Calcium Carbonate: Enthalpy and Entropy Difference at Room Temp & 0K

[stunner5000pt]

CaCO3 (calcium carbonate) has two states

Calcite \Delta_{f}H^\Phi = -1206.9 kJ/mol
and S_{m}^\Phi = 92.9kJ/mol

Aragonite \Delta_{f}H^\Phi = -1207.1 kJ/mol
and S_{m}^\Phi = 88.7 kJ/mol

where H is the standard enthalpy and S means the standard entropy

a) Assuming that \Delta_{trs} H and \Delta_{trs} S are independant of temperature, at what temperature can these two forms exist in equilibrium at one bar

well at equilibrium delta G = 0

and thus 0 = dH + T dS for each one and thus i get two equations

-1206.9 + T 92.9 = 0 and
-1207.1 + T 88.7 = 0

so is that the equilibrium temperature they can both co exist at?

b) Which form is more stable than the other at room temperature (298K) and whcih is more stable as one approaches 0 K??

So both calcite and aragonite i get 1.23 x 10^6 j / mol and then for 0 K i get the standard delta H of fusion so a more negative value would mean that that one is more stable??

c) Give physical reasons in terms of H and S why the two structures have different H and S values.

Well a solid with a lower S (entropy) would mean that it's density is lower and thus it would be easier to break up something that is less dense as opposed to something that is the opposite.

 

[member 553083]

H (enthalpy) is related to the energy of the bonds between the atoms and thus it can be said that the two forms of calcium carbonate have different H values because the bonds between the atoms could be arranged differently in each form.

 

[wais]

The difference in H values could be due to the different crystal structures of calcite and aragonite. Calcite has a trigonal crystal structure while aragonite has an orthorhombic crystal structure. These different structures may result in different bonding energies and therefore different enthalpy values. Additionally, the arrangement of atoms in the crystal lattice can also affect the entropy value, as different arrangements can lead to different degrees of disorder.