Question

Calcium carbonate, CaCO3, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book.

(a) Which is stable at earth's surface, calcite or aragonite?

(b) Calculate the pressure (still at room temperature) at which the other phase

should become stable.

Short answer

The aragonite is more stable than calcite at pressure of 3.6 kbar

Step-by-step solution

Given information

A crystal with a lower Gibbs free energy is more stable at a given temperature and pressure. The Gibbs free energy per mole of two crystals is used to compare their stability.

Explanation 

(a) The Gibbs free energy of calcite and aragonite are

$$\begin{gathered} G_C=-1128·8\mathrm{kJ} \\ {\mathrm{G}}_{\mathrm{a}}=-1127·8\mathrm{kJ}\mathrm{S} \end{gathered}$$

Under typical conditions, the Gibbs free energy of 1 mole of calcite is 1-0KJ lower than that of a mole of aragonite, based on the above numbers. As a result, at room temperature and air pressure, or at the earth's surface, calcite is in a more stable phase.

Concept

The volume of the substance determines the dependency of the Gibbs free energy on pressure.

${\left(\frac{\partial G}{\partial P}\right)}_{T,N}=V$

Where,

G is Gibbs free energy

P is the atmospheric pressure

V is the volume of substance

The volumes of calcite and aragonite are

$$\begin{gathered} V_{\mathrm{c}}=36.93{\mathrm{cm}}^3 \\ {V}_{\mathrm{a}}=34.15{\mathrm{cm}}^3 \end{gathered}$$

Converting it into kJ/kbar

$$\begin{gathered} V_{\mathrm{c}}=36.93{\mathrm{cm}}^3\left(\frac{{10}^{-1}\mathrm{kJ}/\mathrm{kbar}}{1{\mathrm{cm}}^3}\right) \\ =3.693\mathrm{kJ}/\mathrm{kbar} \\ {V}_{\mathrm{a}}=34.15{\mathrm{cm}}^3\left(\frac{{10}^{-1}\mathrm{kJ}/\mathrm{kbar}}{1{\mathrm{cm}}^3}\right) \\ =3.415\mathrm{kJ}/\mathrm{kbar} \end{gathered}$$

Because the Aragonite has a smaller volume, it should be stable at high pressure.

Set $G_c=0$at $P=Ibar$for simplicity; then $G_a=1.0kJatP=1bar$

For calcite and aragonite, the equations relating Gibbs free energy, pressure, and volume are as follows:

$$\begin{gathered} G_C=V_{C}P \\ {G}_a=V_{a}P+1·0\mathrm{kJ} \end{gathered}$$

Calculations

$G_a$should equal $G_c$, at the place where the graphs for Gibbs free energy and pressure for Calcite and Aragonite intersect.

$G_C=G_a$

Substituting the values of $G_{c}and{G}_a$

$V_{c}P=V_{a}P+1·0\mathrm{kJ}$

Substituting the values and solving for P

$$\begin{gathered} P=\frac{1·0\mathrm{kJ}}{V_C-V_a} \\ P=\frac{1.0\mathrm{kJ}}{3·693\mathrm{kJ}/\mathrm{kbar}-3.415\mathrm{kJ}/\mathrm{kbar}} \\ P=3.6\mathrm{kbar} \end{gathered}$$

The aragonite is more stable than calcite at pressure of 3.6 kbar