Thanks for this video, quick question though. For part a) once we found the equilibrium moles we straight away put that into the formula. However for the c part we divided the equilibrium moles from the volume before inserting it into the equation. Could you please elaborate on why that was as It is really confusing me?
Good question - most of the time you'll need to use equilibrium concentration so need to divide equilibrium moles by volume. But if there are the same number of moles on each side of the equation you can use equilibrium moles instead of concentration, since all the volume terms will cancel out.
Well, I teach aqa so I make sure they cover that first and foremost. But I'm careful to make it line up with other exam boards as well. If you do Aqa all of my a level Paper 3 playlist will be useful
hi at 10:53 when you say product concentration increases i know this is wrong but I'm confused as to why eqm wouldn't shift left to oppose that change in eqm apart from that the rest of the video helped a lot thank you! :)
Hi, thanks for the question. So, the equilibrium concentration of products does go up in that situation. This is a result of the temperature increase. It is the effect that results from the temperature increase. If you think about the way the Kc expression is set out, if the value of Kc ever changes, then either the product or reactant concentrations must have gone up as part of that change. Equilibrium doesn't shift as a result of this because its already shifted to get to this position to oppose the decrease in temperature. Good luck for today!
Why we divide by the moles by the total volume of the mixture?Can anyone please explain:)) I understand that we need the concentration but how come dividing by the total volume gives us the individual concentrations?
The moles of X will be spread out over the new total volume. It won't stay as being only in whatever volume it was with at first. Once two solutions mix together, the ions dissolved in them spread over the whole of the volume. This is true whether some of the ions react or not
Great recap just night before the exam. thanks
You're welcome! Good luck 👍
You are the absolute best.
Thank you, that's very kind 😊
Thanks for this video, quick question though. For part a) once we found the equilibrium moles we straight away put that into the formula. However for the c part we divided the equilibrium moles from the volume before inserting it into the equation. Could you please elaborate on why that was as It is really confusing me?
Good question - most of the time you'll need to use equilibrium concentration so need to divide equilibrium moles by volume. But if there are the same number of moles on each side of the equation you can use equilibrium moles instead of concentration, since all the volume terms will cancel out.
Great video sir! Will you be making explain videos for inorganic chemistry?
That's my plan for September. For the exam season I'm focusing on exam question walkthrough style
Urghhh, wish i found you way earlier! This is way more useful than just plain knowledge.
Thank you! Good luck 👍
Thanks so much for these!! last one then I'm fully prepared :)
Best of luck tomorrow!
thankyou!! cramming for tomorrow
Best of luck!
really helpful, thank you!!
That's great to hear! Thanks for the feedback 😀
Please could you make a question walkthrough for electrode potentials and electrochemical cells?
Good idea. I'll put it on the list
Great idea!
will this type of question come up in an edexcel paper?
Definitely it could, yes! 😀
i do the aqa specification. DO your all your videos apply to aqa or all the exam boards?
Well, I teach aqa so I make sure they cover that first and foremost. But I'm careful to make it line up with other exam boards as well.
If you do Aqa all of my a level Paper 3 playlist will be useful
@@chemistrytutor thats very convient. thank you.
hi at 10:53 when you say product concentration increases i know this is wrong but I'm confused as to why eqm wouldn't shift left to oppose that change in eqm apart from that the rest of the video helped a lot thank you! :)
Hi, thanks for the question.
So, the equilibrium concentration of products does go up in that situation. This is a result of the temperature increase. It is the effect that results from the temperature increase. If you think about the way the Kc expression is set out, if the value of Kc ever changes, then either the product or reactant concentrations must have gone up as part of that change. Equilibrium doesn't shift as a result of this because its already shifted to get to this position to oppose the decrease in temperature.
Good luck for today!
@@chemistrytutor Thank youuu it makes sense, a question similar ending up coming up in the exam today I was so gassed 😃
@@mlakmohamed1900 Excellent! Glad it was useful!
Why we divide by the moles by the total volume of the mixture?Can anyone please explain:))
I understand that we need the concentration but how come dividing by the total volume gives us the individual concentrations?
The moles of X will be spread out over the new total volume. It won't stay as being only in whatever volume it was with at first. Once two solutions mix together, the ions dissolved in them spread over the whole of the volume. This is true whether some of the ions react or not
@@chemistrytutor ohhh I now understand it!! thank you so much!!!
@@synthetic_polymer you're very welcome 😀
5:60 Sir, would it just be mol dm-3 since we only have 1 mole of reactants and none of products, where do we get the 1 from? Thanks!🎉
Reactants are on the bottom of the kc expression so mol dm-3 is on the bottom. We need to change it to get it to the top