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# AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry watch

1. i cant seem to calculate values for the born haber cycle
how would you work this out

2. Calculate the lattice enthalpy of sodium chloride from the following data:

Process Enthalpy change/kJmol-1
Na(s) ? Na(g) +109
Na(g) ? Na+(g) + e +494
Cl2(g) ? 2Cl(g) +242
Cl(g) + e ? Cl-(g) -360
Na(s) + 1/2Cl2(g) ? NaCl(s) -411
2. How do we know that the element carbon (diamond) would have an entropy value of 0 when its 0 K / -273C / absolute zero....
3. (Original post by Sparkly-Star)
Okay last chapter, bring it on.
frustrated no?

Top Tip: Be calm
4. does temperature increase always increase entropy
5. (Original post by rrelish)
Haha!!

Yeah, I thought I was being ignored
Thanks
the higher oxidation state is written closest to the middle of the cell... if that makes sense?!
6. (Original post by rrelish)
How do we know that the element carbon (diamond) would have an entropy value of 0 when its 0 K / -273C / absolute zero....
Well at absolute zero, atoms don't have any energy to move about at all. They are therefore in their maximum possible state of order which is why their entropy is 0
7. It depends, entropy increases from Solid>Liquid>Gas so yes, giving the particles more enery to vibrte would increase disorder.
8. Can anyone explain this to me....(from Jan 2011 Chem5 on A Level Chemistry)

1/2N2(g) + 1/2O2(g) --> NO(g)
Enthalpy of formation for N2 and O2 both zero, enthalpy of formation NO +90.4
Surely the enthalpy of the reaction should be -90.4 as no energy is put in but 90.4 is given out, but the mark scheme says the correct answer is +90.4. Am I just being really stupid?
9. (Original post by angel1992)
does temperature increase always increase entropy
well if you have a liquid and raise its temperature and it becomes a gas, then it has a huge increase in entropy.. so i guess it does yeah!
10. (Original post by rrelish)
How do we know that the element carbon (diamond) would have an entropy value of 0 when its 0 K / -273C / absolute zero....
It would have no kinetic energy therefore cannot have any disorder. That's a physics-ish answer, which is what I'd write in the chem exam :/. Not sure for a chem based answer
11. (Original post by king101)
It depends, entropy increases from Solid>Liquid>Gas so yes, giving the particles more enery to vibrte would increase disorder.
hmm, if an increase in temperature for a reversible exothermic reaction brings about a decrease in the yield of product, wouldn't entropy decrease in that case?
12. (Original post by Emily-C-27)
Can anyone explain this to me....(from Jan 2011 Chem5 on A Level Chemistry)

1/2N2(g) + 1/2O2(g) --> NO(g)
Enthalpy of formation for N2 and O2 both zero, enthalpy of formation NO +90.4
Surely the enthalpy of the reaction should be -90.4 as no energy is put in but 90.4 is given out, but the mark scheme says the correct answer is +90.4. Am I just being really stupid?
products - reactants
so +90.4 - 0 = +90.4
13. (Original post by IFondledAGibbon)
I really don't know what we're supposed to know from the 'cells' part of the spec. How much detail do we need?

As in fuel cells.

I would learn the equations for the fuel cell, along with the basic structure and the bit about it not being carbon-neutral (i.e hydrogen is made from crude oil and that it is powered by electricity which uses fossil fuels)

They asked a huge question on this in one of the previous papers and that's basically what they wanted
14. (Original post by Emily-C-27)
Can anyone explain this to me....(from Jan 2011 Chem5 on A Level Chemistry)

1/2N2(g) + 1/2O2(g) --> NO(g)
Enthalpy of formation for N2 and O2 both zero, enthalpy of formation NO +90.4
Surely the enthalpy of the reaction should be -90.4 as no energy is put in but 90.4 is given out, but the mark scheme says the correct answer is +90.4. Am I just being really stupid?
Go back to AS and draw out one of the Hess' law cycles. You'll see you go with the arrow when you go from the elements to NO so it will be a postive enthalpy change
15. (Original post by Sparkly-Star)
Okay last chapter, bring it on.
Lol, I can hear the Rocky music in my head ...
You can do it !
16. (Original post by Emily-C-27)
Can anyone explain this to me....(from Jan 2011 Chem5 on A Level Chemistry)

1/2N2(g) + 1/2O2(g) --> NO(g)
Enthalpy of formation for N2 and O2 both zero, enthalpy of formation NO +90.4
Surely the enthalpy of the reaction should be -90.4 as no energy is put in but 90.4 is given out, but the mark scheme says the correct answer is +90.4. Am I just being really stupid?
Enthalpy of formation = sum of enthalpy of products - sum of enthalpy of reactants. So yeah it would be +90.4
17. (Original post by Stirlo)
Because that mole calculation is for the 25cm3 diluted sample which is taken from the whole 250cm3 sample. In the 250cm3 sample there will be 10x as many moles of the H2O2 than in the 25cm3 sample. These moles will represented the amount of moles that were originally dilluted. As the question asks you to calculate the original conc of the H2O2, you need to get the moles in the 250cm3, not just the25cm3 sample
Ah yeah you're correct. Thank you soooo much
18. (Original post by ChloeElizabeth)
Don't know if anyone's asked this already (toooo many pages) but in the specification it says - "be able to explain why values from mean bond enthalpy calculations differ from those determined from enthalpy cycles".

But don't enthalpy cycles use mean bond enthalpies? Or do they mean Born-Haber cycles which use enthalpies of electron affinity and things like that rather than bonds broken vs bonds made??

anyonee?
19. (Original post by GoodOl'CharlieB)

I would learn the equations for the fuel cell, along with the basic structure and the bit about it not being carbon-neutral (i.e hydrogen is made from crude oil and that it is powered by electricity which uses fossil fuels)

They asked a huge question on this in one of the previous papers and that's basically what they wanted
Ah great, that's pretty much what I know. Thanks and good luck for tomorrow!
20. Does anyone have the jan06,07,08 papers please? I can't find them

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