Hi, I love your videos. Just one question, at 16:13 you used q/n to get delta H, but in the marking scheme, they used qxn to get delta H. Can you explain this for me please. Thanks!
I can't comment on the mark scheme you've got for the question you're using... but delta H is definitely calculated as q/n That makes sense if you think about DH being in kJmol-1 and Q is in kJ and n is in moles, so we calculate doing kJ/mol which is the less scientific way of writing kJmol-1 Qxn would give the units of kJmol (without a -1) You'll never have to do qxn You can do q= n x DH though
I've made this video about enthalpy changes generally, including enthalpy of formation. th-cam.com/video/GD4INHJLPPk/w-d-xo.html Enthalpy of formation starts at about 18 minutes
i dont understand how to determine the limiting factor and one that is in excess - is it just the one that has less moles is the limiting factor, so u use that to work out the enthalpy change
Good question... you're right that is the way. The only difference could come when the equation has a 2:1 ratio (or just not 1:1) and you see which of your moles is too small to make that ratio
Quick question When determining the moles (n) to find the enthalpy change, I know that you’re supposed to find the moles of the limiting factor in all enthalpy changes *except* enthalpy change of neutralisation where you’re supposed to take the number of moles of water formed. Is that right?
That's a great question! You're right about the limiting factor and yes you are right about the acids too... unless they say specifically for example "...per mole of acid". Thanks for the comment! 😀
With Q=mc delta T, you get Q in joules. So you then need to convert it to KJ to divide by the moles to get the enthalpy change. How come in the last example, when basically doing the steps in reverse, once you found Q by using the moles, why didn't you convert it from KJ to J, before dividing by mc? Because when you multiply the enthalpy by the moles, that gives you Q in KJ, so wouldn't you need to convert it into J before putting it into the mct equation? I'm a little bit confused by this part.
Good question. It also depends on the units for mass as well. If mass is in kg then the units of q are kJ. This works in both directions as well, so if you use q=mcT and m is in kg q automatically is in kJ
Just a quick question.. why is it that some enthalpy changes cannot be measured directly? Ive come across this question several times and it never really did make any sense to me 😅
It depends on the situation, but generally it's because you can't be sure that the equation will happen exactly as it says. So for instance, for CuSO4 + 5H2O --> CuSO4.5H2O How can you be sure that the anhydrous copper sulfate has been hydrated by precisely 5 waters per CuSO4? Answer = you can't
hi sir, great vid!! im just confused. I was taught that the line of best fit needs to either touch all of the point or at least have an equal amount of points on each side. can you please clear up any confusions for me :))
Hi, thanks for the feedback. You've not been taught the wrong thing! This is true for the majority of Graphs That's broadly the case here as well. It's just that with an extrapolation graph, there is a certain 'expected shape' and so we never use the points 'on the way up' for our best fit line. Also, watching my video back again, my data wasn't the easiest/most clear cut for a best fit line. Exams use clear and obvious data usually!
Your videos are amazing, you clear up things that are often overlooked, please don’t stop making these.
I am struggling with A level Chemistry but this helped me so much. Please make more of these xx
Thank you so much for the feedback! I'm really pleased they are useful. More this week!
you deserve to be more popular. I'm surprised to see that u only have 910 subs?!! U are very underrated!! Keep up the good work
I'm writing chemistry gce a level practicals tomorrow and this vid just taught me a whole school year study❤
Good luck! 👍
Thank you so much. I was struggling for such a long time because i couldn't understand. But you did it so clear and explained it so well. Thank you.
😁 thank you
Hi, I love your videos. Just one question, at 16:13 you used q/n to get delta H, but in the marking scheme, they used qxn to get delta H. Can you explain this for me please. Thanks!
I can't comment on the mark scheme you've got for the question you're using... but delta H is definitely calculated as q/n
That makes sense if you think about DH being in kJmol-1 and Q is in kJ and n is in moles, so we calculate doing kJ/mol which is the less scientific way of writing kJmol-1
Qxn would give the units of kJmol (without a -1)
You'll never have to do qxn
You can do q= n x DH though
@@chemistrytutor oh that makes sense, thank you so much. Love your videos.
@@ria-ip4ln thanks 😊
Do you have any videos on forming equations, in the energetics topic ?
I've made this video about enthalpy changes generally, including enthalpy of formation. th-cam.com/video/GD4INHJLPPk/w-d-xo.html
Enthalpy of formation starts at about 18 minutes
i dont understand how to determine the limiting factor and one that is in excess - is it just the one that has less moles is the limiting factor, so u use that to work out the enthalpy change
Good question... you're right that is the way.
The only difference could come when the equation has a 2:1 ratio (or just not 1:1) and you see which of your moles is too small to make that ratio
Quick question
When determining the moles (n) to find the enthalpy change, I know that you’re supposed to find the moles of the limiting factor in all enthalpy changes *except* enthalpy change of neutralisation where you’re supposed to take the number of moles of water formed. Is that right?
That's a great question! You're right about the limiting factor and yes you are right about the acids too... unless they say specifically for example "...per mole of acid".
Thanks for the comment! 😀
With Q=mc delta T, you get Q in joules. So you then need to convert it to KJ to divide by the moles to get the enthalpy change. How come in the last example, when basically doing the steps in reverse, once you found Q by using the moles, why didn't you convert it from KJ to J, before dividing by mc? Because when you multiply the enthalpy by the moles, that gives you Q in KJ, so wouldn't you need to convert it into J before putting it into the mct equation?
I'm a little bit confused by this part.
Good question.
It also depends on the units for mass as well. If mass is in kg then the units of q are kJ.
This works in both directions as well, so if you use q=mcT and m is in kg q automatically is in kJ
Hello sir , Is bomb calorimetry included in the spec ?, if yes , could you explain me how to do those types of questions ?
It's not on AQA, is that the exam board you do?
@@chemistrytutor yep , thanks for the reply . Your videos have helped me so much .
@@sjeditzandfootballreviews6297 😀 no worries 👍
Can u solve all past papers of As level practical
Paper 3
Paper 3 Question walkthroughs are on my plan, definitely. With some forcus on practical techniques too 👌
this is sooo good life saver X
Just a quick question.. why is it that some enthalpy changes cannot be measured directly?
Ive come across this question several times and it never really did make any sense to me 😅
It depends on the situation, but generally it's because you can't be sure that the equation will happen exactly as it says. So for instance, for CuSO4 + 5H2O --> CuSO4.5H2O
How can you be sure that the anhydrous copper sulfate has been hydrated by precisely 5 waters per CuSO4? Answer = you can't
@@chemistrytutor Thank you very much for your prompt response.. much appreciated.. 😊
@@umayyahosman2004 happy to help 😊
55:15 sir I don't understand why you don't need to use the negative sign, we didn't ditch it in the previous answers
Energy is a scalar quantity, so Q shouldn't ever really have units.
Since DH was negative that means its exothermic so the DT will be positive
hi sir, great vid!! im just confused. I was taught that the line of best fit needs to either touch all of the point or at least have an equal amount of points on each side. can you please clear up any confusions for me :))
Hi, thanks for the feedback. You've not been taught the wrong thing! This is true for the majority of Graphs
That's broadly the case here as well. It's just that with an extrapolation graph, there is a certain 'expected shape' and so we never use the points 'on the way up' for our best fit line.
Also, watching my video back again, my data wasn't the easiest/most clear cut for a best fit line. Exams use clear and obvious data usually!
Thanks for the help!
No problem 😊 glad it was useful!
fantastic video, thank u so much.
That's lovely to hear, I'm really pleased it's useful 😀