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Equilibria, Energetics and Elements (F325) - June 2011 Exam.

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Original post by jlcf
Someone posted this in this thread, a few pages back,

http://www.youtube.com/watch?v=AgJDp3CrBo8&feature=relmfu


Haha my teacher Brady, he's basically one of us.
Reply 1001
Avatar for jlcf
jlcf
1
Original post by haydyb123
Haha my teacher Brady, he's basically one of us.


You are lucky to have him, he's brillant :smile:
Original post by Flux_Pav
in your example, wasn't the NaOH in excess ?

H2S04 + 2NAOH = NA2S04 2H20

so if u have 0.5 moles of acid and 0.5mol for the alkali, you know that the naoh is in excess so you divide 0.5 by 2 as it will form 1 mol of water

is it possible in neutralization for the no. of moles of acid and alkali to be different?
what would happen then?

LOL this part is doing my head in.


No because H2SO4 dissociates twice, so the H2SO4 was in excess by 0.5, and the moles were different :smile:
Original post by Flux_Pav
in your example, wasn't the NaOH in excess ?

H2S04 + 2NAOH = NA2S04 2H20

so if u have 0.5 moles of acid and 0.5mol for the alkali, you know that the naoh is in excess so you divide 0.5 by 2 as it will form 1 mol of water

is it possible in neutralization for the no. of moles of acid and alkali to be different?
what would happen then?

LOL this part is doing my head in.


say you've got 50cm3 of NaOH reacting with 50cm3 of H2SO4, both are 1 mol dm3
the moles of them both is gonna be 0.05
the stoichiometrical equation is 2NaOH + H2SO4 -> Na2SO4 + 2H2O
say the change in temp is 13C
so Q=100x4.18x13 = -5.343 kJ
but you need to get this into kJ mol-1 so you divide by the amount of moles used, which is 0.025 (because 2:1 for NaOH:h:2SO4) so for every 2 moles of NaOH, 1 mol of H2SO4 is reacted. so even if you have 0.05 moles of both, only 0.025 moles of the sulphuric acid has reacted, so you divide by that.
so its -217.36 kJ mol -1

i don't even know if im right anymore, you've confused me :frown: why can't you just stick with HCl!!!! haha
Original post by jlcf
Your are lucky to have him, he's brillant :smile:


He mostly teaches AS, as a kind-of get them to love Chemistry role.
Reply 1005
Avatar for M_I
M_I
Can someone explain this to me:

If you have this reaction (I put the charge in the bracket):

MnO4(-) + 8H(+) + 5Fe(2+) -------------> Mn(2+) + 4H2O + 5Fe(3+)

What is the colour change?

Fe2+ is pale green and Fe3+ is yellow
but
MnO4- is purple and [Mn(H20)6](2+) is colourless

So whats the colour change?
I think everyone should watch this video:

http://www.youtube.com/watch?v=AgeCzuLVpXA&feature=relmfu
Im gonna seriously start revision now!!!
up till now I was demotivated by F215 exam from yesterday...
Original post by haydyb123
I don't think you would ever get it as 3 moles of water forming as by definition, enthalpy of neutralisation is the formation of one mole of water. :smile:


thats what i wrote
Original post by M_I
Can someone explain this to me:

If you have this reaction (I put the charge in the bracket):

MnO4(-) + 8H(+) + 5Fe(2+) -------------> Mn(2+) + 4H2O + 5Fe(3+)

What is the colour change?

Fe2+ is pale green and Fe3+ is yellow
but
MnO4- is purple and [Mn(H20)6](2+) is colourless

So whats the colour change?


purple to yellow (I could be wrong as i didn't revise this module yet)
Original post by monkeyDace
purple to yellow (I could be wrong as i didn't revise this module yet)


purple to very pale pink.
Original post by Limesasquatch
thats what i wrote


Sorry I was meant to quite flux pav!
Reply 1012
Avatar for Flux_Pav
Flux_Pav
Original post by wilsea05
say you've got 50cm3 of NaOH reacting with 50cm3 of H2SO4, both are 1 mol dm3
the moles of them both is gonna be 0.05
the stoichiometrical equation is 2NaOH + H2SO4 -> Na2SO4 + 2H2O
say the change in temp is 13C
so Q=100x4.18x13 = -5.343 kJ
but you need to get this into kJ mol-1 so you divide by the amount of moles used, which is 0.025 (because 2:1 for NaOH:h:2SO4) so for every 2 moles of NaOH, 1 mol of H2SO4 is reacted. so even if you have 0.05 moles of both, only 0.025 moles of the sulphuric acid has reacted, so you divide by that.
so its -217.36 kJ mol -1

i don't even know if im right anymore, you've confused me :frown: why can't you just stick with HCl!!!! haha


LOL haha - lets just hope this doesn't come up :smile:
anyways if it does - i hope its not as complex as i'm expecting it to be!
thanks for the explanation though - its ever so slowly increasing my understanding lol

TIME TO REVISE WHOLE OF MODULE 3 IN THE NEXT FEW HOURS.
Original post by haydyb123
Sorry I was meant to quite flux pav!


Thanks for the correct answer:smile:
Now at least I wont make that
mistake in the exam:biggrin:
Reply 1014
Avatar for jlcf
jlcf
1
Definitions we have to learn/ useful to learn

Rate of Reaction
Rate constant
Order
Half Life
RD Step
Equilibrium Law
Dynamic Equilibrium
Bronsted Lowry Acid and Base
Conjugate acid-base pairs
pH
[H+]
Strong & weak acid
Acid dissociation constant
Ionic product of water
Buffers
Equivalence point
End point
Standard enthalpy change of neutralisation
Standard enthalpy change of hydration
Standard enthalpy change of formation
Standard enthalpy change of atomisation
Standard enthalpy change of solution
Lattice Enthalpy
Ionisation energy – First and Second
Electron affinity – First and Second
Entropy
Enthalpy
Standard entropy change of reaction
Free energy change
Oxidation & reduction & their agents
Standard electrode potential
Fuel cell
Half cell
Transition element
Precipitation reactions
Complex ion
Coordinate number
Ligand
Stereoisomers
cis-platin
Mono/BI/Hexa – dentate ligands
Ligand substitution
Stability constant


Anymore?
Original post by Rogercbinboy
Anyone know if we're meant to remember the names/pH ranges of indicators? Can't find a spec' sheet anywhere to find out.

Cheers


Re-quoting that, anybody know?

Good luck people anyway, prep your tits off and you'll do fine.
Reply 1016
Avatar for waecskt
waecskt
Original post by haydyb123
I think the rule is, everything other than Pt. complexes (with a co-ordination no. of 4) is tetrahedral, where as the Pt. complexes are square planar :smile:


definitly don't need to know this, only that they could be either
Original post by wilsea05
say you've got 50cm3 of NaOH reacting with 50cm3 of H2SO4, both are 1 mol dm3
the moles of them both is gonna be 0.05
the stoichiometrical equation is 2NaOH + H2SO4 -> Na2SO4 + 2H2O
say the change in temp is 13C
so Q=100x4.18x13 = -5.343 kJ
but you need to get this into kJ mol-1 so you divide by the amount of moles used, which is 0.025 (because 2:1 for NaOH:h:2SO4) so for every 2 moles of NaOH, 1 mol of H2SO4 is reacted. so even if you have 0.05 moles of both, only 0.025 moles of the sulphuric acid has reacted, so you divide by that.
so its -217.36 kJ mol -1

i don't even know if im right anymore, you've confused me :frown: why can't you just stick with HCl!!!! haha

I am not sure now either
Isn't what you have done the enthalpy change of neutralisation per mole of acid?
Since one mole of H2SO4 produces 2H2O I think you may need to divide by 2 to get the enthalpy change of neutralisation per mole of water.
Surely they will just stick to monobasic acids in the exam :frown:
Original post by jlcf

Original post by jlcf
Definitions we have to learn/ useful to learn

Rate of Reaction
Rate constant
Order
Half Life
RD Step
Equilibrium Law
Dynamic Equilibrium
Bronsted Lowry Acid and Base
Conjugate acid-base pairs
pH
[H+]
Strong & weak acid
Acid dissociation constant
Ionic product of water
Buffers
Equivalence point
End point
Standard enthalpy change of neutralisation
Standard enthalpy change of hydration
Standard enthalpy change of formation
Standard enthalpy change of atomisation
Standard enthalpy change of solution
Lattice Enthalpy
Ionisation energy – First and Second
Electron affinity – First and Second
Entropy
Enthalpy
Standard entropy change of reaction
Free energy change
Oxidation & reduction & their agents
Standard electrode potential
Fuel cell
Half cell
Transition element
Precipitation reactions
Complex ion
Coordinate number
Ligand
Stereoisomers
cis-platin
Mono/BI/Hexa – dentate ligands
Ligand substitution
Stability constant


Anymore?


you mentioned half life twice! lol! could you please tell me the difference between equivalence point and end point? Equivalence point is the point where both the acid and base are found in same quantities, whereas end point is when you reach full neutralization? is this right?
Original post by jlcf

Anymore?


Le Chatelier's Principle? and are you making revision cards out of this? :biggrin:

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