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Converting delta G units

if delta G is 29500 J mol-1
would it be 29.5 kJmol-1 ?

If it was -657824.4 Jmol-1
would it be -658kJmol-1


I'm asking this because my revision guide says the first one is 295 and the second one is -65.8

:s-smilie: help please
Reply 1
Avatar for alow
alow
19
29 500 = 29.5 thousand (to 3sf)

657 824.4 = 658 thousand (to 3sf)
Original post by kiiten
if delta G is 29500 J mol-1
would it be 29.5 kJmol-1 ?

If it was -657824.4 Jmol-1
would it be -658kJmol-1


I'm asking this because my revision guide says the first one is 295 and the second one is -65.8

:s-smilie: help please


Revision guide is bogus
Reply 3
Avatar for kiiten
kiiten
OP
18
Original post by AlexFam
Revision guide is bogus



But its a CGP one
Original post by kiiten
But its a CGP one


CGP = Crap Guide, People ...
Original post by kiiten
But its a CGP one


It's also wrong
Reply 6
Avatar for nao123
nao123
7
Nah, CGP ones are kinda good but you have to take the information with a bit of salt; try not to rely on it too much.
Reply 7
Avatar for alow
alow
19
Original post by kiiten
But its a CGP one


Are you sure you read the question correctly? 0.7MJmol-1 is an awfully large energy change.
Original post by alow
Are you sure you read the question correctly? 0.7MJmol-1 is an awfully large energy change.


C=O bond energy in CO2 is about 802 kJ mol-1 :dontknow:
Reply 9
Avatar for kiiten
kiiten
OP
18
Original post by alow
Are you sure you read the question correctly? 0.7MJmol-1 is an awfully large energy change.


Yeah ques 1.a)i) and 2.a)

- sorry its a bit dark :3
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1475008189593.jpg58.7KB
Reply 10
Avatar for kiiten
kiiten
OP
18
Original post by alow
Are you sure you read the question correctly? 0.7MJmol-1 is an awfully large energy change.


I also have another ques - about born haber cycle for KBr.

So if you put KBr (s) at the bottom then on the left hand side (starting from the bottom going up) is it

K (s) + 1/2 Br2 (g)

then

K (g) + 1/2 Br2 (g)

(atomisation of metal)
Reply 11
Avatar for kiiten
kiiten
OP
18
Original post by kiiten
I also have another ques - about born haber cycle for KBr.

So if you put KBr (s) at the bottom then on the left hand side (starting from the bottom going up) is it

K (s) + 1/2 Br2 (g)

then

K (g) + 1/2 Br2 (g)

(atomisation of metal)


Bump :frown:

Posted from TSR Mobile
Original post by kiiten


Any chance you could draw what you're describing in terms of a Born Haber cycle and post a picture? It's quite hard to work out exactly what you mean by your description and a picture speaks 1000 words :biggrin:
Reply 13
Avatar for kiiten
kiiten
OP
18
Original post by MexicanKeith
Any chance you could draw what you're describing in terms of a Born Haber cycle and post a picture? It's quite hard to work out exactly what you mean by your description and a picture speaks 1000 words :biggrin:


Oh yeah of course :smile: - sorry its a bit messy

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1475604392534.jpg38.8KB
Original post by kiiten
Oh yeah of course :smile: - sorry its a bit messy

Posted from TSR Mobile


yes, although with BH it does not matter if you deal with the metal or the non-metal first.

Check out the Born Haber cycles on this page
Reply 15
Avatar for kiiten
kiiten
OP
18
Original post by charco
yes, although with BH it does not matter if you deal with the metal or the non-metal first.

Check out the Born Haber cycles on this page


So was I right to cross out the 1/2 and 2 on Br (see diagram)?
Original post by kiiten
So was I right to cross out the 1/2 and 2 on Br (see diagram)?


No, you were wrong.

Your first step is atomisation of potassium

Your second step is atomisation of bromine.
Reply 17
Avatar for kiiten
kiiten
OP
18
Original post by charco
No, you were wrong.

Your first step is atomisation of potassium

Your second step is atomisation of bromine.


Ah ok, thanks :smile:

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