Stankovic123's chem q's

[zvezda]

When the intermolecular bonds are broken, the kinetic energy of the molecules don't increase; the temperature remains constant.


And effect of pressure on boiling point? Yeah that's not needed in this course

Thats interesting, i wouldve thought that breaking intermolecular bonds wouldve increased kinetic energy. The molecules would be more free to move?

[lzxnl]

When they've broken, you have water vapour, not liquid water anymore.

[Mao]

So why would that have been a different comparison? (The C-O and C-Cl).

C-H vs C-O: the effect, we have agreed, is due to the mass increase

C-Cl vs C-O: also a large mass difference, but the two absorptions are at very similar wavenumbers. Therefore, something else is happening (electronegativity, bond length/bond strength).

As you can see, if we were to independently examine C-H, C-O and C-Cl, without the prior knowledge of the formulas I have dugged up, it will be very difficult if not impossible to decide the importance of mass, bond strength and electronegativity.

This is why it was a very daring question. It only really reward students who did read very deeply into the fundamental principles behind the different spectroscopic techniques (i.e. IR absorption is due to spring-like motion of bonds).

[Mao]

Also, in some neap exam, apparently a sample cell is not made out of glass or plastic because they ansorb infrared radiation. But arent the spectra for IR derived from comparisons between a reference cell and the sample cell? The fact that glass and plastic absorb IR radiation would then have no effect?

If your experiment is only to determine a relative quantity, then what you propose should be acceptable.

But, if you are doing very sensitive experiments (e.g. looking at exotic compounds by studying small changes in their UV absorption), the absolute peak heights become important, because they tell us information about quantum yields and other fancy things, which in turn tell us about orbital overlaps, the stability of excited states, and the relaxation pathways. For questions about developing the next generation solar panels, this is incredibly important.

Comparing two almost-transparent cells would give us the relationship between absorbance vs. concentration, but it wouldn't easily tell us how much light the cell has scattered/absorbed. This is especially tricky to deal with if we don't have a totally transparent solvent (e.g. if we use a special organic solvent that also absorbs IR, it will be hard to separate the attenuation caused by the solvent [more technically referred to as the "chemical matrix"], and the attenuation caused by the cell). Therefore, we try our best to use totally transparent cells, so that we can measure the absolute absorbance without worrying about an unknown amount of cell attenuation (there is always a small amount, but we minimise it where possible).

The reason why we use reference vs sample cell is mainly due to possible absorption by the chemical matrix. As far as water is concerned (effectively transparent in the visible spectrum), it seems puzzling why we have to be so careful. But when we actually want to look at the UV spectrum or the IR spectrum, the effects of the chemical matrix/solvent can be quite large.

Re: unidirectional scattering and linearity in spectroscopy,

Going a fair while back here lol. But does this apply to AAS?

Yes. This applies to all of the "usual" types of spectroscopy.

If you continue on with chemistry into Honours/Masters, you might eventually learn about special kinds of spectroscopy in different kinds of polarised light. There, scattering becomes a bit more tricky to interpret. Rest assured, spectroscopy methods in VCE all belong to the usual and simple kinds of spectroscopy.

[Mao]

Another thing thats bothered me about the 2012 exam. Theres a question where we're asked to calculate the molar enthalpy of methanol in a calorimeter that has already been calibrated. Basically this value is lower than the one in the data book and were asked to explain why. My answer included the uneven distribution of heat energy in the water of the calorimeter, potentially leading to a situation where the thermometer is placed in an area that has a lower temperature. The assessor's report mentioned the lack of insulation asthe reason for a lower delta H value, but i deliberately avoided this because i thought to myself: if the calorimeter is calibrated, we know how much energy is required to increase the temperature of it by one degree, so who cares if its not well insulated? As long as the calorimeter is in the same environment, it shouldnt matter should it?

It's a bit more tricky than that, because the rate of heat loss is not necessarily always the same. How slowly or quickly you lose heat depends on the temperature difference between the calorimeter and its surroundings, slowed by insulation. The amount of heat lost is also time dependent. So, even though you have calibrated the calorimeter using a set amount of energy, the temperatures reached would be different (:. different rates of heat loss), and the time taken would be different (:. different amounts of heat loss).

The calibration factor of a poorly-insulated calorimeter is not well translatable to different temperatures and time scales.

[Mao]

Thats interesting, i wouldve thought that breaking intermolecular bonds wouldve increased kinetic energy. The molecules would be more free to move?

Imagine if you have a group of particles held together by elastic bands, where the elastic bands behave like intermolecular forces. The particles move in random directions with some average kinetic energy corresponding to the temperature.

Now, we stretch out the inter-particle spacing, while maintaining the same random motions with the same speed, and we keep stretching until the elastic bands snap (broken bonds). The particles still move at the same speed, but we have spent a lot of energy on breaking the intermolecular rubber bands. The particles are far away from each other and not interacting anymore, it is now a gas, a slow-moving gas, but a gas nonetheless. If we heat up the system more, the particles will then increase in their velocities/temperature.

This is a crude analogy, but I hope it will show the picture that heating doesn't always have to increase kinetic energy.

[zvezda]

It's a bit more tricky than that, because the rate of heat loss is not necessarily always the same. How slowly or quickly you lose heat depends on the temperature difference between the calorimeter and its surroundings, slowed by insulation. The amount of heat lost is also time dependent. So, even though you have calibrated the calorimeter using a set amount of energy, the temperatures reached would be different (:. different rates of heat loss), and the time taken would be different (:. different amounts of heat loss).

The calibration factor of a poorly-insulated calorimeter is not well translatable to different temperatures and time scales.

That was another reason why i avoided "lack of insulation" as an answer because the stem of the question didnt specify a different environment for the calorimeter and it was almost implied that nothing changed between the time when the calorimeter was calibrated and when the methanol was combusted. Is my reasoning still legitimate though?

[zvezda]

If your experiment is only to determine a relative quantity, then what you propose should be acceptable.

But, if you are doing very sensitive experiments (e.g. looking at exotic compounds by studying small changes in their UV absorption), the absolute peak heights become important, because they tell us information about quantum yields and other fancy things, which in turn tell us about orbital overlaps, the stability of excited states, and the relaxation pathways. For questions about developing the next generation solar panels, this is incredibly important.

Comparing two almost-transparent cells would give us the relationship between absorbance vs. concentration, but it wouldn't easily tell us how much light the cell has scattered/absorbed. This is especially tricky to deal with if we don't have a totally transparent solvent (e.g. if we use a special organic solvent that also absorbs IR, it will be hard to separate the attenuation caused by the solvent [more technically referred to as the "chemical matrix"], and the attenuation caused by the cell). Therefore, we try our best to use totally transparent cells, so that we can measure the absolute absorbance without worrying about an unknown amount of cell attenuation (there is always a small amount, but we minimise it where possible).

The reason why we use reference vs sample cell is mainly due to possible absorption by the chemical matrix. As far as water is concerned (effectively transparent in the visible spectrum), it seems puzzling why we have to be so careful. But when we actually want to look at the UV spectrum or the IR spectrum, the effects of the chemical matrix/solvent can be quite large.

Re: unidirectional scattering and linearity in spectroscopy,Yes. This applies to all of the "usual" types of spectroscopy.

If you continue on with chemistry into Honours/Masters, you might eventually learn about special kinds of spectroscopy in different kinds of polarised light. There, scattering becomes a bit more tricky to interpret. Rest assured, spectroscopy methods in VCE all belong to the usual and simple kinds of spectroscopy.

So what youre saying for that first point is that we want as little attenuation as possible to be able to maximise the possible range of absorbances for the purpose of accuracy in determining concentration and other things that im yet to touch at this stage?

[zvezda]

C-H vs C-O: the effect, we have agreed, is due to the mass increase

C-Cl vs C-O: also a large mass difference, but the two absorptions are at very similar wavenumbers. Therefore, something else is happening (electronegativity, bond length/bond strength).

As you can see, if we were to independently examine C-H, C-O and C-Cl, without the prior knowledge of the formulas I have dugged up, it will be very difficult if not impossible to decide the importance of mass, bond strength and electronegativity.

This is why it was a very daring question. It only really reward students who did read very deeply into the fundamental principles behind the different spectroscopic techniques (i.e. IR absorption is due to spring-like motion of bonds).

Yeah i see now. But thats still bizarre how mass wouldnt be the cause of the differences in wavenumbers between C-Cl and C-O.
Thanks Mao

[zvezda]

Imagine if you have a group of particles held together by elastic bands, where the elastic bands behave like intermolecular forces. The particles move in random directions with some average kinetic energy corresponding to the temperature.

Now, we stretch out the inter-particle spacing, while maintaining the same random motions with the same speed, and we keep stretching until the elastic bands snap (broken bonds). The particles still move at the same speed, but we have spent a lot of energy on breaking the intermolecular rubber bands. The particles are far away from each other and not interacting anymore, it is now a gas, a slow-moving gas, but a gas nonetheless. If we heat up the system more, the particles will then increase in their velocities/temperature.

This is a crude analogy, but I hope it will show the picture that heating doesn't always have to increase kinetic energy.

Ah so youre saying that molecules can still have high kinetic energies even if they are bonded together intermolecularly.
Thanks again Mao.

[zvezda]

Hey,
Ive been searching the internet for this but havent found an answer. In AAS and UV-Vis, as the electrons are promoted to higher energy levels, are these electrons only valence electrons or are they ones from all different energy levels?
Thanks

[lzxnl]

AAS, they're valence electrons; you can't really promote inner shell electrons to higher levels as the energy cost would be too high.
For UV-Vis, they're not valence electrons; they're electrons in molecular orbitals that are promoted, but you don't need to know about those.

[brightsky]

AAS, they're valence electrons; you can't really promote inner shell electrons to higher levels as the energy cost would be too high.
For UV-Vis, they're not valence electrons; they're electrons in molecular orbitals that are promoted, but you don't need to know about those.

as far as I recall, x-rays promote inner shell electrons to higher energy levels...but I don't think x-rays are on the course...

[zvezda]

Thanks guys. At least as far as vce goes theyre valence electrons.

Also, when changing the pH of an enzyme's surroundings, the enzyme denatures obviously. But, if one was to return the pH to where the enzyme is most active, can the enzyme function once more?
I once thought that not they cant, but after doing NEAP 2012, where they had a questions that asked "suggest why trypsin must be stored at low temperatures and a low pH", im begenning to doubt myself.

[jgoudie]

Low Temperature is fine for storage, however excessive increases in temp and changes in pH will break up the 3^O structure and the bonds will re-form in different locations.

Looks like Trypsin is a special case and it does come back to function normally.  This is the reason they gave the info and told you that it "should" be kept at a low pH, thus you would say that it is to inhibit the activity while stored.

Seems like a bit of strange thing to do and will most likely confused people, but hey, its neap and thats what they do best!

Thanks guys. At least as far as vce goes theyre valence electrons.

Also, when changing the pH of an enzyme's surroundings, the enzyme denatures obviously. But, if one was to return the pH to where the enzyme is most active, can the enzyme function once more?
I once thought that not they cant, but after doing NEAP 2012, where they had a questions that asked "suggest why trypsin must be stored at low temperatures and a low pH", im begenning to doubt myself.