15.2 Le Chatelier's Principle | General Chemistry

[@beagonzales8918]

HOW IS CHAD NOT MONETIZED??? He's literally the best! I've been having a hard time understanding what happens if the pressure is changed and this is the only video that helped me. Binging his other videos!

[@ChadsPrep]

Happy Studying!

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i just want to let you know that youre helping out people (like me just now) with your videos to this day

[@ChadsPrep]

Glad to hear it - Happy Studying!

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It is hard to get such quality lectures on KZbin. Very well explained and hope you will keep up the good work...

[@ChadsPrep]

Glad the channel/videos are helping you!

[@ConnorStear]

I had to stop and applaud how beautifully simple you made everything right around the range of 3:40 to 4:40. Everything clicked so well it's like I could feel the neurons firing in my brain lol.

[@ChadsPrep]

I am happy for your brain expansion! Thanks for saying so!

[@saqibkhan3364]

Chad, In the last lecture you said that only temperature can change the equilibrium constant. So it is fair to say that when we are are changing the concentration of reactants and products, we are not changing the equilibrium constant but rather the reaction quotient. So when we change the reaction quotient, this shifts the reaction in a way that allows us to get back to the equilibrium constant which remains the same value because only temperature can change the equilibrium constant? (I hope this makes sense).

[@ChadsPrep]

Exactly right Saqib! I can't tell you how confusing many students find this, but you've got it!

[@ashleem3295]

I am so thankful I found your videos! You are a great teacher, and I wish I had found you in gen chem 1! I love that you take the time to teach each aspect of the section. My professor teaches quickly and skips many small steps, so your videos help me actually learn the material.

[@ChadsPrep]

Glad the channel/videos are helping you - Happy Studying!

[@SakeenaManzoor-u4y]

Sir l know you are the only teacher who can clear away my all confusions relevant to my questions😢

[@life0fjah]

youre the best man, you explain things so well! THANK YOU!

[@ChadsPrep]

You're welcome and Thank You.

[@captainamericawhyso5917]

AMAZING VIDEO ONCE AGAIN ! Just a question though when i add argon won't the collisions between the reactants become fewer? Before, in the same volume there were only these two reactants: solid carbon and diatomic gas oxygen. But now that we added the argon some of the collisions are going to be with the argon rather than between the reactants. So if its harder for the reactants to collide with each other shouldn't the reaction sift to the left ?

[@DavidElks-qh2pn]

You are truly a life savior ❤❤

[@ChadsPrep]

Wow - thank you!

[@devikanair44]

thank you so much for explaining the effects with temperature. I was struggling with that for a long time b/c I was confused how shifting T changes a rxn b/c it affects equilibrium. One quick question! how would you reconcile that delta G = -RTlnK but if you added heat in an exothermic reaction the delta G should be less negative, but according to the formula if you increase temperature the Delta G would be more negative? loving the new playlist!!

[@ChadsPrep]

Hey Devika - The equation ΔG° = -RT ln K is for standard delta G only, and if we increase the temperature we are no longer at standard conditions and this equation doesn't apply. Instead, if you look at the equation ΔG = ΔH-TΔS and consider that an exothermic reaction has negative ΔH then the value of ΔS will determine whether ΔG is positive or negative overall

[@shamsalnaharrajoub8495]

Why adding solid will not affect equilibrium position? Adding solid will produce more products

[@kareemosama785]

The same question I am struggling with

[@saqibkhan3364]

Another Question, What if we decrease the pressure by increasing the volume? How would this effect the shift of the rxn?

[@ChadsPrep]

Saqib, if you increase the volume, you lower the concentration of all species (since molarity is mol/L). Now if you have an equal number of moles of gas on both sides of the reaction, it won't shift as Q would not change. But if they are not equal, then Q will change and the reaction will shift to the point where Q is equal to K again. Overall, by increasing the volume to lower the pressure it will shift toward whichever side has the greater number of moles of gas. Hope this helps!

[@destinycross7977]

why would we shift Co to right and not left to decrease Co?

[@ذَكاءذكاء-ذ3ز]

Do pressure and volume affect the equilibrium constant? If they don't affect it, how? I got confused about it Especially when I was watching the clip at minute 19

[@SakeenaManzoor-u4y]

Sir can a Q become greater than Kc before we reach eq

[@dawithabteselassie7116]

is the universe in equilibrium applying Le Chatlier principle?

[@ChadsPrep]

If it were, then you and I wouldn't be here to discuss it!

[@destinycross7977]

Hi, you said shifting to the right causes the Q value to go up, I thought it causes it to go down since it Q

[@SakeenaManzoor-u4y]

Dear sir you asked that change in conc of either reactants or products at eq doesn't affect the value of Kc but if we increase the conc of O2 here the reaction will go in forward direction and reverse reaction remains unchanged and after a second the consumption of extra O2 again rate of forward reaction also becomes slow down and both rates again become equal . In other hand if at eq we decrease the conc of Co the reaction will go in forward direction but in this case we have no extra addition of reactants or products but our normally present reactants at eq are consuming and in this case the rate of reverse reaction slow down and after a second both the rates again become equal sir how both these reactions could have equal value of Kc😢😢😢😢

[@kimberlysanchez6693]

Thank you so much!!!! Been watching all of your chem videos

[@ChadsPrep]

Glad to hear - Happy Studying!

[@alihasnain4045]

good video thank you

[@ChadsPrep]

You're welcome!

[@RelaxRailRide]

Sir, what is your age and marital status?

[@ChadsPrep]

I'm pretty old regarding the first question and happily married regarding the second.😊

[@RelaxRailRide]

@@ChadsPrep 😇

[@ذَكاءذكاء-ذ3ز]

All three changes: 1- Withdrawing or adding to the interaction 2-Change the volume or pressure 3- Changing the reaction temperature It will change with the concentrations of substances for the equilibrium reaction Why is it only temperature that changes the value of the equilibrium constant?

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Thanks!

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thank you

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you're welcome

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Job well done ✅✅✅ God bless you

[@ChadsPrep]

Thank you!

[@SakeenaManzoor-u4y]

Sir Q is greater than Kc means that we are talking about before eq and if Q is smaller than Kc means that we are talking about after eq because after eq l think we should have a lower conc of products and higher conc of reactants but sir we also say that once the reaction reached eq with the passage of time there is a constant change in the conc of products and reactants how this happens ???? I am totally confused sir😢😢😢😢

[@SakeenaManzoor-u4y]

Sir how can I understand it????

[@jeremyalex7678]

best lecture✊

[@ChadsPrep]

Thank you!

[@mabotiyn]

Chad I have a question

[@SakeenaManzoor-u4y]

Sir you also said that stochimetric cofficients do effect the value of Kc but now you are saying that if we increase or decrease the conc of reactants or products at eq mean we increase or decrease the number of moles of reactants or products the value of Kc will not change ????? 😢😢😢😢how sir?????

[@ChadsPrep]

The idea is that the ratio of products to reactants is only equal to Kc if the reaction is at equilibrium, whereas we can always say that the ratio is equal to Q, whether the reaction is at equilibrium or not. So if the reaction is at equilibrium, then the ratio of products to reactants will equal both Kc and Q (as Q=Kc at equilibrium). But if change the concentration of a reactant or a product, then at that instant the reaction will no longer be at equilibrium, and the ratio of products to reactants will still equal Q by definition, but it will not equal Kc (since the reaction is not at equilibrium). But you could also look at the situation mathematically and say that this is why the reaction will shift...in order to get back to equilibrium. If the altering of the concentration resulted in Q < Kc, then the reaction will shift to the right until we once again reach equilibrium and Q = Kc again. Or If the altering of the concentration resulted in Q > Kc, then the reaction will shift to the left until we once again reach equilibrium and Q = Kc again. Ultimately, changing a concentration does change the ratio and this changes Q, but not Kc (as a reminder the only thing that results in a change in Kc is a change in temperature). Hope this helps!

[@SakeenaManzoor-u4y]

@@ChadsPrep Thanks alot dear professor l just understand it

[@ChadsPrep]

@@SakeenaManzoor-u4y Glad to hear!

[@raeesadhorat2170]

GOD BLESS YOU YOU BRILLIANTSIR

[@ChadsPrep]

Thank you

[@SakeenaManzoor-u4y]

I just cut off your answer before looking at it unfortunately

[@etodd1234]

I was under the assumption that an increase in temperature always caused a rightward shift and vice versa for a decrease in temperature...

[@ChadsPrep]

Not true; it absolutely depends upon whether the reaction is endo- or exothermic. However, you may be thinking about the RATE of reaction as the reaction will be faster at higher temperatures...but so will the reverse reaction. So generally a reaction will proceed to equilibrium more quickly at higher temperatures, but whether or not it shifts left or right again depends upon whether it is endo- or exothermic. Hope this helps!

[@destinycross7977]

Im sorry I am so confused. I really need help

[@SakeenaManzoor-u4y]

Waiting for your reply sir??????

[@SakeenaManzoor-u4y]

Sir g sorry plz reply again 😮

[@ChadsPrep]

what are you struggling with?

[@SakeenaManzoor-u4y]

Sir I want to become a doctor and l don't understand your question and sir lam waiting for your answers about my questions sir did you see my questions which I asked About your topic?

[@mathewhorodner2000]

@@ChadsPrep @user-vl1bk9jh4h 3 months ago Dear sir you asked that change in conc of either reactants or products at eq doesn't affect the value of Kc but if we increase the conc of O2 here the reaction will go in forward direction and reverse reaction remains unchanged and after a second the consumption of extra O2 again rate of forward reaction also becomes slow down and both rates again become equal . In other hand if at eq we decrease the conc of Co the reaction will go in forward direction but in this case we have no extra addition of reactants or products but our normally present reactants at eq are consuming and in this case the rate of reverse reaction slow down and after a second both the rates again become equal sir how both these reactions could have equal value of Kc.......... @user-vl1bk9jh4h 3 months ago Sir Q is greater than Kc means that we are talking about before eq and if Q is smaller than Kc means that we are talking about after eq because after eq l think we should have a lower conc of products and higher conc of reactants but sir we also say that once the reaction reached eq with the passage of time there is a constant change in the conc of products and reactants how this happens ???? I am totally confused sir This is from @user-vl1bk9jh4h