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Calcite [tex] \Delta_{f}H^\Phi [/tex] = -1206.9 kJ/mol

and [tex] S_{m}^\Phi [/tex] = 92.9kJ/mol

Aragonite [tex] \Delta_{f}H^\Phi [/tex] = -1207.1 kJ/mol

and [tex] S_{m}^\Phi [/tex] = 88.7 kJ/mol

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

**a) Assuming that [tex] \Delta_{trs} H [/tex] and [tex] \Delta_{trs} S [/tex] 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??

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.