Stankovic123's chem q's

[lzxnl]

If it makes you feel better, it's actually so amazing when you start doing stuff with symmetry - really eye-opening (though pretty hard to get your head around if you don't have the mathematical background...).

With maths? Sigh. And I thought qualitatively working out molecular orbitals of FHF- was hard. This will be fun.
Bring it. Second year chemistry plus Accelerated Maths next year. Bring it.

Yay we agree now (:

[Aurelian]

With maths? Sigh. And I thought qualitatively working out molecular orbitals of FHF- was hard. This will be fun.
Bring it. Second year chemistry plus Accelerated Maths next year. Bring it.

Look up "group theory" if you're feeling procrastinatey - it's the mathematical basis of discussions about molecular symmetry. Fortunately you don't need to know much about it (I know more or less nothing) to do lots of cool operational chemistry-related stuff. However actually properly understanding what you're doing is another story...

[lzxnl]

I've heard of that. Makes no sense to me; I don't think I'll actually come across group theory hahahaha I don't plan on doing lots of maths at uni.

[zvezda]

PS: Sorry for us semi-hijacking your thread, stankovic!

Yes. Shame on you for trying to clear things up...

Lol....
Totally fine. This was a good read.
Cheers guys

[Mao]

YES. You guys forgot something in your intense debate about IR and mass. Allow me to chime in:

The easiest way to look at the effect of mass on IR absorption is by looking at isotopic effects. That is, given the same electronic/dipole structure, how does IR absorption depend on mass?

The answer? http://en.wikipedia.org/wiki/Infrared_spectroscopy#Isotope_effects Heavier atoms gives lower wavenumbers.
Since the IR absorption frequency correspond to the frequency of vibration of bonds, it is then natural to think that heavier atoms would damp the frequency of oscillation, and so result in lower wavenumbers. You will learn later that IR absorption is a result of the fluctuation of dipole moments, not a result of (at least not directly) the magnitude of dipole moments.

Also, as a side note, the notion that atoms have a clearly defined radius is a bit of a taboo as you dive into the more precise quantum chemistry. There is no hard spherical limit that early visualisation would have you believe. You may be able to compare relative size, but never using the a precise definition of "radius". When dealing with quantum effects (IR, UV-vis, and any kind of spectroscopy), it is far more relevant to talk about orbital overlaps than atomic radii.


PS, @ stankovic123, apologies about the continued derailment.

[lzxnl]

Well, mentioning orbital overlap is hardly going to work in a vce context. I mean multi choices will have atomic radius  as an option

[zvezda]

Nah dont worry about it Mao.
So basically, whilst electronegativities may have some effect on the absorption of IR, atomic mass is more influential?

[Mao]

There is no good way to answer that. Let me quote the question here for clarity:



The exam writers were very bold to have asked this. I don't believe there is a clear cut answer (at least not within the context of VCE), as the theory and mechanism of IR absorption is not discussed in detail at all. Knowledge required to eliminate the "incorrect" option are not accessible to year 12 students.

Firstly, we ask ourselves, does a simple, general pattern exist? The answer is not so clear-cut, because more complex molecules behave in even more exotic ways. There is no simple, clear trend determining IR absorption wavenumbers from available data. Sometimes, the fluctuation of the absorption wavenumber is greater than the trend itself (e.g. C-C vs C-Cl, O-H vs N-H).

What if we limit ourselves to very simple molecules? This runs into another problem, and that is the limited number of "simple" molecules. All the simple molecules have the same bonds, C-H, C-O, C-C, etc etc. It is impossible to untangle the different effects just by looking at these simple molecules, because all the trends follow each other. Down the group, increasing mass increases the radius and decreases the electronegativity. Across the period, increasing mass decreases the radius and increases the electronegativity. Radius and electronegativity seems to always be opposite each other. As we constrict ourselves to the non-metallic part of the periodic table, we can only really make systematic comparisons down the halide group (4 data points) and across period 2 (3 or 4 data points). It is very hard to separate any of the effects since we have so little data.

The reality is, these effects all contribute significantly, and we must develop theoretical, mathematical models that capture all of the effects. With a mathematical model, we can then look at the "hypothetical" contribution of each individual factor.

Amazingly, the actual formulas for this is surprisingly simple. http://scitation.aip.org/content/aip/journal/jcp/14/5/10.1063/1.1724138 To understand how it is derived would be much more difficult (takes years or months), but we can use the formula to breakdown the contributions here.

Using this formula, it is actually possible to breakdown the different contribution of electronegativity, atomic radius and mass to the change in bond wavenumber. They are:
- electronegativity (~20% decrease from C-O to C-H)
- atomic radius/van der Waals radius (~20% increase from C-O to C-H)
- mass (~170% increase from C-O to C-H)
So, as nature would have it, mass is the most improtant thing in this particular case. A different set of comparisons (e.g. C-O vs C-Cl) would yield a very different answer.

What do we learn from this? Nature is complicated. There is not always simple trends that can be explained by concepts like electronegativity and atomic radius. The more chemistry you know, the less sense it makes.

/rant.
ps. apologies for bad grammar. It's late here <.< I am le tired.

[zvezda]

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

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?

Thanks again guys

[zvezda]

To a very small extent. It mostly the original beam which did not get absorbed.

E.g. light at a illuminance is shown at the cell, 30% gets absorbed, the light that enters the detector would then have an illuminance of . The figure of 70% is the 'transmittance'. This makes up most of the light entering the detector.

Note that light is not necessarily 'emitted' by the molecules after absorption. The absorbed light can be dissipated via other pathways that do not emit light (e.g. via bending/vibrations of the molecules), or emit light at a different frequency (e.g. fluorescence/phosphorescence).

Sometimes, light is emitted by the excited molecule at the same frequency. However, because light is emitted in all directions, it typically contributes very little to the total transmittance. Furthermore, even if the emitted light is significant, the linearity of the absorbance-vs-concentration relationship would be preserved, so our usual calibration curve method still works. Note that the measured transmittance includes the effect of light emitted at the same frequency.

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

[zvezda]

Hey,
Just a question with the theory behind q15 on the multi choice of last years exam 2. I always thought that when we calculate the energy required to change the temperature of x grams of water by y degrees, this covered everything such as change of states etc. Because in this question, a separate amount of energy is required to convert liquid water to a gaseous state, aside from heating it to 100 degrees.
Would anyone be able to further explain this? Because doesnt this imply that at 100 degrees and even higher, water can still be a liquid?
Thanks

[brightsky]

Hey,
Just a question with the theory behind q15 on the multi choice of last years exam 2. I always thought that when we calculate the energy required to change the temperature of x grams of water by y degrees, this covered everything such as change of states etc. Because in this question, a separate amount of energy is required to convert liquid water to a gaseous state, aside from heating it to 100 degrees.
Would anyone be able to further explain this? Because doesnt this imply that at 100 degrees and even higher, water can still be a liquid?
Thanks

it has to do with this thing called latent heat of vaporisation which we learned in year 11 and which I can only vaguely remember. latent heat is invisible to the thermometer, and is basically the heat required to actually change the state of something at the threshold temperature/pressure. I don't really fully understand it myself...maybe nliu can offer a better explanation...

[Dismounted]

Hey,
Just a question with the theory behind q15 on the multi choice of last years exam 2. I always thought that when we calculate the energy required to change the temperature of x grams of water by y degrees, this covered everything such as change of states etc. Because in this question, a separate amount of energy is required to convert liquid water to a gaseous state, aside from heating it to 100 degrees.
Would anyone be able to further explain this? Because doesnt this imply that at 100 degrees and even higher, water can still be a liquid?
Thanks

Let's get water from, say 30 degrees C to 110 degrees C.

First, we use its specific heat capacity (c=4.18 J/g/K) and to go from 30 to 100. At this stage, the water is still in liquid form at 100 degrees C.

Further energy is required to actually pull the molecules apart (we call this its latent heat of vaporisation in J/g or similar units, and denoted L_v). So we use to work out how much energy it takes to pull the molecules apart (liquid to gas). At this stage, the gaseous water remains at 100 degrees C.

Lastly, we continue on our merry way, again using to get the gaseous water from 100 to 110 degrees C.

Specific heat capacities are the same across all states (usually), but to change states, we require an additional load of energy to either pull and push molecules together.

To go from solid to liquid (and vice versa), we use something called the latent heat of fusion, denoted L_f.

(Trying not to confuse you any further, but it is only at 1 atm (pressure) that water boils at 100 degrees C. If we change the pressure of the environment, water boils and freezes at different temperatures. In fact, at higher elevations (lower pressure), water boils at below 100 degrees C, causing issues for hikers trying to cook their food!)

[lzxnl]

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

[zvezda]

Ahh right i see.
Thanks guys. Nice finishing touch to that explanation, dismounted lol.

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?