Calculating Enthalpy Changes at Different Temperatures To calculate $\Delta H$, the change in enthalpy at $\mathrm{100^\circ C}$ for the reaction below, one needs what addition information?

  $$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$$

  $\Delta H^\circ = \pu{-92.0 kJ}$ at $\mathrm{25^\circ C}$

  (a) The equilibrium constant for the reaction at $\mathrm{100^\circ
 C}$

  (b) The molar heat capacities of the reaction and the products as a
   function of temperature

  (c) $\Delta E^\circ$, the standard internal energy change for the
   reaction

  (d) The partial pressures of the reactants and products at
   $\mathrm{100^\circ C}$

  (e) The entropies of formation for the reactants and products at
   $\mathrm{100^\circ C}$

The correct answer is (b), but I'm not entirely sure why.

I know it can't be (a) because I think we need $\Delta S$ as well.  Also, (c) won't be helpful because it is at standard temperature, and we are wanting information at $\mathrm{100^\circ C}$.  Finally, (e), seems correct but they are actually talking about entropy instead of enthalpy.  However, I'm not sure how to rule out (d) or justify (b).

@Chet already gave the qualitative answer but the actual equation it is based on is:

$$
\Delta H^\circ(T)=\Delta H{^\circ}_{T_1} + \int_{T_1}^{T} \Delta C_p{^\circ}(T^\prime ) \,\mathrm{d}T^\prime$$

If you know the enthalpy at a reference temperature and the heat capacity as a function of temperature you can calculate the enthalpy change at any other temperature.

Actually d) and e) together are not wrong. We all know these equations:
$$\Delta G= - RT\ln (K) \qquad \Delta G= \Delta H - T\Delta S $$

With d) you can calculate $\Delta G$ and with e) you can calculate $\Delta H$.

Lets look deeper into: $\Delta G^\circ$, $\Delta G$, $K$ and $Q$.
To answer your second question completely I would recommend to look again into a textbook of your choice. Especially the definitions of activity and chemical potential.

If you know the chemical potential $\mu^\circ$ of a pure substance you can calculate the chemical potential of mixtures if you know the activities:
$$\mu= \mu^\circ + RT \ln(a)$$

Now we make two assumptions:

We have constant temperature and pressure. ($\mu = \Delta G$)
We can calculate the activities simply over the partial pressures or concentrations.

This leads then to your equation:
$$
\Delta G= \Delta G^\circ + RT \ln(Q)
$$
where $Q$ is the reaction quotient.
Since we assume equilibrium,  we know that $\Delta G=0$ and $Q = K$. This leads to:
$$ 0= \Delta G^\circ + RT \ln(K) \Rightarrow \Delta G^\circ = -RT \ln(K) $$

The last equation is the handy one you use for calculating equilibrium constants from energies and vice versa.
So to be really precise one has to write:
$$ \Delta G^\circ= \Delta H^\circ - T\Delta S^\circ $$

But one usually skips the superscript $^\circ$.
At equilibrium $Q$ and $K$ are essentially the same.


Is there anything like "Saturated Fullerene"? This question arose from my previous question: Can benzyne be trapped inside a polycyclic hydrocarbon?. While dodecahedrane was too small to fit an anthranilate ion or even a benzyne ion, there are lots of other cage molecules, notably fullerenes. Unfortunately fullerenes themselves have partial double bond characters in them, making them liable to undergo a Diels-Alder reaction with benzyne.

However, if it were possible to "saturate" the fullerene, it could serve as a molecular container to hold the benzyne molecule.

Is it possible to saturated fullerenes? After being saturated,  can we use it as a molecular container?

I'm limiting my answer to $\ce{C60}$-fullerene, but the same idea applies to the higher fullerenes as well.

In short, it's not possible to fully hydrogenate $\ce{C60}$ to $\ce{C60H60}$. This is a result of the bond strain that would result upon hydrogenation, as the addition of hydrogen across an alkene in $\ce{C60}$ produces carbon atoms that would optimally be tetrahedral with angles near $109.5^\circ$, while the bond angles in $\ce{C60}$ are $108^\circ$ and $120^\circ$ for the pentagonal and hexagonal rings, respectively.

Partial hydrogenation of $\ce{C60}$ with zinc and hydrochloric acid has been done to form $\ce{C60H36}$,1 but this compound degrades quickly in the presence of oxygen to form hydroxylated groups; degradation of the fullerene cage and formation of ketones also occurs. Efforts to hydrogenate further under high pressure hydrogenation conditions results in complete breakdown of the fullerene cage, with pretty interesting results, including ejection of carbon atoms and $\ce{C2}$ loss, producing hydrogenated species with fewer than 60 carbon atoms.2 Collapse of the fullerene cage was also noted, forming hydrogenated compounds with 25 - 45 carbon atoms.

Cataldo, Franco. Fullerenes, Nanotubes and Carbon Nanostructures 2003, 11 (4), 295-316. DOI: 10.1081/FST-120025852
Talyzin, Alexandr V., Yury O. Tsybin, Jeremiah M. Purcell, Tanner M. Schaub, Yury M. Shulga, Dag Noréus, Toyoto Sato, Andrzej Dzwilewski, Bertil Sundqvist, and Alan G. Marshall. J. Phys. Chem. A 2006, 110 (27), 8528–8534. DOI: 10.1021/jp0557971


Synthesis; effect of hydrogen gas and platinium I'm pretty sure the last step in there is just something to throw us off. From what I understand, the first step will give us a Grignard, which will attack the carbonyl in formaldehyde to give us a primary alcohol, and the workup is just to neutralize any remaining Grignard and of course protonate the oxide ion that was just formed. I don't see any possibility of an intramolecular attack either ......

The H2/Pt step looks like a serious red herring from here; I know it doesn't touch the phenyl substituent. And there are no other pi bonds in this system for me to even consider. So is the last step just something to throw me off?

The selective debenzylation of benzyl ethers can be carried out under very mild conditions, leaving other functional groups intact. Commonly carried out under H2 using a heterogeneous catalyst such as Pd/C or Raney Ni, Pt can also be used. Solvent choice is critical, and reactions times are longer, but it is still effective.

I don't like red herrings.


Is cobalt (II) chloride in silica gel dangerous when heated? Today I decided to reheat 3 x 40 gram silica gel canisters with the blue/pink silica gel for the first time, and after I decided to order some more. While shopping, I noticed people say how it is a carcinogen, and that my family and I could now be at risk for cancer. Is this true?

It was in the toaster oven of the kitchen for 5 minutes, 2x each since by the time the canisters cooled off they became pink again before I put them where I needed to.

So am I looking at a risk, or is what I have done too minimal to worry about?

I have to assume things to answer this question. It would be great if you could include a picture of the actual stuff you have so that I may assess whether my assumption was correct or not.

The only aggregate state of desiccant silica gel, whether it be blue gel, orange gel or a colourless one, that I have seen to date is best described as a very coarse granulate. From these, unless you vividly shake the canisters repeatedly, there should be no dust evolution. Stuff like that is routinely stored and handled in chemical laboratories without anybody ever giving a damn because of dust — much unlike the very fine silica gel used for chromatography which can very much, as sixtytrees points out, cause silicosis.

The carcinogenic stuff inside blue gel is a cobalt compound. The salts of cobalt are only carcinogenic if you indigest them, and they do not permeate well through skin. They are even less prone to produce dust or carcinogenic volatiles even when heated to a hundred or so degrees. What they can do is stay on the skin after you touched the gel and then be transferred onwards, for example in your mouth on the crunchy apple you decided to eat — so wash your hands thoroughly if you accidentally handled the actual gel and did not use gloves.

Orange gel is slightly different, it uses organic moisture indicators. These have a much more significant vapour pressure than inorganic salts do. On the upside, these are also less carcinogenic. So the overall effect is probably similar.

Aside from that, you are at a very low risk. Sitting by a campfire is very likely more carcinogenic than putting blue gel desiccant into the oven for a few minutes.


Do Li4 or Li8 molecules exist? I've read that alkali metals form ionic bonds; $\ce{Li}$ is an exception which majorly forms covalent bonds. Wikipedia says dilithium exists. This makes me wonder why $\ce{Li_8}$ doesn't exist. It seems having all 8 electrons shared produces a low energy state (because the valence shell in each of the 8 atoms will be completely filled) ?

Yes, they do exist and were characterised spectroscopically, see here (and there is a note on similar clusters for sodium):

Blanc, J.; Bonačić‐Koutecký, V.; Broyer, M.; Chevaleyre, J.; Dugourd, P.; Koutecký, J.; Scheuch, C.; Wolf, J. P.; Wöste, L. Evolution of the electronic structure of lithium clusters between four and eight atoms. J. Chem. Phys. 1992, 96 (3), 1793–1809. DOI: 10.1063/1.462846.

They are not, however, "stable" enough to exist in solid state or even to obtain a pure gas, rather complicated arcane trickery is used to work with (very small amount of) them.

Lithium has rather complicated chemistry, easily forming polyhedral structures when bound to proper partner. For example, methyllithium is a tetramer. The simplest rationalization is that its s- and p-orbitals have very close energies and thus low-energy vacant p-orbitals participate in formation of multicentre electron-deficient bonds.