Calcite is a stable form of calcium carbonate (CaCO3). The solubility product (Ksp) is decreased wit — Analytical Chemistry Chemistry Question
Solubility of Calcite
Calcite is a stable form of calcium carbonate (CaCO3). The solubility product (Ksp) is decreased with increasing temperature; Ksp are 9.50 × 10^-9 and 2.30 × 10^-9 at 0 °C and 50 °C, respectively.
2.1.
Estimate the enthalpy change for the solubility process of calcite.
Model Answer
According to ΔG°sol = –RT ln Ksp = ΔH°sol – TΔS°sol,
we have ln Ksp = –ΔH°sol / (RT) + ΔS°sol / R.
Assuming that ΔH°sol and ΔS°sol are temperature independent,
ln (Ksp1 / Ksp2) = –ΔH°sol / R (1/T1 – 1/T2)
ln (9.50 / 2.30) = –ΔH°sol / R (1/273 – 1/323)
Solving this equation to get ΔH°sol = –21 kJ mol^-1.
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