Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: xshadow on January 09, 2021, 03:16:04 PM
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Hi
I've read that in nature we usually have HgS rather than HgO...this can be explained looking at the hard-soft theory.
Hg+ is soft
S is softer than O.
But why if I look at the standard formation enthalpy I see that :
Hg + 1/2O2 ---> HgO
Hg + S --> HgS
The standard formation enthalpy is more negative for HgO!!
This mean that HgO is more stable than HgS (in contraddition with the statement that it's HgS the most stable) or I am wrong??
Thanks
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At the very least, you should be using a lattice energy rather than an enthalpy of formation as a gauge of the thermodynamic stability of different crystalline solids.
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At the very least, you should be using a lattice energy rather than an enthalpy of formation as a gauge of the thermodynamic stability of different crystalline solids.
My teacher used a table where there were the standard formation enthalpy for HgO and HgS.... ???
But in fact using that value I have that HgO is more stable than HgS!! In contraddition with the explanation using the hard soft theory
...so are lattice energy better? But IF they both have the same type cell I should get a much more negative value for HgO (O small ionic radius than S)..so anyway HgO is much more stabilized but I know that the more stable in nature is HgS (I don't find lattice energy for HgO)
Perhaps is not a good way using these things instead the hard -soft theory....?