1st year chem question thread

[psyxwar]

ok so it's kind of dawned on me how underprepared I am for chemistry atm, so I'm gonna be asking some qs in this thread.

Can someone explain this:

I thought the entropy of the surroundings could be calculated using -deltaH/T. How'd they use the negative of that to calculate the entropy of the system?

[lzxnl]

This is the entropy change of the REACTION. Not the surroundings.
Now, the reason is because at the boiling point, the system is at equilibrium. Boiling and condensing acetone are both reversible. Therefore dG = dH - T dS = 0
dH = T dS
etc

[psyxwar]

This is the entropy change of the REACTION. Not the surroundings.
Now, the reason is because at the boiling point, the system is at equilibrium. Boiling and condensing acetone are both reversible. Therefore dG = dH - T dS = 0
dH = T dS
etc

oh yeah I see... thanks

You wouldn't be able to use dH/T for dS of system unless the dG is zero right, but you'd be able to use dS surroundings = -dH/T no matter what?

[lzxnl]

Yes. -dH/T works for the entropy change of the surroundings as dH is the amount of heat absorbed by the system, so -dH is the amount of heat absorbed by the surroundings, and by the definition of entropy, dS = d(heat)/T or -dH/T

[psyxwar]

The question is:

"As part of some experiments on the effects of acid rain on lake water pH, an aquatic chemist prepares a synthetic water sample by mixing equal volumes of 2.5mM HCO3 - and 0.25 mM CO3 2-. Calculate the pH of the synthetic water sample.

Data:

H2CO3 + H2O <-> H3O+ + HCO3 -     K1=4.47x10^-7 at 298K
HCO3- + H2O <-> H3O+ + CO32-       K2=4.68x10^-11 at 298K"

How would I go about solving this?

[lzxnl]

The question is:

"As part of some experiments on the effects of acid rain on lake water pH, an aquatic chemist prepares a synthetic water sample by mixing equal volumes of 2.5mM HCO3 - and 0.25 mM CO3 2-. Calculate the pH of the synthetic water sample.

Data:

H2CO3 + H2O <-> H3O+ + HCO3 -     K1=4.47x10^-7 at 298K
HCO3- + H2O <-> H3O+ + CO32-       K2=4.68x10^-11 at 298K"

How would I go about solving this?

Well, you know that [H+][HCO3]-/[H2CO3] = 4.47*10^-7 (which I think is an incorrect value; this is more the Ka of CO2, but whatever)
and [H+][CO3 2-]/[HCO3 -] = 4.68 * 10^-11
You can work out the concentrations of carbonate and hydrogen carbonate ions easily because you're essentially diluting each solution by a factor of 2, to give 1.25 mM HCO 3- and 0.125 mM CO3 2-
Plug into the second equation to find H+ concentration. First equation only helps you find H2CO3 concentration

Note: you can do this question this simply only because the acids and bases are weak.

[psyxwar]

Ok im here to salvage my uni chem grade again LOL

Why is the hydroxy group ortho and para directing?

[Random_Acts_of_Kindness]

Ok im here to salvage my uni chem grade again LOL

It's ok man. I believe in you.

[mahler004]

Ok im here to salvage my uni chem grade again LOL

Why is the hydroxy group ortho and para directing?

It's best explained with a picture (check your lecture notes/textbook,) but deprotonating at the ortho/para positions opens up four resonance forms, while deprotenating at the meta position leaves only three. This is true for all substituents where there's an electron donating group (and the halides.) =

[psyxwar]

It's best explained with a picture (check your lecture notes/textbook,) but deprotonating at the ortho/para positions opens up four resonance forms, while deprotenating at the meta position leaves only three. This is true for all substituents where there's an electron donating group (and the halides.) =

Wait so like the resonance stabilisation of the phenoxide ion?

Sorry what do you mean by deprotonating at the meta and deprotonating at the ortho/para? I'm pretty rusty with all this

[lzxnl]

A hydroxyl group is able to, via the lone pairs on the oxygen, donate an electron to the aromatic ring. Only at the ortho and para positions can the hydroxyl group stabilise a carbocation intermediate in this way.

[psyxwar]

Thanks.

Does high spin in the context of transition metal complexes refer to the presence of a weak field ligand -> low delta oct -> unpaired electrons, or does it refer to simply when there are unpaired electrons? I thought it'd be the former but exam answers are throwing me in doubt.

[lzxnl]

Thanks.

Does high spin in the context of transition metal complexes refer to the presence of a weak field ligand -> low delta oct -> unpaired electrons, or does it refer to simply when there are unpaired electrons? I thought it'd be the former but exam answers are throwing me in doubt.

High spin = weak field ligand
Generally second row and further transition metals are high spin as the crystal field stabilisation energy becomes so small

[nerdgasm]

Hmm, I would have thought the 'strict' definition of high-spin is when a transition metal complex can have more than one possible d-electron distribution (e.g. d4 octahedral); then the 'high spin' form is the one with more unpaired electrons, while the 'low spin' has fewer unpaired electrons.

I agree that weak field ligands mean a complex is more likely to be high spin (this is because the lowered energy gap between the orbitals means it's more energetically favourable to put an electron into a 'higher energy' orbital, instead of trying to put two electrons into a 'lower energy' orbital, as the energy required to do this (the spin-pairing energy) is greater than the energy gap (crystal-field splitting energy).

[lzxnl]

Hmm, I would have thought the 'strict' definition of high-spin is when a transition metal complex can have more than one possible d-electron distribution (e.g. d4 octahedral); then the 'high spin' form is the one with more unpaired electrons, while the 'low spin' has fewer unpaired electrons.

I agree that weak field ligands mean a complex is more likely to be high spin (this is because the lowered energy gap between the orbitals means it's more energetically favourable to put an electron into a 'higher energy' orbital, instead of trying to put two electrons into a 'lower energy' orbital, as the energy required to do this (the spin-pairing energy) is greater than the energy gap (crystal-field splitting energy).

Well yeah, I made that generalisation because for the purposes of first-year chemistry, it's sufficient. But you're right. High vs low spin configurations are really competitions between the stabilisation of the LFSE and the electron repulsions by being in the same orbital. 'High' spin, literally, would mean pairing as few electrons as possible.

The problem about what you've given there though is that d5 octahedral, if high spin, has only one possible d configuration. In addition, you can have low spin electron configurations in which you can fill the orbitals differently like d4, in which any of the t2g orbitals can accommodate the extra electron.

Of course, things get a bit more complicated when you consider different geometries and introduce ligand field theory