Calculating Moles at Equilibrium

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  • Опубліковано 17 гру 2024

КОМЕНТАРІ • 45

  • @simranradia2438
    @simranradia2438 Рік тому +16

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  • @briannafuchs8145
    @briannafuchs8145 2 роки тому +5

    Thank you so much for such a clear understanding! After watching your video, my HW became a breeze.

  • @aaryamangupta
    @aaryamangupta 2 місяці тому

    Thank you so much mate. This cleared up my problem so quickly. I was struggling to think of how it would work. You earned a new sub.

  • @iiiii235
    @iiiii235 3 роки тому +2

    This guy should get more likes and subscribers

  • @Zara-xk1xj
    @Zara-xk1xj Рік тому +1

    Thank u smm for such a good and clear explanation!! My teachers couldn't teach me this despite many tries but u did in a single vid!!

  • @kevinpolpitiya3200
    @kevinpolpitiya3200 3 роки тому +7

    Thankyou so much!!! This was VERY helpful!!

  • @robzee2598
    @robzee2598 3 роки тому +3

    Perfectly explained, thanks 👍

  • @miam.g
    @miam.g 3 роки тому +2

    You explained it really well, thank youuuu

  • @ararix3722
    @ararix3722 3 роки тому +1

    Very many thanks I was unsure on when you use the constants in front of the substance but now I know thx

  • @lordofpots9947
    @lordofpots9947 2 роки тому +2

    👍 cleared things up, thank you

  • @molliegraham3329
    @molliegraham3329 3 роки тому +1

    SO well explained! thank you

  • @miguel0350
    @miguel0350 Рік тому

    Perfectly and clearly explained. Thank you sm!!

  • @vandasemedolopesfernandes3007
    @vandasemedolopesfernandes3007 2 роки тому

    Thank you so much I have been struggling with this for so long you made it easier and understandable in 6mim. there video with 15/20 minute or more but it was so difficult to get it at once. Great teacher

    • @mrmurraygreen
      @mrmurraygreen  2 роки тому

      I’m so glad you found it useful! Good luck with your studies 👍🏻

  • @ritugoyal8501
    @ritugoyal8501 Рік тому

    Very good explanation
    😊😊

  • @N.xox25
    @N.xox25 2 роки тому +1

    Perfect! Thank you

  • @trainerzard7
    @trainerzard7 3 роки тому +1

    Thank you, super duper helpful

  • @nicholaswong1359
    @nicholaswong1359 2 роки тому +1

    Bloody good video this

  • @alphaclunk
    @alphaclunk 2 роки тому +1

    the goat.

  • @DINESHPAL-xi5fk
    @DINESHPAL-xi5fk 3 роки тому +2

    Here, from india ❤️

  • @shimshonboreri
    @shimshonboreri 6 місяців тому

    absolutely saved me!

  • @yoyon3774
    @yoyon3774 3 роки тому +1

    thanks man

  • @karupsvivek992
    @karupsvivek992 Місяць тому

    Thanks a lot

  • @winproduction7585
    @winproduction7585 Рік тому

    THank you sir

  • @durjoysen3866
    @durjoysen3866 Рік тому

    Thank you sir

  • @mmaleki
    @mmaleki 2 роки тому +1

    Shouldn't all the moles in equilibrium add up to exactly 6 moles (started with 6 moles) though ? But it doesn't ...

    • @mrmurraygreen
      @mrmurraygreen  2 роки тому +1

      Hi, this is a common misconception - the total moles at equilibrium do NOT have to equal the total starting moles. If you look at some of the exam questions as a quick reference, you will find this to be the case.

    • @mmaleki
      @mmaleki 2 роки тому +1

      Mr Murray-Green Chemistry Tutorials is there maybe an explanation to it? I don't quite understand why it shouldn't and to what the rest of the moles convert to...this is confusing me a little ...
      still thanks for the video it was very well explained

    • @mrmurraygreen
      @mrmurraygreen  2 роки тому

      @@mmaleki sure thing.
      The concept of the moles at the start not having to be equal to the moles at equilibrium is the same principle as why total moles of the products in any chemical reaction do not have to equal the total moles of the reactants.
      For example, if 1 mol of Barium Carbonate thermally decomposes, you create 1 mol of Barium Oxide and 1 mol of Carbon Dioxide. That's an example where 1 mole of a reactant gives 2 moles of products in total. The reason it happens is because the 'mol' is a measure of the amount of a substance and if you change what the substances are, then the total amount of substances may now be different. Here, the 'mol' does NOT comment on the total number of atoms (that cannot change in a reaction) but what it does comment on is how those atoms are perhaps organised into different substances.
      With an equilibrium, you are using SOME of the reactant moles to make different products which are different substances so therefore may have different amounts because they may contain more/less atoms each than the reactants did. As a result the amounts of those new substances may be different and so we should not expect moles at equilibrium to always be the same as moles at the start. Just like any other chemical reaction.
      Does this help?

    • @mmaleki
      @mmaleki 2 роки тому

      Mr Murray-Green Chemistry Tutorials so if I understand it right... the moles are used to describe ratios and starting with 1 mole doesn't always result in 1 mole of product. But how is it with the limiting reactant: I learned that the limiting reactant moles limit the reaction's product moles. So let's say I've substance A (limiting, 3 moles) that reacts with substance B (6 moles) I always end up with 3 moles excess B. So what you said is it has to do with the stochiometry and that's why it doesn't always mean that the starting amount of moles have to be equal to the sum of the equilibrium moles of reactants and products.

    • @mrmurraygreen
      @mrmurraygreen  2 роки тому +2

      @@mmaleki Hi, so I feel we have wandered into what the stoichiometry means here but I shall still explain.
      This is going to be a long answer so brace yourself haha.
      Firstly, in a regular chemical reaction, the stoichiometry tells your the ratio of moles required for complete reaction (reactants side) and the amount of products you can expect to be formed (products side). Because of the way molecules are structured (assembled), you should not always expect TOTAL moles of reactants to equal TOTAL moles of products. For example in my barium carbonate decomposition example, 1 mol of barium carbonate (the only reactant) will decompose to form 1 mol of barium oxide and 1 mol of carbon dioxide (2 moles of products in total).
      Secondly, you mentioned the "limiting reagent" - this would then apply to a regular chemical reaction which does NOT reach completion i.e. something is left behind in excess as it didn't react. In this scenario, the deduce the number of moles of the limiting reagent (the reactant that does react completely) and use this to deduce the moles of the other reactant(s) used and how much product(s) are formed, again using the ratios.
      Finally (thirdly), "limiting reagent" doesn't really apply towards an equilibrium as it is an example of a reaction which does not 'reach completion' in a traditional sense. Instead we reach a balance (not an equal one) between reactants and products. In lots of equilibrium exam questions you may find that there is lots more of one reactant than another despite them being 1:1 in the equation but this does not really factor into the determination of moles at equilibrium.

  • @AlanAlan-r3d
    @AlanAlan-r3d 11 місяців тому

    Thanks🌹✨

  • @shampoo4273
    @shampoo4273 2 роки тому

    Thanks bro

  • @AshishGupta-vo1ws
    @AshishGupta-vo1ws 2 роки тому

    In a container of 1 L Which contains 4 mol o NH3​ and 6 moles of H2​S at equilibrium. Calculate the number of moles of H2​S that should be added such that at new equilibrium, moles of NH3​ decrease by 50% o its value at previous equilibrium.
    Kc = 12
    Plzz solve this question

    • @missachin5054
      @missachin5054 Рік тому

      Work out concentrations of each using MOL/VOL, plug into Kc formula and re-arrange for conc of NH3, multiply by volume for moles of NH3 at equilibrium and I assume work out 50% of that

  • @joginder294
    @joginder294 3 роки тому +2

    Here from india

  • @NWordBoy
    @NWordBoy 7 місяців тому

    I love you