CaCO3 (calcium carbonate) has two states(adsbygoogle = window.adsbygoogle || []).push({});

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??

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.

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# Homework Help: Chemistry Question

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