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

1. (Original post by Stirlo)
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
Thats what weve been taught for chem
2. (Original post by Stirlo)
Just spent the last 2 hours revising them. Think I may finally understand them :/. Now to do a last minute check of the colours
I might have to do that. Gah, I can see a very late night coming on... I've got English tomorrow too x_x
3. (Original post by angel1992)
can someone explain in words why a reaction with a negative entropy and which is exothermic won't be feasible at higher temperatures?
For this reaction, at high temperatures, delta S T will be very positive as the two negatives make a positive. Exothermic means delta H is negative, so as the temperature gets higher, delta S T will exceed delta H and therefore delta G will be Positive

Hope this helped
4. (Original post by Rahul_V)
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
any1
5. can someone explain in words why a reaction with a negative entropy and which is exothermic won't be feasible at higher temperatures
6. (Original post by Rahul_V)
any1
Have you drawn out a born-haber cycle for it?
7. (Original post by angel1992)
can someone explain in words why a reaction with a negative entropy and which is exothermic won't be feasible at higher temperatures?
delta G = delta H - (T delta S)
exothermic therefore delta h = negative
delta s = negative
therefore:
delta G = negative H minus (- S X T)
so delta G = negative H + ST
So at high temperature ST will be greater than delta H so delta G wont be -ve. So the reaction will no longer be feasible.
8. (Original post by angel1992)
can someone explain in words why a reaction with a negative entropy and which is exothermic won't be feasible at higher temperatures?
G=H-TS (don't know how to do deltas )

If the entropy is negative, then TS is always negative, so -TS is always positive.
H is negative, so H-TS will be negative when TS is small (low temp) and positive when TS is big (high temp)
Since a positive value of G means it isn't feasible, this reaction would have a maximum temperature
9. (Original post by king101)
The reaction would always be feasible since no matter the value of T, it will always be negative.

Also since Change in H is always negative for an exothermic reaction then Delta G is always negative hence is always feasible at whatever temperature.
The quoted said a negative entropy. When you minus a negative you get a positive. What you've explained it for a positive entropy
10. (Original post by angel1992)
can someone explain in words why a reaction with a negative entropy and which is exothermic won't be feasible at higher temperatures
Explained above?
11. What equation do we need to know from chapter 14.4 anyone? It has rechargeable - non rechargeable and fuel cells.
12. (Original post by Stirlo)
The quoted said a negative entropy. When you minus a negative you get a positive. What you've explained it for a positive entropy
Yeah i realsied that after i posted :P
13. (Original post by Rahul_V)
any1
I got lattice formation = +47 KJmol-1
14. (Original post by Stirlo)
Have you drawn out a born-haber cycle for it?
yeah but im not getting the right answer
keep getting -896 and the answer is -775

i can never get the right answer for these types
of questions
15. (Original post by KissMyArtichoke)
G=H-TS (don't know how to do deltas )

If the entropy is negative, then TS is always negative, so -TS is always positive.
H is negative, so H-TS will be negative when TS is small (low temp) and positive when TS is big (high temp)
Since a positive value of G means it isn't feasible, this reaction would have a maximum temperature

thankyou i just had a mental block on the mathsy bit, so if -ts is positive and h is negative , can you give an example using values, like what would your calculation be

say overall h= -20 and -ts= 140 so what would you do with these two values?
16. (Original post by angel1992)
thankyou i just had a mental block on the mathsy bit, so if -ts is positive and h is negative , can you give an example using values, like what would your calculation be

say overall h= -20 and -ts= 140 so what would you do with these two values?
17. (Original post by Rahul_V)
yeah but im not getting the right answer
keep getting -896 and the answer is -775

i can never get the right answer for these types
of questions
The value for the atomisation of Chlorine is for Cl2, not 1/2Cl2 which is what you need. You need to half the value of that
18. (Original post by Rahul_V)
yeah but im not getting the right answer
keep getting -896 and the answer is -775

i can never get the right answer for these types
of questions
i got -774. Did you remember to half the cl2 --->2cl?
19. (Original post by Stirlo)
The value for the atomisation of Chlorine is for Cl2, not 1/2Cl2 which is what you need. You need to half the value of that
thanks done that and got the right answer
20. (Original post by Rahul_V)
thanks done that and got the right answer
remember that atomisation results in one mole of product, but bond dissociation is to do with breaking one mole of bonds, so you can end up with 2 moles of product. In the exam they may just write enthalpy of atomisation of Cl2 for example instead of how you set it out.

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