* Chemistry * Predictions / discussion / aftermath * (1 thread please)

EverybodyLies

Quick question, in 2002 Q8 (e) you have to divide by 2, but I have no idea why. Can anyone help me out?

ClaireMarie

EverybodyLies wrote: »

Quick question, in 2002 Q8 (e) you have to divide by 2, but I have no idea why. Can anyone help me out?

If you look at the equation, there is 2 moles of butane. You have to have ONE mole to do Hess's Law - so divide be 2, at the start or the end
That goes for any Hess's Law that has 2 or more moles of something, they try to trick you out that way!

RMD

"Do you have mole problems? If so, call Avogadro at 602-1023."

pajero12

Haha, I'm so proud I started these...One more., Why did the white bear dissolve in water??? He was a polar bear...BOOM

Exothermic

Helium walks up to the bar. The barman says "We do not serve noble gases".
... Helium doesn't react.

Well, at least we're somewhat on topic :P

._.

When Chemists die, we Barium.

RMD

One atom says to another "I think I've lost an electron". The other atom replies shocked "are you sure", to which the atom replies "I'm positive".

Harharharharhar.

ClaireMarie

Why is the world so diverse?
It's made up of alkynes of people!
Hehe.. geddit?
Why amn't I studying....

._.

RMD wrote: »

One atom says to another "I think I've lost an electron". The other atom replies shocked "are you sure", to which the atom replies "I'm positive".

Harharharharhar.

A guy on the last page got there before you

Also, €10 says they'll repeat something they asked last year on this paper. After all the other papers I've sat, I've noticed a general increase in the trickiness compared to other years, and asking the same thing 2 years in a row is something they've never done.. So most people won't be studying the experiments asked last year because they aren't expecting them to come up in any capacity.
I'm also dreading looking at Q6 incase they've decided to not ask fuels in it.. It's always a handy 50 marker for me and many other people.

Expect the unexpected people..

Gavarn

._. wrote: »

A guy on the last page got there before you

Also, €10 says they'll repeat something they asked last year on this paper. After all the other papers I've sat, I've noticed a general increase in the trickiness compared to other years, and asking the same thing 2 years in a row is something they've never done.. So most people won't be studying the experiments asked last year because they aren't expecting them to come up in any capacity.
I'm also dreading looking at Q6 incase they've decided to not ask fuels in it.. It's always a handy 50 marker for me and many other people.

Expect the unexpected people..

I hope you are wrong! Please let the paper be nice!

partyatmygaff

ClaireMarie wrote: »

If you look at the equation, there is 2 moles of butane. You have to have ONE mole to do Hess's Law - so divide be 2, at the start or the end
That goes for any Hess's Law that has 2 or more moles of something, they try to trick you out that way!

I don't think the reason for that has anything to do with Hess' law itself. The question says that the balanced chemical equation for the combustion of butane is:
2(C4H10) + 13(O2) -> 8(CO2) + 10(H20)

As you can see, there are two moles of butane used in that reaction. But the question is asking you to find the heat of combustion of butane as per definition. Not the heat of reaction of that specific reaction. As per definition, the heat of combustion of a substance is the heat change when one mole of the substance is burned completely in excess oxygen. Seeing as the reaction we were given contains two moles of butane we have to divide the heat change by two to get the heat of combustion of butane.

Another far less error prone method would be to halve the equation to C4H10 + 6.5(O2) - > 4(CO2) + 5(H20) and then proceed as normal.

Edit: I just tried that out. Is the answer -2881 Kj/Mol-1?

macskanadrag

._. wrote: »

Also, €10 says they'll repeat something they asked last year on this paper. After all the other papers I've sat, I've noticed a general increase in the trickiness compared to other years, and asking the same thing 2 years in a row is something they've never done.. So most people won't be studying the experiments asked last year because they aren't expecting them to come up in any capacity.
I'm also dreading looking at Q6 incase they've decided to not ask fuels in it.. It's always a handy 50 marker for me and many other people.

Expect the unexpected people..

Agreed. Last year's paper seemed tricky already (maybe only to me?) and then this year's biology was a bit of everything, see if we could apply a wide range of knowledge from various topics AND they also didn't bother asking stuff that came up every year since 2004, when the new syllabus was into place. Stuff that people spent looong days to get their head around :rolleyes:

I expect similar tricks for chemistry. After all, there is no rule what each question should be, is there? Only that Part A is practicals, the rest: anything!
The more i hear that the paper is pretty predictable, the more freaked out i am. What if the SEC people read Boards and think "HAH! We'll get you this time, smugfaces!" Brrr...

macskanadrag

partyatmygaff wrote: »

Edit: I just tried that out. Is the answer -2881 Kj/Mol-1?

Just looked it up, it is indeed.

ClaireMarie

partyatmygaff wrote: »

I don't think the reason for that has anything to do with Hess' law itself. The question says that the balanced chemical equation for the combustion of butane is:
2(C4H10) + 13(O2) -> 8(CO2) + 10(H20)

As you can see, there are two moles of butane used in that reaction. But the question is asking you to find the heat of combustion of butane as per definition. Not the heat of reaction of that specific reaction. As per definition, the heat of combustion of a substance is the heat change when one mole of the substance is burned completely in excess oxygen. Seeing as the reaction we were given contains two moles of butane we have to divide the heat change by two to get the heat of combustion of butane.

Another far less error prone method would be to halve the equation to C4H10 + 6.5(O2) - > 4(CO2) + 5(H20) and then proceed as normal.

Edit: I just tried that out. Is the answer -2881 Kj/Mol-1?

Maybe I didn't explain it well, but that's what I meant.
I always halve the equation at the beginning, didn't mean to say that Hess' Law was the reason for it...

jumpguy

For the preparation of ethanal, why is the ethanol + sodium dichromate from the tap funnel added so slowly? I have in my notes that it's to prevent a vigorous reaction from occuring and the contents from spurting out, but I've read elsewhere that it's to ensure there's no excess of sodium dichromate in the pear-shaped flask at any time (as ethanol must always be kept in excess to maximise the yield of ethanal).

The latter answer seems to be more sensible to me, but can anyone confirm are both correct or which is correct?

Exothermic

The answer is to prevent a vigorous exothermic (Tee hee hee :P, my username) reaction from occurring.
The sodium dichromate will always be the limiting reagent/ethanol in excess. The quantity of it in the flask doesn't affect that.

stealinhorses

Just had a look at it there.
2007 paper, Q2, a): "State two precautions which should be taken when carrying out this addition in order to avoid excessive heat production.", and the answers in the marking scheme are "add in small quantities", "add slowly", "stir gently", etc.

So yeah, to prevent a vigorous reaction.

._.

Exothermic wrote: »

The answer is to prevent a vigorous exothermic (Tee hee hee :P, my username) reaction from occurring.
The sodium dichromate will always be the linking reagent/ethanol in excess. The quantity of it in the flask doesn't affect that.

At least people have the decency to use your full username, people just keep using parts of mine over and over again

AG2R

Anyone else dropping down on the day or am I the only failure?

._.

AG2R wrote: »

Anyone else dropping down on the day or am I the only failure?

I'm going for an A1 in HL myself :pac:

resistantdoor

Hey guys I have a question about the volatile liquids experiment. In '03 they asked you to identify how the temperature of the sample was recorded. Was it recorded by putting the thermometer in the water or in the volatile liquid?

._.

resistantdoor wrote: »

Hey guys I have a question about the volatile liquids experiment. In '03 they asked you to identify how the temperature of the sample was recorded. Was it recorded by putting the thermometer in the water or in the volatile liquid?

Water

AG2R

resistantdoor wrote: »

Hey guys I have a question about the volatile liquids experiment. In '03 they asked you to identify how the temperature of the sample was recorded. Was it recorded by putting the thermometer in the water or in the volatile liquid?

temperature: use thermometer (probe, sensor) to read temperature of water (or got from diagram). (

trotsky93

hey, so i've done all pass papers, definitions for this subject
experiments seem to be piss easy questions, just wondering which ones i should do?? any help would be appreciated!:)

SoTek72

For the titration to work out the percentage of hypochlorite in bleach,

why is the Potassium iodide needed in excess, cant seem to find a reason in my book ?

ClaireMarie

SoTek72 wrote: »

For the titration to work out the percentage of hypochlorite in bleach,

why is the Potassium iodide needed in excess, cant seem to find a reason in my book ?

To make sure that all the CLO- ions have reacted fully, I think..

mpdg

SoTek72 wrote: »

For the titration to work out the percentage of hypochlorite in bleach,

why is the Potassium iodide needed in excess, cant seem to find a reason in my book ?

This question applies to both the bleach titration and the other one that subs KMnO4 for Bleach.

It is important that the KI is in excess so that all the oxidising agent (bleach/KMnO4) is reduced so that the maximum amount of iodine is released due to the amount of KMnO4/Bleach used. Essentially: because you want KMnO4/Bleach to be the limited reactant, so the amount of KMnO4/Bleach is the reactant determining how much I2 is produced.

If that makes sense?

Also: Why has Chemistry suddenly become so difficult in the last two days? I love the subject! Am I alone in this? Has anyone else forgotten everything over the last two weeks? Boo.

partyatmygaff

SoTek72 wrote: »

For the titration to work out the percentage of hypochlorite in bleach,

why is the Potassium iodide needed in excess, cant seem to find a reason in my book ?

Well first of all you need to realise the purpose of the titration. We are looking for the concentration of sodium hypochlorite in a sample of bleach. To do this we must titrate the sample of bleach against another chemical with a known concentration.

The problem however is that there are no reagents readily available for a direct titration against Sodium Hypochlorite. Therefore we use the fact that when Potassium Iodide is added to Sodium Hypochlorite elemental iodine is liberated in a ratio of one mole of elemental iodine for every one mole of sodium hypochlorite.

We add KI in excess to ensure that all the Sodium Hypochlorite reacts to liberate elemental Iodine. Once we have liberated the elemental iodine we begin the titration against Sodium Thiosulfate.

[LATEX]CLO^- + 2I^- + 2H^+ --> CL^- + I_2 + H_2O[/LATEX]
[LATEX]2S_2O_3^2- + I_2 --> S_4O_6^2- + 2I^-[/LATEX]

To explain the above two reactions...
In reaction one we add two salts together. These two salts disassociate in acidified solution to form CLO, I and H ions. These ions react together to form a CL ion, an [LATEX]I_2[/LATEX] molecule and a [LATEX]H_2O[/LATEX] molecule.

Of reaction one's products the only thing we are interested in is the ratio of [LATEX]CL^-[/LATEX] ions to [LATEX]I_2[/LATEX] molecules. They are in a ratio of 1:1

We then proceed to reaction two. We are adding a salt (Sodium Thiosulfate) to what is essentially a solution of iodine. The salt disassociates in to Na and the thiosulfate ions. Two of these thiosulfate ions then proceed to react with the Iodine molecules evolved in reaction one. We use the amount of thiosulfate used to calculate the amount of iodine that was evolved in reaction one. Seeing as in reaction one we established that one mole of iodine is liberated for each mole of hypochlorite the concentration of elemental iodine is then used as the molarity of hypochlorite.

cobear

I'm really hoping that fuels etc comes up especially after the BP oil spill - as some other poster said. Any ideas if you have to know the rough amount of carbons and boiling points for each fraction or is knowing the names, examples and uses enough?

Chris Martin

Went to UCC lecture and they said chances of Ethanol/Ethanal/Ethanoic Acid were probably not going to come up and they're taking it off the Mandatory Practical Experiment Course next year as Chemicals in their solid forms are seen to be dangerous when airborne.
What I heard anyway...
Wouldn't put money on it but he sounded pretty certain.