Equilibrium Graphs: Le Chatelier's Principle (Chemical Equilibrium).

[@instructorjohnny]

@ 4:53 - An easy way to know the graph is adding/removing reactants or products is to look for a "different" line from the rest. The line for CO2 is different from the rest and it particularly stands out. The only cause for one line to be different, is due to adding or removing it. @ 5:18 - Temperature changes on a graph will always be the most normal looking. All substances on one side will either increase or decrease and all substances on the other side will do the opposite. This happens since you only graph the effect of the shift. - sorry for the lack of explanation in the video, I won't do that again :D 🍎 𝐑𝐞𝐝𝐨𝐱 𝐚𝐧𝐝 𝐄𝐥𝐞𝐜𝐭𝐫𝐨𝐜𝐡𝐞𝐦𝐢𝐬𝐭𝐫𝐲 𝐓𝐮𝐭𝐨𝐫𝐢𝐚𝐥𝐬: bit.ly/RedoxElectrochemistry ⚗️ 𝐂𝐡𝐞𝐦𝐢𝐜𝐚𝐥 𝐄𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 𝐚𝐧𝐝 𝐀𝐜𝐢𝐝-𝐁𝐚𝐬𝐞 𝐒𝐲𝐬𝐭𝐞𝐦𝐬 𝐓𝐮𝐭𝐨𝐫𝐢𝐚𝐥𝐬: bit.ly/EquilibriumAcidsBases 🧪 𝐂𝐡𝐞𝐦𝐢𝐬𝐭𝐫𝐲 𝟑𝟎 (𝐆𝐫𝐚𝐝𝐞 𝟏𝟐) 𝐓𝐮𝐭𝐨𝐫𝐢𝐚𝐥𝐬: bit.ly/Johnny_Chemistry30

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homie explained these graphs better in a 7 minute video than my teacher explaining for 3 hours

[@instructorjohnny]

Thank you homie!

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At least your teacher gave an explanation

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[@instructorjohnny]

That means a lot to me! Thank you as well

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[@instructorjohnny]

Thank you as well! Seeing this comment makes me smile. I'm glad I was able to help you

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[@instructorjohnny]

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[@instructorjohnny]

Anytime my friend!

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[@sandyhamilton5394]

please stop moving the red dot continually. Good info, but the constant swirling of the dot is very distracting.

[@instructorjohnny]

I 100% agree my friend. I've already planned to remove the red dot for all future videos. :D

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[@instructorjohnny]

Thank you for the generous words. I will, indeed, try my best to make more awesome videos.

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[@instructorjohnny]

Good luck with your studies, I'm glad I could help you!

[@Ethherr]

Hello, I have a question in this reaction (A+B---->C+heat) by removing (C) the shift will be to the right and that increases temperature does that change the value of (K/equilibrium constant)?

[@instructorjohnny]

So, in the reaction (A+B---->C+heat) heat isn't really considered a product. When heat is written on the product side, it indicates the reaction is exothermic. If heat was written on the reactant side, it would indicate the reaction is endothermic. Basically, you can ignore heat in the reaction for concentration changes and pressure changes. When (C) is removed, the reaction will shift right and a shift right increases product concentration and decreases reactant concentration. Remember that heat isn't actually a product. It's written on the product side to show that the equilibrium reaction is exothermic. This means temperature doesn't increase, since there were no changes to temperature/heat. An important thing to remember: only temperature can change (K/equilibrium constant). In your example above, removing (C) would not change the value of (K/equilibrium constant), because only temperature can change (K/equilibrium constant). These 2 videos from me may help if you are still confused: kzbin.info/www/bejne/kJisanueasuWg6M kzbin.info/www/bejne/qV7XaHWfr9-pasU (5:21 onwards)

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[@instructorjohnny]

Good luck my friend, you got this!

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[@instructorjohnny]

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[@abbiesxnshine]

I understand how to graph it after the shifts but how would I know the concentration of the reactants and products before the shift when only the equation is given?

[@instructorjohnny]

So this is an an example of a reaction at equilibrium: N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) + energy. This equation written in words would mean: when a mixture of N₂(g) and H₂(g) is placed into a sealed off container with no NH₃(g) initially, over time, N₂(g) and H₂(g) concentration decrease or is consumed to produce NH₃(g). Equilibrium would then be reached, when all the substances remained at a constant concentration (represented by a straight line). At the beginning of the video, I explain a little bit about this though the concentration over time graph. For the equilibrium shift on a graph at 1:00, the graph actually only represents the equation when all the substances are at equilibrium. Essentially, it doesn't include the part where reactants decrease in concentration while product increase in concentration as time goes on. Notice that N₂(g) and H₂(g) are higher up in concentration then NH₃(g). This is because the reaction initially starts with N₂(g) and H₂(g), but over time NH₃(g) is produced. So, we don't technically know the exact numerical concentration of the reactants and products before the shift when only the equation is given, but we know we have reactants initially which are then being consumed to produce products over time. Just remember whatever is there initially will have more concentration, which in most cases will be the reactants as they are consumed to produce products. Numbers for exact concentration are given for problem solving, such as ICE table or equilibrium constant problems.

[@ellezarglitched7108]

Thank you so much! You explained this thoroughly and fast enough that it kept me engaged. I was really struggling with graphing these this unit, but I have a way better understanding thanks to you!

[@instructorjohnny]

Thank you for your comment my friend! It means a lot. I'm glad I was able to help you understand this concept in equilibrium.

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[@instructorjohnny]

Good luck on your finals! :D

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[@instructorjohnny]

I'm glad to hear my video was helpful!

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[@instructorjohnny]

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[@neamnmussie5319]

This video is helpful but I have one question as we know the number of mole and pressure are directly proportioned. when we increase the pressure why the reaction shift to the smallest mole number?

[@instructorjohnny]

Sorry about the extremely late response. I hope my response can still be helpful for you. This question involves Le Chatelier's principle -- If stress is applied to a system at equilibrium, the system shifts to relieve the stress. When a gaseous equilibrium reaction experiences a pressure increase (volume decrease), the system needs to relieve the stress; therefore, the system would shift in a manner to offset the excess pressure. This would be a shift to the side of the reaction with less gas particles (gas mols). Mol number and pressure are related because pressure exerted by gases are due to continuous molecular motion and collisions with the sides of the container. Basically, more particles (gas mol number) in a set volume, more molecular motion and collisions, and thus a higher pressure. So in our example at 3:03, the equilibrium shifts right to the product side -- side with the smallest mol number. Since there are fewer particles in the container, there are less collisions, and thus a lower pressure.

[@fatemehpourmohammad1020]

I have a question. When there's a addition or removal of a product or reactant, after the shift occurs, why doesn't concentration of that product or reactant return to its initial amount. For example, In the video above, hydrogen was added (reactant) but its concentration never returned to its original amount. At the newly established equilibrium, the concentration was slightly higher. Why?

[@instructorjohnny]

Sorry for such a late response back. Concentration changes (addition or removal of a product or reactant) is an example of stress that can cause an equilibrium to shift. When we add a reactant or product, the equilibrium system will shift to consume some of the added substance. When we remove a reactant or product, the equilibrium system will shift to produce more of the removed substance. When concentrations changes occur, the system will then achieve a new equilibrium. Remember that Kc is unaffected. Remember that Le Chatelier′s Principle states: if stress is applied to a system at equilibrium, the system changes (shifts) to relieve that stress. At 1:45 in the video above: The reason hydrogen concentration is slightly higher is because we added hydrogen to the equilibrium system, which will cause the hydrogen concentration to increase immediately. The system now responds to the stress by shifting right to consume the extra hydrogen that was added and in doing so causes N₂ and H₂ to decrease in concentration and NH₃ to increase -- all proportional to their mole ratios (coefficients). The system will then reach a new equilibrium. The concentration of hydrogen will never return to its original amount because first, the initial amount was when the system was at a different equilibrium. After responding to the stress, the system comes to a new equilibrium. Second, the system shifts to relieve the stress and will stop when the stress is relieved (a new equilibrium). In our example, the system shifts to consume the extra hydrogen that was added; thus, producing more products (increase) and as a byproduct consuming the reactants (decrease), but stops when it reaches a new equilibrium. The system's response to stress is to relieve that stress, so it will only consume the extra hydrogen that was added until a new equilibrium can be reached.

[@fatemehpourmohammad1020]

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[@instructorjohnny]

Happy to help my friend :D

[@Quayside2127]

what about if it was endothermic will the temperature increase or decrease?

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[@BakangTshubyane]

Will an increase in concentration of a solid affect equilibrium

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[@instructorjohnny]

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[@instructorjohnny]

That's good to hear my boy. You're welcome and good luck in all your classes! :D

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[@instructorjohnny]

no problem my friend!

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[@instructorjohnny]

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[@instructorjohnny]

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[@instructorjohnny]

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[@instructorjohnny]

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[@instructorjohnny]

No problem! I'm glad you found the video helpful

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[@instructorjohnny]

You're welcome!

[@adammaxwell4784]

When ammonia is taken out, the NH3 concentration at the new equilibrium is less than at the original concentration so why do they remove the ammonia in the haber process??

[@instructorjohnny]

Le Chatelier's Principle is used in The Haber process to maximize the production of ammonia. Like you said, removing ammonia causes the NH3 concentration at the new equilibrium to be less than the original concentration; therefore, it would not be practical to remove ammonia in The Haber Process. Sorry, about the confusion I caused by removing NH3 in my video. My intent was to show how the stress of removing would be graphed. I'll make sure not to do that again. If you search "Le Chatelier's Principle and The Haber Process," you can find articles and texts that go into this topic in more details. I recommend the text from: chem.libretexts.org

[@adammaxwell4784]

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[@instructorjohnny]

Anytime Adam! :)

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[@instructorjohnny]

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[@instructorjohnny]

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[@staceyii7366]

when comparison between the rate-time graph and concentration-time graph for example: when removing (ammonia) what will happen? will it be the same shape as the ammonia curve in conc-time graph? i mean the reverse rate will sharply decrease and then will be back in time? or the forward rate will sharply increase as le chatelier's principle.... please answer, i'm very confused... and i hv the exam next week..

[@instructorjohnny]

If I'm interpreting your question correctly, you're asking "what does the rate-time graph look like when removing ammonia?" If I'm not, please clarify your question once more. Use this image as reference to help visualize what I'm saying: imgur.com/a/JforCRx The rate-time graph mimics what happens on the concentration-time graph -- in its own unique way. The forward reaction represents the reactants, while the reverse reaction represents the product(s). On the rate-time graph, the line for N₂ and H₂ would become "forward reaction" to represent them both, and the line for NH₃ would be represented by "reverse reaction." The lines were straight (unchanging) on the concentration-time graph, so that means everything was initially at equilibrium. Remember. Dynamic equilibrium states that: the rate of forward reaction is equal to rate of reverse reaction. Therefore, to represent a system at equilibrium on the rate-time graph, the lines are both on top of each other (equal). From here, we follow the same steps we did to graph stress on the concentration-time graph. [1] First, graph the stress. [2] Second, graph the effect of the shift. [1] Remove NH₃ (product), so the "reverse reaction" line decreases. Removing NH₃ causes the reaction to shift right; therefore, increasing the concentration of the product, and decreasing the concentration of the reactants. [2] product increases, so "reverse reaction" line increases. [2] reactants decrease, so "forward reaction line decreases. I hoped this helped! I wish you the best on your exam.

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[@instructorjohnny]

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