OMG thank you! I was struggling so much on this and now I've finally got it. Will take a bit for me to remember this but now that I've got it in my notes I can keep going back to it. So helpful!
the x on the other side is technically 1 x , so when you are moving 2.65X you are adding 2.65X to 1 X which would equal 3.65X , as the x's arent squared so they are just added so its essentially 2.65 + 1 but you need to add the x too. I hope that makes sense i was confused too but just think about it.
I believe if x is the molar ratio for the products, then it is -x (not 1-x). This would then also make the equations work at the end, e.g. 1 mole less x (being 0.73) is 1-0.73=.27. Otherwise, the video suggests x is 0.73 and the change is 0.27 leaving 0.27 which does not work
Hi sir, I have a question here, surely, the value of Kc is generated from the concentrations at equilibrium. However, when i plug in the values at 'E' the answer does not give Kc as 7. I did (0.73)(0.73)/(0.27)(4.27)and Kc = 0.46
My understanding is that, no matter how many moles of h2o there are, only one mole will REACT in the reaction because of the given stochiomoetry, so the 1-x for the CHANGE in moles is correct
@@AlleryChemistry If you follow this link and go to page 250, this is contradicted. I am quite confused but i do think it should be (5-x) instead of (1-x). Thoughts? books.google.co.uk/books?id=QtiZA9ZyuUMC&pg=PA253&lpg=PA253&dq=The+esterification+reaction+has+an+equilibrium+constant+of+10.0+at+25+If+1+mol+of+ester+is+dissolved+in+5+mol+of+water,+what+are+the+equilibrium+amounts+of+a)&source=bl&ots=0PhsK3gnDk&sig=ACfU3U0D8PrZuZILAtt08bgsYonIqCI7GQ&hl=fr&sa=X&ved=2ahUKEwidx5Wx0OjpAhXVuXEKHevnA40Q6AEwAHoECAoQAQ#v=onepage&q=The%20esterification%20reaction%20has%20an%20equilibrium%20constant%20of%2010.0%20at%2025%20If%201%20mol%20of%20ester%20is%20dissolved%20in%205%20mol%20of%20water%2C%20what%20are%20the%20equilibrium%20amounts%20of%20a)&f=false
At about 5:30 , when we square root both sides, why don’t we account for the + and - when square rooting? Surely this results in two values? Thanks in advance!
Hi, I have a question: 4.0 mole of H2 reacts with 4.0 mole of I2 and allowed to acquire equilibrium. What is the composition of the equilibrium mixture? Kc=25For the change in mole, how do I represent the 2HI. (H2 + I2 = 2HI) I know that H2 = (1-x) and I2 = (1-x) but what about the HI since the molar ratio of HI = 2. Thank you!
+Elizabeth EYKL Good question! All you do is put 2x as your product instead of just x in the video. When you get to writing out your Kc expression it should be [HI]2 / [H2][I2] you have to square [HI] as you have 2 moles of it in the molar ratio. I worked out that x = 2.86 doing it this way. This leaves you with 1.14 of H2, 1.14 I2 and 5.72 HI. Hope this helps!
+A Level Chemistry Revision Videos by Allery Tutors Thank you for your reply. I have a slight problem, cant seem to get 2.86. 25 = (2x)^2 / (1-x)^2 Then I root both sides to get 5 = 2x / 1-x which I then get x = 0.714, can you tell me where I'm going wrong? Oh and your videos are a real lifesaver. Thank you for your hard work!
+Elizabeth EYKL Sorry. I should have said that I used the "E" part of ICE as 4-x for H2, 4-x for I2 and 2x for HI. I did this as your equation doesn't show a 1:1 ratio so 1-x can't be used as a general algebraic term. Hope this helps!
+A Level Chemistry by Allery Tutors Can you please explain which numbers we should use as I am soo confused :( In the video you said that we should you use the ration but in this question you used the number of moles?
at 3:53 you say changes, do you mean equilibrium amounts? , because you have already taken away x from the reactants, what's left is the equilibrium amount.
Why are you able to use the algebraic expressions of (1-x) and 'x' from 'C' when the Kc expression always uses the values at "E"? Also, inputting those values you obtained at the end for E does not give 7 as the Kc value. Is there something we're missing?
Hi, Can you please explain this question? Kc for the equilibrium below is 10 at 450 K. If 1 mole of the ester is mixed with 5 moles of water and the mixture allowed to reach equilibrium, how many moles of each species will be present at equilibrium? HCOOCH3(g) + H2O(g) ¾ HCOOH(g) + CH3OH(g)
Really don’t understand this at all, if there was 5 moles of water to begin with, and x moles have been used up, the equilibrium moles of water should be 5-x So why has 1-x been used instead?
actually if you use (5-x) for H2O you will get the concentrations 0.03,4.03,0.967,0.967 which will give you the correct value for Kc which is 7
what if the ratio between the reactions or between reactions or product is not 1:1, e.g. 1:3
Then you would do 3-x instead of 1-x
OMG thank you! I was struggling so much on this and now I've finally got it. Will take a bit for me to remember this but now that I've got it in my notes I can keep going back to it. So helpful!
Hey, At 8:00, why do you change -2.65x to +3.65x when you move it across?
because its 2.65x + x which inherently has a value of 1 hence its 2.65 x +1x= 3.65x.
How do you go from 2.65 - 2.65x= x to 2.65 = 3.65x why does it increase from 2.65 to 3.65??????
the x on the other side is technically 1 x , so when you are moving 2.65X you are adding 2.65X to 1 X which would equal 3.65X , as the x's arent squared so they are just added so its essentially 2.65 + 1 but you need to add the x too. I hope that makes sense i was confused too but just think about it.
I believe if x is the molar ratio for the products, then it is -x (not 1-x). This would then also make the equations work at the end, e.g. 1 mole less x (being 0.73) is 1-0.73=.27. Otherwise, the video suggests x is 0.73 and the change is 0.27 leaving 0.27 which does not work
Hi sir, I have a question here, surely, the value of Kc is generated from the concentrations at equilibrium. However, when i plug in the values at 'E' the answer does not give Kc as 7. I did (0.73)(0.73)/(0.27)(4.27)and Kc = 0.46
Those are the number of moles at equilibrium not the concentrations
If you plug in x to your algebraic Kc (x)(x)/(x-1)(x-1) you get (0.73)(0.73)/(0.23)(0.23) which does equal around 7 (7.31 but this is due to rounding)
America Explain
Hey, why 1-x instead of 5-x for h20?
+Sohaa Zahid As the molar ratio in the equation is a 1:1 ratio. We don't use the actual moles when using this algebraic form. Hope this helps
If it wasn't a 1:1 ratio will we then use it
that was water in excess, not all 5 mol has to necessarily react
My understanding is that, no matter how many moles of h2o there are, only one mole will REACT in the reaction because of the given stochiomoetry, so the 1-x for the CHANGE in moles is correct
@@AlleryChemistry If you follow this link and go to page 250, this is contradicted. I am quite confused but i do think it should be (5-x) instead of (1-x). Thoughts?
books.google.co.uk/books?id=QtiZA9ZyuUMC&pg=PA253&lpg=PA253&dq=The+esterification+reaction+has+an+equilibrium+constant+of+10.0+at+25+If+1+mol+of+ester+is+dissolved+in+5+mol+of+water,+what+are+the+equilibrium+amounts+of+a)&source=bl&ots=0PhsK3gnDk&sig=ACfU3U0D8PrZuZILAtt08bgsYonIqCI7GQ&hl=fr&sa=X&ved=2ahUKEwidx5Wx0OjpAhXVuXEKHevnA40Q6AEwAHoECAoQAQ#v=onepage&q=The%20esterification%20reaction%20has%20an%20equilibrium%20constant%20of%2010.0%20at%2025%20If%201%20mol%20of%20ester%20is%20dissolved%20in%205%20mol%20of%20water%2C%20what%20are%20the%20equilibrium%20amounts%20of%20a)&f=false
At 2:30, why do you put 1-x for water, surely it would be 5-x?
Because the '1' bit represents the molar ratio in the equation not the 'actual' value of the moles.
You just saved me a day to my Final exam. Thank you buddy. Content reloaded
All the best for your exam. 👍
How did you do?
At about 5:30 , when we square root both sides, why don’t we account for the + and - when square rooting? Surely this results in two values?
Thanks in advance!
Oh, thank you!
I was actually looking for a way to help me solve a mock test question and I thought I would never find it.
You're very welcome!
What if it isn’t a 1:1 ratio so what would we do if it was like a 1:3 or 1:4 ratio?
Hi, I have a question: 4.0 mole of H2 reacts with 4.0 mole of I2 and allowed to acquire equilibrium. What is the composition of the equilibrium mixture? Kc=25For the change in mole, how do I represent the 2HI. (H2 + I2 = 2HI) I know that H2 = (1-x) and I2 = (1-x) but what about the HI since the molar ratio of HI = 2. Thank you!
+Elizabeth EYKL Good question! All you do is put 2x as your product instead of just x in the video. When you get to writing out your Kc expression it should be [HI]2 / [H2][I2] you have to square [HI] as you have 2 moles of it in the molar ratio. I worked out that x = 2.86 doing it this way. This leaves you with 1.14 of H2, 1.14 I2 and 5.72 HI. Hope this helps!
+A Level Chemistry Revision Videos by Allery Tutors Thank you for your reply. I have a slight problem, cant seem to get 2.86.
25 = (2x)^2 / (1-x)^2
Then I root both sides to get 5 = 2x / 1-x which I then get x = 0.714, can you tell me where I'm going wrong?
Oh and your videos are a real lifesaver. Thank you for your hard work!
+Elizabeth EYKL Sorry. I should have said that I used the "E" part of ICE as 4-x for H2, 4-x for I2 and 2x for HI. I did this as your equation doesn't show a 1:1 ratio so 1-x can't be used as a general algebraic term. Hope this helps!
+A Level Chemistry by Allery Tutors Can you please explain which numbers we should use as I am soo confused :( In the video you said that we should you use the ration but in this question you used the number of moles?
where did the 3,65 come from
2.65x + x = 3.65x
x is actually 1x but in maths u never write 1 with an algebraic letter (i think)
@@Fairtrademi1 how?
would the method change if there were more moles in the product according to the molar ratio for example
H2+ I2 = 2HI
Hi. If you look at the first comment on this video from EYKL 96 you will find I have answered the same question as you have asked. Hope this helps!
thanks !!!
No problem!
Can you do 5-x instead of 1-x?
So the method used is correct but wrong numbers.. (5-x) has to be used or alse the answer will be wrong.
at 3:53 you say changes, do you mean equilibrium amounts? , because you have already taken away x from the reactants, what's left is the equilibrium amount.
how to i get = in my casio fx-83gt plus calculator for the algebra part?
why we don't subtract (x-5)when getting change of h2o
+wardah pervez Yes you could use (5-x) too. Just an alternative way of working it out.
OK thanks alot
Why are you able to use the algebraic expressions of (1-x) and 'x' from 'C' when the Kc expression always uses the values at "E"? Also, inputting those values you obtained at the end for E does not give 7 as the Kc value. Is there something we're missing?
This was really helpful! THANKS ! 😊
Well done Bro. Excellent.
Thank you! Cheers!
can you do an example for a different molar ratio for example: 2SO4 +O2 = 2SO3
Is this relevant for old spec A2 AQA? I don't think I've come across it before
Yes this is in A2 old spec.
Why isn't there power 5 for water on the kc expression??
Can I check - should the Change not be -x and the equilibrium amount be 1-x?
Hi,
Can you please explain this question?
Kc for the equilibrium below is 10 at 450 K. If 1 mole of the ester is mixed
with 5 moles of water and the mixture allowed to reach equilibrium, how
many moles of each species will be present at equilibrium?
HCOOCH3(g) + H2O(g) ¾ HCOOH(g) + CH3OH(g)
5 years later... 💀
Why do u use molar ratios to work out number of moles and not actual moles?
How would you do it if it isn’t a 1:1 ratio?
Hi, What paper/ spec was this from?
Many Thanks
What if its N2 + O2 = 2NO
Would it be 2x for 2NO because we have two moles of no?
What if the volume is also given in the question?
Brilliant, thank you!!!
Glad it was helpful!
Could we get asked this at AS? Since new spec has Kc in it
Yes I know some specs now have Kc in at AS / Year 1.
Very helpful, thank you.
what is x called?
Thank you!
Really don’t understand this at all, if there was 5 moles of water to begin with, and x moles have been used up, the equilibrium moles of water should be 5-x
So why has 1-x been used instead?
can I have a shout out for allergy chemistry and MysticDyce
i have a test in 2 days and my teacher has never taught this
now what if you aren’t given the value of kc😩😩😩
I thank you Sir
Thank you!!
thank you so much u lifesaver u
your tha man
Think I need glasses even if I don't wear them cause i can't see much. Its too far
Legend
so...many..shhmerrr
Seriously what a stupid question. So incredibly confusing. It should be 5-x
Not 1-x