Question

The Fe-Mg exchange reaction between olivine and spinel is the following:

\[ \mathrm{Mg_2SiO_4 + 2FeAl_2O_4 \rightleftharpoons Fe_2SiO_4 + 2MgAl_2O_4} \]
(Forsterite) (Fe-spinel) (Fayalite) (Mg-spinel)

Assuming ideal solid solutions, the temperature of equilibration between olivine and spinel using the following data is \(^{\circ}\text{C}\) (round off to nearest integer).

[Assume that \(\Delta H_r^{0}\) and \(\Delta S_r^{0}\) are independent of P and T, and \(K_d\) is independent of P]

\[ K_d = \left[\frac{(X_{Fe}^{Ol})(X_{Mg}^{Sp})}{(X_{Mg}^{Ol})(X_{Fe}^{Sp})}\right]^2 \]

ΔH0r (kJ/mol)	ΔS0r (J/mol.K)	Kd	R (J/mol.K)
-10.91	-3.90	3.024	8.314

Hint:

Combine \(\Delta G=\Delta H-T\Delta S\) with \(\Delta G=-RT\ln K_d\), solve for T in Kelvin, then convert to Celsius.

Answer 1

Correct Answer: 560

Step 1: Set up the thermodynamic relation.
For a mineral exchange reaction at equilibrium, the standard Gibbs free energy change connects to enthalpy and entropy through
\[ \Delta G_r^{0} = \Delta H_r^{0} - T\Delta S_r^{0} \]
The same free energy change is also tied to the distribution coefficient \(K_d\) by
\[ \Delta G_r^{0} = -RT\ln K_d \]
Since both expressions equal the same \(\Delta G_r^{0}\), we can set them equal to each other.

Step 2: Combine the two expressions and solve for T.
\[ \Delta H_r^{0} - T\Delta S_r^{0} = -RT\ln K_d \]
Move every term that carries \(T\) to one side:
\[ \Delta H_r^{0} = T\Delta S_r^{0} - RT\ln K_d = T\left(\Delta S_r^{0} - R\ln K_d\right) \]
So the equilibration temperature is
\[ T = \frac{\Delta H_r^{0}}{\Delta S_r^{0} - R\ln K_d} \]

Step 3: Put the data in consistent units.
The table gives \(\Delta H_r^{0} = -10.91\) kJ/mol, but \(\Delta S_r^{0}\) and \(R\) are in J/mol.K. Convert \(\Delta H_r^{0}\) to J/mol:
\[ \Delta H_r^{0} = -10.91 \times 1000 = -10910 \text{ J/mol} \]
The other values from the table are \(\Delta S_r^{0} = -3.90\) J/mol.K, \(K_d = 3.024\), and \(R = 8.314\) J/mol.K.

Step 4: Work out ln Kd.
\[ \ln(3.024) = 1.1066 \]

Step 5: Compute R ln Kd and the denominator.
\[ R\ln K_d = 8.314 \times 1.1066 = 9.200 \text{ J/mol.K} \]
\[ \Delta S_r^{0} - R\ln K_d = -3.90 - 9.200 = -13.100 \text{ J/mol.K} \]

Step 6: Solve for T in Kelvin.
\[ T = \frac{-10910}{-13.100} = 832.8 \text{ K} \]
Both the numerator and the denominator are negative, so the ratio comes out positive. Watch this sign carefully: flip either sign by mistake and the answer turns negative or a very different number.

Step 7: Convert to degree Celsius.
\[ T(^{\circ}\text{C}) = T(K) - 273.15 = 832.8 - 273.15 = 559.7^{\circ}\text{C} \]

Final Answer:
Rounded to the nearest integer, the temperature of equilibration between olivine and spinel is
\[ \boxed{560\,^{\circ}\text{C}} \]