Chemistry 3/4 2013 Thread

[Edward21]

Has anyone done the STAV 2011 U4? I just did it, it's really lazy! They ask you to calculate a temperature change for 6 marks with a thermochemical equation, then a 5 mark question calculate avogadro's number. Sure I know how to do it, but would VCAA ask such point blank questions for heaps of marks with just one designated area for the entire working? A part of me thinks they'd prefer separated steps..

[ECheong]

Typically, (using past exams as a guide) VCAA will 'guide' you (separate out the marks) through a question that's based on either a new process or calculating something that you're unlikely to have calculated before (most things out of left-field). However, I do want to hedge my bets and say that even though that's been the trend, they are perfectly within their rights to turn around and do the opposite though. Just keep in mind that their aim is to test the students' knowledge of chemistry and they do want to stretch that to a degree, but on the flip side, it wouldn't be a very good question if all they did was chuck everyone in the deep end.

P.S. it's a fellow Edward! haha

[lzxnl]

Thanks nliu1995

And just checking, it'd still be possible to oxidise water to hydrogen peroxide if we bubbled fluorine gas into it, yeah?

So we'd essentially have two reactions occurring (one with water being oxidised to hydrogen peroxide and the other with water being oxidised to oxygen and H+), but oxidation to oxygen and H+ would occur to a much greater extent?

You've forgotten something: the fluorine gas would react with the hydrogen peroxide as well. That thing is thermodynamically unstable; it reacts with itself, if you look at the electrochemical series. That's why H2O2 isn't formed this way.

[Edward21]

I'm a little confused with regards to the catalysts used in esterification processes. I know H2SO4(l) is used for mixing an alcohol and acid to make an ester plus water in a general lab experiment like at school for a SAC. I'm confused when it comes to transesterification of biodiesel. You have a triglyceride with 3 fatty acids that you'd like to make into methyl esters. So you add methanol where the H atom not used in the OH group becomes added to glycerol's O to become OH again. This produces water and your methyl ester(s). WHY do we add a strong base of KOH in the process to catalyse this!? Like it's more or less another esterification reaction, why do we now use a base instead of a strong acid? Is it because we need to separate the fatty acids from the the glycerol in the first place so the methanol can condense with the carboxyl groups? It's just confusing how we go from strong acid to strong base and the textbook doesn't explicitly state why.

[Scooby]

I'm a little confused with regards to the catalysts used in esterification processes. I know H2SO4(l) is used for mixing an alcohol and acid to make an ester plus water in a general lab experiment like at school for a SAC. I'm confused when it comes to transesterification of biodiesel. You have a triglyceride with 3 fatty acids that you'd like to make into methyl esters. So you add methanol where the H atom not used in the OH group becomes added to glycerol's O to become OH again. This produces water and your methyl ester(s). WHY do we add a strong base of KOH in the process to catalyse this!? Like it's more or less another esterification reaction, why do we now use a base instead of a strong acid? Is it because we need to separate the fatty acids from the the glycerol in the first place so the methanol can condense with the carboxyl groups? It's just confusing how we go from strong acid to strong base and the textbook doesn't explicitly state why.

Yeah, I'm a bit confused about this too. Is transesterification just the direct substitution of the glycerol for a methyl/ethyl group, or does it involve first hydrolysing the triglyceride and then reacting each of the free fatty acids with an alcohol?

[Edward21]

Yeah, I'm a bit confused about this too. Is transesterification just the direct substitution of the glycerol for a methyl/ethyl group, or does it involve first hydrolysing the triglyceride and then reacting each of the free fatty acids with an alcohol?

Mmmm I'm calling Nilu1995 on this one! It just seems like there's 2 ways of looking at it. Firstly the base is used as a catalyst to hydrolyse the oil to have separate glycerol and fatty acids, then you add your alcohol to make the methyl ester and glycerol is left behind, with 3 mol. of methanol used in the equation, to produce 3 mol of fatty acids as you have to consider this happening 3 times per one mole of glycerol... However there's other representations omitting the glycerol as useless (almost like a spectator ion) and you just see the fatty acid + methanol --> water + ester, so that's like a symbolic representation (NOT LITERALLY!) but like an ionic equation that shows you what's really happening of importance in the chemical process.

[thushan]

Mmmm I'm calling Nilu1995 on this one! It just seems like there's 2 ways of looking at it. Firstly the base is used as a catalyst to hydrolyse the oil to have separate glycerol and fatty acids, then you add your alcohol to make the methyl ester and glycerol is left behind, with 3 mol. of methanol used in the equation, to produce 3 mol of fatty acids as you have to consider this happening 3 times per one mole of glycerol... However there's other representations omitting the glycerol as useless (almost like a spectator ion) and you just see the fatty acid + methanol --> water + ester, so that's like a symbolic representation (NOT LITERALLY!) but like an ionic equation that shows you what's really happening of importance in the chemical process.

No, no. Transesterification basically is where you have your triglyceride, and you mix it with solid KOH dissolved in methanol, it's not particularly good to have much water around.

(You dont need to know this) - the KOH basically deprotonates the methanol to form CH3O-, which can basically directly react with the triglyceride to form glycerol and the methyl ester in one step.

[xenial]

I believe methanol is deprotonated by KOH in this reaction - so it acts as an acid. Then the CH3CO- ion attacks the carbonyl carbon which breaks apart the ester bond and produces the methyl ester Then the H2O redonates a proton to the deprotonated glycerol. Hence you end up with the glycerol molecule, 3 methyl esters and three hydroxide ions at the end of the reaction. So the KOH is a catalyst - it isn't actually used up in the reaction.

Edit: What Thushan said haha.

[lzxnl]

From what I know, you can use either an acid or base catalyst to decompose and form esters. The mechanisms are different, but both can do the job in theory

http://www.google.com.au/imgres?imgurl=http://www.fsj.ualberta.ca/chimie/chem161/bacids/img074.gif&imgrefurl=http://www.fsj.ualberta.ca/chimie/chem161/bacids/sld074.htm&h=360&w=480&sz=13&tbnid=pccTCJgxWwp09M:&tbnh=103&tbnw=137&zoom=1&usg=__Tq5_fCUsMMcnq_6NkvWhsmTGTxU=&docid=yKJ0ed2MCoRq0M&sa=X&ei=DTRzUrKzMoSRkQWF_oFY&ved=0CFIQ9QEwBQ

Makes my point clearer. Although admittedly, the reaction mechanism doesn't seem to be as fast as the one using base. More steps than the base-catalysed version.

The problem with water is that the ion formed when KOH reacts with methanol, the methoxide ion, is probably stronger than aqueous hydroxide ion or close to it, so the methoxide ion can't form as readily.

I believe methanol is deprotonated by KOH in this reaction - so it acts as an acid. Then the CH3CO- ion attacks the carbonyl carbon which produces the methyl ester, breaking apart the ester bond and producing the methyl ester Then the H2O redonates a proton to the deprotonated glycerol.

Edit: What Thushan said haha.

Doesn't happen in water; CH3O- can't exist in water.

[xenial]

Doesn't happen in water; CH3O- can't exist in water.

What's produced when KOH reacts with methanol then? Does the ester not undergo nucleophilic attack by CH3O-?

Soz, you confused me.

[lzxnl]

You can't conduct the experiment in water. You'd have liquid methanol or something.

[xenial]

You can't conduct the experiment in water. You'd have liquid methanol or something.

Yeah yeah I know! But what's produced when the KOH reacts with methanol? Doesn't it produce water, which then deprotonates again to produce the glycerol? Do I have that part wrong?

[lzxnl]

Think of it this way.
You have KOH, which consists of solvated hydroxide and potassium ions in liquid methanol.
When this reacts with methanol to form methoxide and water, this reaction will be in equilibrium with the reverse reaction
With such tiny amounts of water available I'm not entirely sure how effective the water is as an acid.

[xenial]

Think of it this way.
You have KOH, which consists of solvated hydroxide and potassium ions in liquid methanol.
When this reacts with methanol to form methoxide and water, this reaction will be in equilibrium with the reverse reaction
With such tiny amounts of water available I'm not entirely sure how effective the water is as an acid.

Okay thanks!

[Stick]

What is the likelihood that we will have to draw actual semi-structural formulae for DNA (either a strand of nucleotides or the specific hydrogen bonding between complementary bases)? I don't really remember them that specifically, but it's come up on a couple of papers.