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a few inorganic q's

Hi,
I did the following paper :


http://www.a-levelchemistry.co.uk/AQA%20A2%20Chemistry/Unit%206W%20-%20multiple%20choice/Unit%206%20June%202004.pdf


For question 3, i thought u assume that the number of moles of O2 is 6.5, u know what the Mr of O2 is and work out the mass. However that gives a huge answer!


Question 9
Apparently on the RHS, the oxidation state of H is -2. How can it be -2? I thought it is always +1

Question 12 : no idea!

Question 34 : The answer is C. I dont get where the chiral centre is?



Thanks!
wizz_kid
Hi,
I did the following paper :


http://www.a-levelchemistry.co.uk/AQA%20A2%20Chemistry/Unit%206W%20-%20multiple%20choice/Unit%206%20June%202004.pdf


For question 3, i thought u assume that the number of moles of O2 is 6.5, u know what the Mr of O2 is and work out the mass. However that gives a huge answer!


Question 9
Apparently on the RHS, the oxidation state of H is -2. How can it be -2? I thought it is always +1

Question 12 : no idea!

Question 34 : The answer is C. I dont get where the chiral centre is?



Thanks!


3. In the balanced equation LHS total moles = 7.5. Moles of butane = 1 therefore mole fraction = 1/7.5

9. Ca is +2 and each hydride is -1

12. Use E = E(red) - E(ox)

34. ??? there is no mention of optical isomerism in q34 ??
Reply 2
Avatar for wizz_kid
wizz_kid
OP
16
charco
3. In the balanced equation LHS total moles = 7.5. Moles of butane = 1 therefore mole fraction = 1/7.5

9. Ca is +2 and each hydride is -1

12. Use E = E(red) - E(ox)

34. ??? there is no mention of optical isomerism in q34 ??

OOh right! I get everything that u said apart from E = E(red) - E(ox).
How do u know which which one is the reducing agent and which 1 is the oxidising agent + y would they tell 0.012moles of bromine?


Thank u very much.
Reply 3
Avatar for -Kav-
-Kav-
12
You're overcomplicating Q3 a bit. All you need to work out is that for complete combustion with minimum oxygen you need 1 mole of butane per 6.5 moles oxygen, or 1 mole of butane per 7.5 total moles. The mole fraction is then just 1/7.5 = 0.1333. (Hope that's right.)

Q9: Calcium's in a +2 oxidation state (as ever), so each hydrogen is -1. This is the hydride ion (H-) and although much less common than H+, you'll see it from time to time. Really good base.

Q12: You might need someone else, I'm notoriously useless with reduction potentials.

Q34: Seems to be the wrong question number, I get a work-out-the-empirical-formula type question.
Reply 4
Avatar for wizz_kid
wizz_kid
OP
16
-Kav-
You're overcomplicating Q3 a bit. All you need to work out is that for complete combustion with minimum oxygen you need 1 mole of butane per 6.5 moles oxygen, or 1 mole of butane per 7.5 total moles. The mole fraction is then just 1/7.5 = 0.1333. (Hope that's right.)

Q9: Calcium's in a +2 oxidation state (as ever), so each hydrogen is -1. This is the hydride ion (H-) and although much less common than H+, you'll see it from time to time. Really good base.

Q12: You might need someone else, I'm notoriously useless with reduction potentials.

Q34: Seems to be the wrong question number, I get a work-out-the-empirical-formula type question.


Oh yea sorry i meant question 35!

Thanks for everything else. I get everything apart from 12.
wizz_kid
OOh right! I get everything that u said apart from E = E(red) - E(ox).
How do u know which which one is the reducing agent and which 1 is the oxidising agent + y would they tell 0.012moles of bromine?


Thank u very much.


E(red) is the electrode potential of the species that gets reduced and E(ox) that the gets oxidised. If the value is greater than +0.3 the reaction is spontaneous.
Reply 6
Avatar for wizz_kid
wizz_kid
OP
16
charco
E(red) is the electrode potential of the species that gets reduced and E(ox) that the gets oxidised. If the value is greater than +0.3 the reaction is spontaneous.


Srry, how do u know that the value of E has to be greates that +0.3 for the reaction to be spontaneous?

Thnks
wizz_kid
Srry, how do u know that the value of E has to be greates that +0.3 for the reaction to be spontaneous?

Thnks


The conditions for spontaneity are related to Gibbs Free Energy.
For a reaction to be spontaneous the Gibbs free energy must be < zero, i.e. negative (in this way the overall entropy of the universe is increasing)

Gibbs free energy change = -nFEº for a reaction meaning that a reaction is feasible if Eº for the reaction is positive.

It is generally stated that a reaction which has Eº between 0 and +0.3 is an equilibrium while values greater than +0.3 the reaction goes to completion.

I admit it seems a rather arbitary number, but one that is used by most exam boards
Reply 8
Avatar for wizz_kid
wizz_kid
OP
16
charco
The conditions for spontaneity are related to Gibbs Free Energy.
For a reaction to be spontaneous the Gibbs free energy must be < zero, i.e. negative (in this way the overall entropy of the universe is increasing)

Gibbs free energy change = -nFEº for a reaction meaning that a reaction is feasible if Eº for the reaction is positive.

It is generally stated that a reaction which has Eº between 0 and +0.3 is an equilibrium while values greater than +0.3 the reaction goes to completion.

I admit it seems a rather arbitary number, but one that is used by most exam boards


Right! So we want the E0 to be >= +0.3
The E value of Bromine is 1.09 which is the specie that is going to get reduced.

Since E = E(red) - E(ox)
0.3 = 1.09 - x
so x = 0.79 which is closest to Fe3+. However the answer is D! I'm baffled lol!

Thanks.
wizz_kid
Right! So we want the E0 to be >= +0.3
The E value of Bromine is 1.09 which is the specie that is going to get reduced.

Since E = E(red) - E(ox)
0.3 = 1.09 - x
so x = 0.79 which is closest to Fe3+. However the answer is D! I'm baffled lol!

Thanks.


BUT Fe3+ is not a reducing agent!!! Bromine would have to react with a species from the RHS i.e. Fe2+

An oxidising agent must always react with a reducing agent, i.e. something from the LHS with something from the RHS

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