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Edexcel A2 Chemistry 6ch04/05 JUNE 2015

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Say if the Enthalpy of Solution is +3kJmol^-1
So surroundings would be ( -3/298)x1000 Jmol^-1K^-1
is that right?
Original post by MeeraP07
Hiya!
What do we actually need to know about 'dilution of acids'?


strong ones increase by 1 per 10 fold
weak ones increase by 0.5
and why they do that
Original post by frozo123
strong ones increase by 1 per 10 fold
weak ones increase by 0.5
and why they do that


can you explain that please?
I saw that ina question white a while ago, still dont get it :tongue:
I know that when you reduce nitriles to primary amines using LiAlH4 you treat the product with HCl but I didnt know you do with carbonyls too?


Original post by FMSN
Can anyone please explain why acid (e.g. dilute HCl) is added after LiAlH4 is used to reduce an aldehyde or ketone? In a question I did they also added HCl after adding NaOH to something (can't remember the reaction). Hope this makes sense.
How similar are the IAL January papers compared to the summer ones?

Also, can somebody please explain this question to me?
It's the titration 6 marker from IAL Jan 2015

http://gyazo.com/12f45a888138d9653bebc8403d5d53b6
http://gyazo.com/08520eae9d080fad5803ac0a9ee5eaa8
Reply 605
Hi, can someone post GCE paper of January 2014 with the marking scheme?
Reply 606
Original post by MeeraP07
So...
the most important part of the question is to note that the NaOH reacts with BOTH acids- HCl and ethanoic acid

to find the moles of ethanoic acid at eqm you have to calculate the amount of moles of HCl and take it away from the moles of NaOH..

moles of NaOH: 45x1÷1000= 0.045
moles of HCl: 5x1÷1000= 0.005
moles of ethanoic acid at eqm: 0.045-0.005= 0.04

ethanoic acid: initial=0.120 moles
eqm=0.04 moles
0.120-0.04=0.08 which is the difference in intial and eqm

ethanol: inital:0.220 moles
eqm:0.220-0.08=0.140 moles

ester: inital: 0 moles
eqm: 0.08moles

water: intial: 0.278 moles
eqm: 0.278+0.08=0.358 moles

then...
Kc= (water)(ester) ÷ (alcohol)(carboxylic acid)
=(0.358)(0.08) ÷ (0.140)(0.04)
Kc= 5.11


Hope this makes sense! If not, lemme know


Edit: The reason the moles of ethanoic acid at eqm is 0.04 is because..
that the no. of moles of NaOH that remain after reacting with the HCl is 0.04 and because its a 1:1 ratio between ethanoic acid and NaOH the moles of ethanoic acid are also 0.04. Thats how Ive understood it anyway! Someone correct me if im wrong please!!


Hi could you as please explain this question to me?

For a given initial reacant pressure, the half-life for a first-order gaseous reaction was found to be 30 minutes. If the eperiment were repeated at half the initial reactant pressure, the half-life would be
a) 15 minutes
b) 30 minutes
c) 45 minutes
d) 60 minutes

The answer is B, but I don't understand why since it's first order?
Original post by sj97
Hi could you as please explain this question to me?

For a given initial reacant pressure, the half-life for a first-order gaseous reaction was found to be 30 minutes. If the eperiment were repeated at half the initial reactant pressure, the half-life would be
a) 15 minutes
b) 30 minutes
c) 45 minutes
d) 60 minutes

The answer is B, but I don't understand why since it's first order?


Yess its because its first order! The half life is constant for a first order reaction so its 30mins


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Original post by sj97
Hi could you as please explain this question to me?

For a given initial reacant pressure, the half-life for a first-order gaseous reaction was found to be 30 minutes. If the eperiment were repeated at half the initial reactant pressure, the half-life would be
a) 15 minutes
b) 30 minutes
c) 45 minutes
d) 60 minutes

The answer is B, but I don't understand why since it's first order?

The half life for a first order reaction is constant at a fixed temp , this means if it takes 30 min for the concentration of a reactant to fall from lets say 8 to 4 units then it will take 30 min for it to fall from 4 to 2 units and so on.
Could someone please explain how to calculate the most likely pH of:
a) 0.20 mol/dm^3 strontium hydroxide?

b) A mixture of 20cm^3 of 1.0 mol/dm^3 nitric acid and 10cm^3 of 1.0 mol/dm^3 sodium hydroxide?

This is from the June 2014 1R paper, questions 4b and 4c. The mark scheme gives the answers as 13.6 and 0.48 respectively.
Reply 610
Original post by Maham88
The half life for a first order reaction is constant at a fixed temp , this means if it takes 30 min for the concentration of a reactant to fall from lets say 8 to 4 units then it will take 30 min for it to fall from 4 to 2 units and so on.


Original post by MeeraP07
Yess its because its first order! The half life is constant for a first order reaction so its 30mins


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Oh right wasn't thinking properly! One last question: I get so confused on calculating the equilibrium moles. I think I'm making it more complicated than it actually is. COuld someone please explain to me how they do it?
(edited 8 years ago)
Original post by jshah1997
Could someone please explain how to calculate the most likely pH of:
a) 0.20 mol/dm^3 strontium hydroxide?

b) A mixture of 20cm^3 of 1.0 mol/dm^3 nitric acid and 10cm^3 of 1.0 mol/dm^3 sodium hydroxide?

This is from the June 2014 1R paper, questions 4b and 4c. The mark scheme gives the answers as 13.6 and 0.48 respectively.


For part a...
using Kw --> 10^-14÷0.2x2= 2.5x10^-14
-log(2.5x10^-14)=13.6

you times the conc of strontium hydroxide by 2 because theres two OHs in the formula (Sr(OH)2)

for part b...
moles of nitric acid= 20x1÷1000=0.02
moles of sodium hydroxide= 10x1÷1000=0.01
moles of excess acid= 0.02-0.01=0.01

conc of excess acid= 0.01x1000÷30=1/3 (30 is the total volume btw)
-log(1/3)=0.48
(edited 8 years ago)
Original post by sj97
Oh right wasn't thinking properly! One last question: I get so confused on calculating the equilibrium moles. I think I'm making it more complicated than it actually is. COuld someone please explain to me how they do it?


do you happen to have a question to make the explanation a bit easier?
Original post by aisha17
Hi, can someone post GCE paper of January 2014 with the marking scheme?


http://www.physicsandmathstutor.com/chemistry-revision/edexcel-unit-4/#past_papers
Original post by sj97
Oh right wasn't thinking properly! One last question: I get so confused on calculating the equilibrium moles. I think I'm making it more complicated than it actually is. COuld someone please explain to me how they do it?

this might help https://chemrevise.files.wordpress.com/2014/04/3-equilibria.pdf
Hi guys,
I get that for this question you'd use 50x22.5x4.2 because theyve given a formula to use but dont you usually times the mass of water in kg by temp change and 4.2 and not the vol of solution formed?
Reply 616


THank you, that was what I was looking for!
just to confirm disordered or less ordered means ions are more randomly arranged right ?
Original post by Maham88
just to confirm disordered or less ordered means ions are more randomly arranged right ?


Yes!


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