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Chemistry P41 How was it?

How hard was it?

  • Easy

    Votes: 1 2.1%

  • medium

    Votes: 27 57.4%

  • Hard!!!

    Votes: 19 40.4%
  • Total voters
    47
marz_katy

marz_katy

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141
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Marks&Spencers said:
I think I might have wrote it as ligand exchange, but I honestly can't remember exactly. How many marks was it out of? That whole table I mean? It was really annoying. For the one about the gases, I wrote one CO2, the first row all H2 I think, and I think that was it. I don't think water would be given off as a gas, so I didn't write it down.
Click to expand...
that table worth 8 marks :'( and yea even i rote dos gases =D
 
Thampi4

Thampi4

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214
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51
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Oh the one with the organic

Yea first row has all H2 gas
Second row has no gas
Third row has only one gas and that was with acid only, the phenols don't react with carbonate
 
marz_katy

marz_katy

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141
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Thampi4 said:
Oh the one with the organic

Yea first row has all H2 gas
Second row has no gas
Third row has only one gas and that was with acid only, the phenols don't react with carbonate
Click to expand...
u mean to say naoh does not gv h2???..well naoh do react with phenol and alcohol ryt? atleast with phenol that im sure of...cz phenol is acidic in nature
 
Thampi4

Thampi4

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214
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51
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marz_katy said:
u mean to say naoh does not gv h2???..well naoh do react with phenol and alcohol ryt? atleast with phenol that im sure of...cz phenol is acidic in nature
Click to expand...


Yea Noah gives no h2 because its a base
Phenol reacts with it tho but no h2 produced
 
V

vivo990

Messages
53
Reaction score
18
Points
18
Marks&Spencers said:
Well then we would never be sure whether a reaction would be to the right or not. I mean, we never took a certain threshold which, if you go above it, the forward reaction would be favoured, and if you go below it, the reverse reaction would be favoured. Besides, if the reverse reaction was favoured, then you would be contradicting that Kc value you proved to everyone you got right. If you look at the values of the concentrations of the products, you would find that they were much greater than those of the reactants, so the forward reaction is favoured. Isn't that true?
Click to expand...


First of all, the relative values of Kc between reactions with different orders cannot be compared by just matching the numbers, since the units differs!. I.E., Kc, of esterification for example has no units, while the Kc in that question has a unit. This means that you cannot say in that situation whether it is high or low. As the Kc value with that unit would usually be in the order of 10^12 for example, which is higher than 10^9 that I got.

Secondly, in the chemistry course book (if you have it), it talks about how unlikely the reaction of Ecell below 0.30V to occur if it is not done under standard conditions. This means this value is low, when coming to talk about E° values.

Here is an example, Sn4+ + 2e- <---------------------> Sn4+ E°= +0.15V
You have been told that Sn+4 is slightly stable than +2 which means the position of equilibrium lies a little bit on the left. The value of that E° is close to 0.23V , so similar situation. ;)
 
Jaf

Jaf

Messages
321
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232
Points
53
About the rate of reaction question... I just didn't think half-life was the method to do it. (at our level, especially using only a graph; you would need differentiation if it were a second order reaction and since we didn't know, we shouldn't use half life)
I drew a tangent for the first curve at the first point (0,0.2). Found the gradient which = rate. Then I drew a tangent at half the conc of ethyl ethanoate and found the rate again. The ratio of concentrations was 1/2 and so was the ratio of rate of reaction. So order = 1.

I drew a third tangent at the same point (0,0.2) but this time for the curve I drew. I showed the gradient to be half of the gradient of my very first tangent (which was for the first curve, and drawn at the same point). The ratio was 0.7 something for the rate (and I'm pretty sure it was because the inaccuracy of the tangents I drew). I stated that was approximately 0.5, so order is 1.

I haven't seen anyone mention this method, so I'm getting a bit worried now... :/
 
1

1357913579

Messages
466
Reaction score
101
Points
53
Jaf said:
About the rate of reaction question... I just didn't think half-life was the method to do it. (at our level, especially using only a graph; you would need differentiation if it were a second order reaction and since we didn't know, we should use half life)
I drew a tangent for the first curve at the first point (0,0.2). Found the gradient which = rate. Then I drew a tangent at half the conc of ethyl ethanoate and found the rate again. The ratio of concentrations was 1/2 and so was the ratio of rate of reaction. So order = 1.

I drew a third tangent at the same point (0,0.2) but this time for the curve I drew. I showed the gradient to be half of the gradient of my very first tangent (which was for the first curve, and drawn at the same point). The ratio was 0.7 something for the rate (and I'm pretty sure it was because the inaccuracy of the tangents I drew). I stated that was approximately 0.5, so order is 1.

I haven't seen anyone mention this method, so I'm getting a bit worried now... :/
Click to expand...

why not the half life bro?
arent we taught that? and even its in the cie course book isnt it? like for finding the concentration of the of ethey etanoate we calculate the successive half life if they are the same then its of the order=1 and when i did i got the same half life hence it was the 1st order
now for order of hcl i drew tangent on the initial points of both the graphs and then calculated the gradient(rate) and it was doubled in the second experiment when higher concentration(which was also double) now it shows hcl rate=1
so over rate=k(hcl)(ehyl etanoate)
iam not sure if iam right but this is the way i did. tell me if iam right becuase these type of question have come n the past papers seperately this time they gave together i think they came before in paper before 2007 if i remember right.
 
1

1357913579

Messages
466
Reaction score
101
Points
53
Jaf said:
About the rate of reaction question... I just didn't think half-life was the method to do it. (at our level, especially using only a graph; you would need differentiation if it were a second order reaction and since we didn't know, we should use half life)
I drew a tangent for the first curve at the first point (0,0.2). Found the gradient which = rate. Then I drew a tangent at half the conc of ethyl ethanoate and found the rate again. The ratio of concentrations was 1/2 and so was the ratio of rate of reaction. So order = 1.

I drew a third tangent at the same point (0,0.2) but this time for the curve I drew. I showed the gradient to be half of the gradient of my very first tangent (which was for the first curve, and drawn at the same point). The ratio was 0.7 something for the rate (and I'm pretty sure it was because the inaccuracy of the tangents I drew). I stated that was approximately 0.5, so order is 1.

I haven't seen anyone mention this method, so I'm getting a bit worried now... :/
Click to expand...

why not the half life bro?
arent we taught that? and even its in the cie course book isnt it? like for finding the concentration of the of ethey etanoate we calculate the successive half life if they are the same then its of the order=1 and when i did i got the same half life hence it was the 1st order
now for order of hcl i drew tangent on the initial points of both the graphs and then calculated the gradient(rate) and it was doubled in the second experiment when higher concentration(which was also double) now it shows hcl rate=1
so over rate=k(hcl)(ehyl etanoate)
iam not sure if iam right but this is the way i did. tell me if iam right becuase these type of question have come n the past papers seperately this time they gave together i think they came before in paper before 2007 if i remember right.
 
1

1357913579

Messages
466
Reaction score
101
Points
53
Ashique said:
Same as yours. Used the half life of the line given to find the order of ester, and drew tangents at t=0 on both the graphs, to get the order of HCl =1.

Anyway, what I mean by k, is the rate constant (Yes, there was a question regarding k, right after you found the order w.r.t HCl).
It's like
Rate=k[ester][HCl]
What was your k?
Click to expand...

if ur asking regarding unit of k
then i wrote
mol^-1dm^3.s^-1
 
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