To find the rate equation, you have to find out how the concentration of each reactant independently affects the rate of reaction. This means that to find the order of reaction with respect to reactant A, only A's concentration can change and all other concentrations must stay constant.

1.

Let's start with BrO₃^-. The only change in concentration here is between 0.01 and 0.02 moldm^-3. Now you just need to select the row: you can see that the concentration of the other reactants don't change between rows 2 and 3 and only the concentration of bromate ions doubles. As concentration doubles, rate doubles. Therefore it is first order with respect to bromate ions. (2 [the factor by which the concentration increases]^{1 [the order]}=2 [the factor by which the rate increases])

2.

Next is Br^-. You can see that bromide ions' concentrations double between the 2nd and 4th rows but all other concentrations stay the same. Doubling the concentration quadruples the rate of reaction, meaning the reaction is second order with respect to bromide ions (2²=4)

3.

Using the same method for the protons, as concentration quadruples, rate quadruples from 1 to 4 so the reaction is first order here.

OVERALL: rate = k[BrO₃^-_(aq)]²[Br^-_(aq)][H⁺_(aq)]

First you need to calculate the oxidation state of vanadium in VCl₂. Cl has an oxidation state of -1 so vanadium's must be +2. This means that the half-equation we are starting at will be the one involving V²⁺. The solution started off lilac.

The vanadium was oxidised by air, showing that the second half-equation involved here is the one involving O₂.

In electrochemical cells, the half equation with a greater standard electrode potential (E^Θ) is the one that is favoured so the other equation will be reversed. We can see that the value of the V³⁺/V²⁺ half cell is more negative than that involving oxygen, so vanadium(II) ions will be oxidised by the electrons lost from oxygen's half cell in preference to that half-reaction. This means that V³⁺will form, which is green from the table.

Next we have to determine what will happen to V³⁺, looking at the second half equation. The half-cell of this and VO²⁺ is more negative than the oxygen half cell and so it V³⁺will oxidise further in preference to the oxygen half cell. VO²⁺is now formed, which is blue.

Now to find out what happens to VO²⁺. We can see that the next equation involving this has a more positive standard electrode potential. This means that vanadium in this half cell will not oxidise in preference to the oxygen half cell, but will in fact reduce if VO₂⁺is present, which it's not. Therefore, VO²⁺will not oxidise further due to its more positive standard electrode potential so the solution remains blue and not yellow.

I hope this made sense! :smile:

Thanks this was really helpful but could you please explain why the solution does not turn green instead of blue as I really couldn't understand this!

Reply 44

Harantony

10

Original post by chocolate102030

Thanks this was really helpful but could you please explain why the solution does not turn green instead of blue as I really couldn't understand this!

Green solutions are formed from V³⁺ which is shown in the table. We know that the second equation down in Table 5.2 involves V³⁺, and still has a more negative standard electrode potential than the O₂/OH^- half-reaction. This means that the second equation down will also go the other way (V³⁺will lose electrons and so is oxidised) in preference to the oxygen half-equation. This means that any V³⁺formed from V²⁺ in the first equation will be converted to VO²⁺ using the second equation.

Let me know if this still doesn't make sense. :smile:

Reply 45

jackerharder

I have been stuck with how you know if something is a homogeneous or a heterogeous catalyste.
I have seen questions come up where you had to explain why a transition metal in this case is heterogeneous and in another is homogenious.
Can somebody please help?

Reply 46

Reyoru

1

Original post by jackerharder

I have been stuck with how you know if something is a homogeneous or a heterogeous catalyste.
I have seen questions come up where you had to explain why a transition metal in this case is heterogeneous and in another is homogenious.
Can somebody please help?

Heterogeneous: catalyst is in a different physical state to the reactants.
Homogenous: catalyst is in the same physical state as the reactants.

With regard to transition metals, they are in solution so if the reactants are also in solution, it's homogeneous catalysis.

(edited 11 years ago)

Reply 47

chocolate102030

Original post by Harantony

Green solutions are formed from V³⁺ which is shown in the table. We know that the second equation down in Table 5.2 involves V³⁺, and still has a more negative standard electrode potential than the O₂/OH^- half-reaction. This means that the second equation down will also go the other way (V³⁺will lose electrons and so is oxidised) in preference to the oxygen half-equation. This means that any V³⁺formed from V²⁺ in the first equation will be converted to VO²⁺ using the second equation.

Let me know if this still doesn't make sense. :smile:

Thanks that really helped :smile:

Reply 48

hyy

Just make sure you use the time that you have left wisely. Maybe make a list of all the topics you want to go over and divide it between the time that you have left.

Also, make some time for past papers ... they are really helpful in preparing for the exam. You get used to the wording and the question kinda repeat themselves each year.

Anyways, at least you will do great in your retake exams cos you've spent load of time revising for them :smile:

Reply 49

hyy

Thanks a lot. I mean it ... I've been looking for this paper for ages. :smile:

Do you know if you can get the mark scheme?

Reply 50

super121

5

Could someone pls explain to me how to write the formula for Copper(II) complex ion formed with EDTA^4- (Last question on the June 2010 paper).

Reply 51

abzy1234

14

Original post by super121

Could someone pls explain to me how to write the formula for Copper(II) complex ion formed with EDTA^4- (Last question on the June 2010 paper).

You should know the rules for writing the formula of complexes, make sure you go over it.

Simply you write the metal first, then the ligand, then close the brackets; and write the overall charge (in this case 2-4=-2)

So in this case it'll be:
[Cu(EDTA)]-2 :smile:

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Still no Ms fellas?

Original post by abzy1234

You should know the rules for writing the formula of complexes, make sure you go over it.

Simply you write the metal first, then the ligand, then close the brackets; and write the overall charge (in this case 2-4=-2)

So in this case it'll be:
[Cu(EDTA)]-2 :smile:

Posted from TSR Mobile

I get the first bit, just not the charge. Where did 2-4 come from?

Reply 54

thegreenchildren

9

Original post by super121

I get the first bit, just not the charge. Where did 2-4 come from?

The copper ion was already Cu 2+ so when you add 4 Chlorine ligands with a charge of -1 each you get 2-4. So the overall charge was previously +2, now that you have added the 4 chlorines it is -2. :smile:

Get it?

Reply 55

abzy1234

14

Original post by super121

I get the first bit, just not the charge. Where did 2-4 come from?

Copper (II) has a charge of 2+ and EDTA has a charge of 4-

So it's simply the metal minus the ligand here, as this ligand has a negative charge. Thus 2 minus 4 equals -2.

Remember some ligands have no charge, such as water or diaminoethane. You have to deduce if the ligand has a charge :smile:

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Reply 56

abzy1234

14

Original post by stealth_writer

Still no Ms fellas?

Only teachers have it am guessing at the moment :/

My teachers are on holiday so they aren't checking my emails pleading for the markscheme lol :tongue:

We should get it by next week hopefully :smile:

Posted from TSR Mobile

Reply 57

super121

5

Original post by thegreenchildren

The copper ion was already Cu 2+ so when you add 4 Chlorine ligands with a charge of -1 each you get 2-4. So the overall charge was previously +2, now that you have added the 4 chlorines it is -2. :smile:

Get it?

That makes sense, thanks :smile:

But now i'm confused about something else, if there are 4 coppers, shouldn't the shape be tetrahedral and have a coordinate number of 4? (Sorry to be a pain, but our crap teacher didn't teach us anything about ligands)

Reply 58

thegreenchildren

9

Original post by super121

That makes sense, thanks :smile:

But now i'm confused about something else, if there are 4 coppers, shouldn't the shape be tetrahedral and have a coordinate number of 4? (Sorry to be a pain, but our crap teacher didn't teach us anything about ligands)

No worries! My teacher hasn't taught us any of steel story because he says we can just learn everything from books at home so he makes us do stupid experiments in class instead (not even important coursework ones).

Anyway, you don't need 4 coppers, you need 1 copper,this is the central metal ion, then you add 4 chlorine ligands on to it. This means it is tetrahedral in shape. Chlorine ligands are large so only 4 can fit around the central metal ion. This means the coordinate number of the complex ion is 4. H2o and OH- ligands are small so 6 of these can fit around the central metal ion, making the complex ion octahedral and the coordination number is 6.

Reply 59

Harantony

10

Original post by super121

That makes sense, thanks :smile:

But now i'm confused about something else, if there are 4 coppers, shouldn't the shape be tetrahedral and have a coordinate number of 4? (Sorry to be a pain, but our crap teacher didn't teach us anything about ligands)

I don't think it's possible to have 4 copper ions in the centre of a complex. Do you mean chlorine ligands?

If there is a coordination number of 4, then yes, the shape could be either tetrahedral or square planar. :smile:

F334 (14/01/2013 1:30PM) - Salters Chemistry (Chemistry of Materials) Revision Thread

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