Rate Law Problems

[@corbinbiology]

thanks for the practice problems. Units are a little more explained here which is cool! awesome

[@Ur.Stargirl]

Ur a life saver!!!!!!! btw I love ur voice huhuhuhu I might come back to ur videos more often hehehehe

[@Tomas-zv2js]

Youre a life saver i understand this so much better now

[@oneeljaba463]

Thank you i watched atleast like 5 diff videos and none of them made sense this one did

[@Chipi.vance16]

fr! finally found a vid like this! youre alife saver!!!!!!!!!!!!

[@KhalidMehmood-vo7pl]

Excellent, first time I understood this concept

[@HelloFreedy]

Really loved it❤

[@raziabano9693]

Amazing ❤

[@favourajala423]

I enjoyed this video. Thank you... Very explanatory

[@henryschmitt4170]

I believe you accidentally made a calculation error. .003/.001 = 3, not 0.3.

[@susanahernandez3531]

Thank you for your explanation!

[@iloveunicorns123]

Thank you so much for this, this helped so much.

[@chrisash6473]

Very helpful 🙏🏼

[@muralisudileti3694]

Nice explanation

[@kzeunIII]

tysm!

[@madisonfleck8433]

do you ignore the coefficients?

[@tmc_marie]

Thank you

[@Dilshani420]

Thanks 👏

[@macaramiller5946]

Thank you
But what happens to the order of reaction when the concentration decrease?

[@amenterekegn7485]

Omg you are my best

[@Chemcrown]

great madam

[@ruthwairimu1623]

Thenks alot1😊

[@ebrahimmoto]

I am going to teach my lecturer 💪

[@estherm877]

haaaha

[@macaramiller5946]

When the concentration doesn't stay consistent what happens to the order of reaction?

[@Duggle_]

If your question refers to a basic law, then the order of the reaction will stay consistent regardless of what concentration values you may plug in. Concentration is almost never constant in actuality. Remember that based on which graph forms a linear relationship, the order can be determined (it will not change). for example, [Cl] vs time forming a straight line on a graph is zero order, ln[Cl] would correspond to first order, and 1/[Cl] would demonstrate 2nd order. Hopefully this helps answer your question ;)