glad you found the videos - thanks for the comment.

[@deetheman8522]

I know it's been years since you posted this video and I never did chemical kinetics so please bare with me. When you calculated the order of reaction from the experimental data, you took the "k" to be the same for the two experiments (and k/k = 1). My question is how did you know that both experiments had the same "k"? I understand that the concentration of one of the reactants is constant between the two experiments, but I don't get how the "k"s remain the same. Apologies if I'm not seeing something obvious here.

[@ChadsPrep]

Hello Dee! If you look at the Arrhenius equation, k = Ae^(-Ea/RT), you can see that the rate constant will change with a change in temperature or with a change in activation energy (which is lowered if you add a catalyst). But it is not affected by a change in concentration of the reactants. Unfortunately, this equation isn't covered until later in the chapter so I was really simply making a declarative statement that the rate constant would be the same in both trials (as long as they are both run at the same temperature and that we don't add a catalyst in one but not the other). If you actually did some kinetics experiments in the lab you would verify that the rate typically changes when you change the concentrations of the reactants, but the rate constant does not (which is why they call it the rate CONSTANT fyi). Hope this helps!

[@deetheman8522]

@@ChadsPrep Cheers Chad, that does. And thank you for taking the time to write back. I assumed that all other conditions in the two expriments remained the same and thus the same "k", but I wasn't sure..you know, with assumption being the mother of all ### and all. Anyways, thanks again and keep up the wonderful work!

[@mohitsain0706]

This is the all syllabus in entrance too

[@cathooode2260]

thank you!!! :)

[@ChadsPrep]

You're welcome!

[@hardikbisht2419]

So sir it means order of reaction depend on the sum of stochiometric coefficient of the product in the reaction

[@ChadsPrep]

The order of the reaction does not necessarily depend on the sum of the stoichiometric coefficients. Sometimes it coincidentally does and it does if a reaction is elementary (can't be broken into simpler steps), but you can't make an absolute statement regarding this. If a reaction is not elementary the orders must be determined from experimental data and they don't have to have any correspondence to any of the coefficients in the balanced reaction. Hope this helps!

[@hardikbisht2419]

@@ChadsPrep so sir in elementary reaction if the reactant element is in access form then is it will affect its order?

[@CeciaEss]

The way he explains things!! 🙌🏽🙌🏽🙌🏽 HALLELUJAH! I’m actually gonna graduate

[@ChadsPrep]

Glad the videos are helping, Cece S.

[@codygruberg7294]

wish I found you sooner! google needs to put you front page

[@ChadsPrep]

Thanks Cody! Your comment elicited a smile!

[@brianaschino4903]

thank you for breaking everything down, i'm retaking chem 2 after 15 years and it's been a struggle

[@ChadsPrep]

You are welcome, Briana. Good luck with your future educational challenges!

[@obaapapapabi3015]

Great

[@ChadsPrep]

Thanks!

[@albalry]

You’re awesome

[@ChadsPrep]

Thanks!

[@tulongashowa1420]

GOAT 🐐🐐

[@ChadsPrep]

Thanks!

[@noorulhadi4891]

The native people will enjoy your lectures but most of the words are out of our range can you please slow down your speed

[@ChadsPrep]

Hey there, Noor Ul Hadi - you can actually choose a speed. At the bottom of videos is the Settings button that looks like a gear. Click that, then select Playback speed and then choose whatever speed works for you. Happy Studying!

[@mohitsain0706]

Look like criminal

[@mohitsain0706]

Jailor give him some time to teach

[@ChadsPrep]

Ha! That's a new one! I guess you could say I'm a prisoner to the love of Chemistry!!!

[@mohitsain0706]

@@ChadsPrep 🙄🙄