HSC 2012-2015 Chemistry Marathon (archive) (5 Viewers)

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Sy123

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skillstriker

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monomer = ethylene, polymerised by gas phase process, 1000-3000 atm, 300C, initiation (organic peroxide --> free radicals --> attacks double bond), propagation (chain lengthens as intiator-ethylene radical continues to attack ethylene; backbiting --> branching), termination (activated chains collide)
 

HeroicPandas

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monomer = ethylene, polymerised by gas phase process, 1000-3000 atm, 300C, initiation (organic peroxide --> free radicals --> attacks double bond --> produces ethylene radical), propagation (chain lengthens as intiator-ethylene radical continues to attack ethylene, hence forming a polymer radical; backbiting --> branching), termination (activated chains collide and is a random process, small polynmer radicals can collide with small polymer radicals (same thing for longer polymer radicals), so chain lengths can vary)
bolded - extra info (if im wrong tell me)

no catalyst + high temp and pressure = LDPE
catalyst (Ziegler-natta) + lower temp and pressure = HDPE
 
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albertcamus

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Not sure if they can ask you to explain since the verb's only 'outline'.

But as a general gist of what to write:

Step 1 initiation: Peroxide molecule splits into 2 peroxide radicals --> react with ethylene monomer in an addition reaction --> forms ethylene radical --> give structural formula.

Step 2 propagation: Activated ethylene radical reacts with 'n' ethylene monomers to form 'n' polyethylene radicals ---> give structural formula and mention that 'n' is a positive integer.

Step 3 termination: 2 polyethylene radicals (afaik 1 from the first peroxide radical and other from 2nd peroxide radical) collide at to form one large polyethylene chain --> give structural formula and mention that it's a random process so length of polyethylene chain varies.

Say that HDPE is produced through Ziegler Natta Process with Ziegler Natta Catalyst @ 60 deg C. in a hydrocarbon solvent.

And then that's pretty much the 'chemistry' behind it that you needa know.
 

someth1ng

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re: HSC Chemistry Marathon Archive



nice try mate
a. Change skillstriker's (aq) to (g).
b. Yes, nitrogen is much more electronegative than hydrogen - when they bond, the electron is closer to the nitrogen atom than the hydrogen atom. Effectively, this results in electron transfers from hydrogen to nitrogen.
 

HeroicPandas

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a. Change skillstriker's (aq) to (g).
b. Yes, nitrogen is much more electronegative than hydrogen - when they bond, the electron is closer to the nitrogen atom than the hydrogen atom. Effectively, this results in electron transfers from hydrogen to nitrogen.


a) ur missing 1 important thing
b) yeh...i'll accept that even though i expected a different answer
 
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doesn't the syllabus only specify to identify neutralization as an electron-transfer reaction?
 

HeroicPandas

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doesn't the syllabus only specify to identify neutralization as an electron-transfer reaction?
this question is within the syllabus, in part (b) if u can recall and apply previously learnt concepts, u can easily do it (if u understand what an electron-transfer reaction is)

overall, u are not required to go in the exam knowing that the Haber process is an electron-transfer reaction


SO, if anyone else wants to try out part (b) using a DIFFERENT explanation, nothing is stopping u
 

bleakarcher

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this question is within the syllabus, in part (b) if u can recall and apply previously learnt concepts, u can easily do it (if u understand what an electron-transfer reaction is)

overall, u are not required to go in the exam knowing that the Haber process is an electron-transfer reaction


SO, if anyone else wants to try out part (b) using a DIFFERENT explanation, nothing is stopping u
Haber process: N2(g)+3H2(g)->2NH3(g) delta(H)=-92 kJ/mol
Since nitrogen on the left hand side of the equation is in its elemental form its oxidation state is zero as is the oxidation state of hydrogen for the same reason. However, on the right hand side hydrogen has an oxidation state of +1 since it is part of a compound and nitrogen has an oxidation state of -3 in order to allow for a neutral NH3 molecule. Hence, redox has occurred and the Haber process is an electron-transfer reaction.
Oxidation half-equation: 3H2(g)->6(H+)+6e
Reduction half-equation: N2(g)+6e->2N^(3-)
Wasn't sure what to use as the states for the ions.
 
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someth1ng

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Haber process: N2(g)+3H2(g)->2NH3(g) delta(H)=-92 kJ/mol
Since nitrogen on the left hand side of the equation is in its elemental form its oxidation state is zero as is the oxidation state of hydrogen for the same reason. However, on the right hand side hydrogen has an oxidation state of +1 since it is part of a compound and nitrogen has an oxidation state of -3 in order to allow for a neutral NH3 molecule. Hence, redox has occurred and the Haber process is an electron-transfer reaction.
Oxidation half-equation: 3H2(g)->6(H+)+6e
Reduction half-equation: N2(g)+6e->2N^(3-)
I would be very careful when saying this. The oxidation state of hydrogen is not always +1 - even when it's in a compound.
 

bleakarcher

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I would be very careful when saying this. The oxidation state of hydrogen is not always +1 - even when it's in a compound.
Yeah true, you could have the hydride ion H- with an oxidation state of -1. So what do you recommend I say instead?
 
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