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Calcium Carbonate: Enthalpy and Entropy Difference at Room Temp & 0K
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[QUOTE="stunner5000pt, post: 377984, member: 6325"] CaCO3 (calcium carbonate) has two states 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 [B]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[/B] 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 [I]so is that the equilibrium temperature they can both co exist at?[/I] [B] b) Which form is more stable than the other at room temperature (298K) and whcih is more stable as one approaches 0 K??[/B] 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?? [B]c) Give physical reasons in terms of H and S why the two structures have different H and S values.[/B] 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. [/QUOTE]
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Calcium Carbonate: Enthalpy and Entropy Difference at Room Temp & 0K
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