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

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SUMMARY

Calcium carbonate (CaCO3) exists in two forms: calcite and aragonite, with distinct thermodynamic properties. The standard enthalpy of formation for calcite is -1206.9 kJ/mol and for aragonite is -1207.1 kJ/mol, while their standard entropy values are 92.9 J/(mol·K) and 88.7 J/(mol·K), respectively. At equilibrium, the two forms coexist at a temperature where the Gibbs free energy change (ΔG) equals zero, leading to specific equations for calculating the equilibrium temperature. The stability of each form varies with temperature, with calcite being more stable at room temperature (298K) and aragonite becoming more stable as the temperature approaches 0 K due to differences in their crystal structures and bonding energies.

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  • Understanding of thermodynamic concepts such as enthalpy (H) and entropy (S)
  • Familiarity with Gibbs free energy and equilibrium conditions
  • Knowledge of crystal structures, specifically trigonal and orthorhombic systems
  • Basic principles of solid-state chemistry and phase transitions
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Chemists, materials scientists, and geologists interested in the thermodynamic properties of minerals, particularly those studying calcium carbonate and its applications in various fields.

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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

a) Assuming that [tex]\Delta_{trs} H[/tex] and [tex]\Delta_{trs} S[/tex] are independent 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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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.
 
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.
 

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