[Chemistry 2] Questions

[Ballerina]

Better late than never! If anyone has questions, feel free to post them and I (and hopefully others) will try and answer them.



1. Why isn't D = products? It seems the most stable...



2. Why does the Nernst equation switch back and forth in these examples between E = Eo  + RT/F ln(conc) and E = Eo  - RT/F ln(conc)?





3. For the second one, couldn't the monodentate ligands be arranged so that they had non-superimposable enantiomers, even though there are not 4 different groups? If not, is it true that all octahedral complexes with only 3 or less different types of monodentate ligands don't display optical isomerism?



4. But all have n=2, which means l=1, which means ml = -1...0...1. Why wouldn't all their ml values be the same?



Thank you! 

[lzxnl]

Better late than never! If anyone has questions, feel free to post them and I (and hopefully others) will try and answer them.



1. Why isn't D = products? It seems the most stable...

(Image removed from quote.)

2. Why does the Nernst equation switch back and forth in these examples between E = Eo  + RT/F ln(conc) and E = Eo  - RT/F ln(conc)?


(Image removed from quote.)


3. For the second one, couldn't the monodentate ligands be arranged so that they had non-superimposable enantiomers, even though there are not 4 different groups? If not, is it true that all octahedral complexes with only 3 or less different types of monodentate ligands don't display optical isomerism?

(Image removed from quote.)

4. But all have n=2, which means l=1, which means ml = -1...0...1. Why wouldn't all their ml values be the same?

(Image removed from quote.)

Thank you! 

For your Nernst equation question...I think the system is dodgy in not being able to recognise two mathematically equivalent forms. I don't think you're wrong there. You've just used ln(p(Cl2)/[Cl-]^2)==ln([Cl-]^2/p(Cl2)), which is fine.

For the ligands question, if you have six ligands, three of them being the same and the other three being equivalent to each other, in an octahedral complex you'll always have a plane of symmetry.
I think you can arrange something of the form MA₂B₂C₂ in a way to create optical isomers. I'm not sure if there is a general rule.

n=2 => second shell. "l" denotes the subshell; there are 2s and 2p subshells. "l" can take the value of 0 as well, don't forget. Otherwise how does hydrogen work?
"ml" denotes the orbital in the subshell. As the program said.

I'm clueless about the first one though.

[Ballerina]

That was well explained. Thank you!

[MelonBar]

The rate constant for the hydrolysis of sucrose is 1.85 x 10^-4 M-1 s-1 @ 25C. The activation energy is 108kJ.  What is rate constant at 37C.

ln(K / 1.85  10^-4) = 0.108 / 8.31 x (1/298 - 1/310)

solve for K

someone kindly tell me what i'm screwing up here?


edit: Shit! That was silly

[LeviLamp]

Okay, so you know the temperature, gas constant and activation energy (and we're making the assumption that Ea and A don't change with temperature, which is wrong, but that doesn't matter) for two values of k.
k = Ae^-Ea/RT
OR
ln(k) = ln(A)-Ea/R(1/T)

Pick your poison and fill this equation in for both rate constants, then just divide one by the other.
Substitute the known value of k in for k1 or whatever you want to call it.

k2                    Ae^-108x10^3/ 8.314 x 310
__________      ________________________
1.85 x 10^-4 = Ae^-108x10^3 / 8.314 x 298

k2 = 1.85 x 10^-4 * 5.406 (that's the value I got for the fraction on the right, I could have made a calculator error :s)
     = 1.00 x 10^-3

I didn't include units but you should probably work those out on the side using the same fractions

Is that the right answer???

[MelonBar]

^ That is indeed the right answer. thanks mate

[lzxnl]

My only objection is the usage of 108*10^-3 somewhere in your working; always use SI units

[MelonBar]

Use the spin-only formula to calculate the magnetic moment of a square planar complex with 4
electrons in the valence d orbitals:

Answer is 4.9 BM (4 unpaired electrons).

But I thought because it's a square planar complex it would have 0 unpaired electrons. sup wit dat

[LeviLamp]

For square planar complexes the four bottom orbitals fill as though degenerate, so you should get one electron in each of dz^2 dxy dxz and dyz - adding an electron to one of them pushes its energy above another orbital so you go to another empty one - either that or complicated quantum shenanigans that Dr. Donnelly simplified down to "just fill them as though degenerate".

also nliu1995, it eliminates the conversion error for me (I always put the decimal place in the wrong place for no reason at all) so I'll abuse the non-SI units until I die >:3

[lzxnl]

also nliu1995, it eliminates the conversion error for me (I always put the decimal place in the wrong place for no reason at all) so I'll abuse the non-SI units until I die >:3

No no, you missed the point; the top of the fraction has *10^-3, the bottom part has *10^3

For square planar complexes the four bottom orbitals fill as though degenerate, so you should get one electron in each of dz^2 dxy dxz and dyz - adding an electron to one of them pushes its energy above another orbital so you go to another empty one - either that or complicated quantum shenanigans that Dr. Donnelly simplified down to "just fill them as though degenerate".

Really...I have to know that as well...oh dear. Time to flick through my summer school notes majorly after my UMEP physics exam. I have a chemistry exam a week after my UMEP physics exam to get an exemption from first-year chemistry, and it looks like I have a LOT to cover again.

[MelonBar]

thanks levilamp my hero

now how do I do this again?

Calculate the wavelength of light (in nanometres) required to ionise a hydrogen atom from its
ground state

quite frankly i have never felt so fkd for an exam in my life 

[lzxnl]

thanks levilamp my hero

now how do I do this again?

Calculate the wavelength of light (in nanometres) required to ionise a hydrogen atom from its
ground state

quite frankly i have never felt so fkd for an exam in my life 

Ionisation energy is 13.6 eV
Use E=hc/wavelength => wavelength = hc/E

[LeviLamp]

E = -2.178 (Z^2 / n^2 (final) - Z^2/n^2 (initial) J (I MIGHT HAVE THE FINAL AND INITIAL THE WRONG WAY ROUND I'M SORRY)
Z^2 = 1 since it's hydrogen, n^2 final should be 4 and initial is 1 (single-energy-level ionisation assumed)
From the energy you can find the wavelength with E = hc/wavelength )))
We don't get/aren't expected to know that energy for chem, nliu
Remember, physics and chem do things -very- differently!

EDIT: Picked up my silly mistake in the earlier post, nliu

[lzxnl]

Wait...wouldn't you have final n be infinity? Because it's ionising?

And yeah, we get given this number in VCE and UMEP physics

[LeviLamp]

I thought it was just to the next energy level. Trevor usually means that when he says "ionise the atom" (same guy who asked how many electrons an atom with quantum number n=2 and l=1 could have -.- the answer was 6 because he 'meant to say orbital').