An important ore of lithium is spodumene, . To extract the metal, the form of spodumene is first converted into the less dense form in preparation for subsequent leaching and washing steps. Use the following data to calculate the lowest temperature at which the conversion is feasible:

It is used in chemistry to indicate a change in enthalpy as well as the addition of heat to the reaction. A triangle symbol is used to represent it.

Solve the temperature at equilibrium of the process below to find the T at which $\alpha \to \beta$ conversion is possible.

$S\left(\text{rhombic}\right)\rightleftharpoons S\text{(monoclinic})$

To begin, use the enthalpy constants shown in the table to solve for $\Delta H^{°}$

$\begin{array}{c}\Delta H^{°}=\sum n_p\Delta H_f^{°}\left(\text{product}\right)-\sum n_r\Delta H_f^{°}\left(\text{reactant}\right)\left(1\right)\\ =\left[n\Delta H_f^{°}\left(\beta \right)\right]-\left[n\Delta H_f^{°}\left(\alpha \right)\right]\left(2\right)\\ =\left(\right[1(-3027)]-[1(-3055)\left]\right)kJ\left(3\right)\\ \Delta H^{°}=28kJ.\left(4\right)\end{array}$

Then, using the entropy constants in Appendix B, solve for $\Delta S^{°}$

$\begin{array}{c}\Delta S^{°}=\sum n_p{S}^{°}\left(\text{product}\right)-\sum n_r{S}^{°}\left(\text{reactant}\right)\left(5\right)\\ =\left[n{S}^{°}\left(\beta \right)\right]-\left[n{S}^{°}\left(\alpha \right)\right]\left(6\right)\\ =\left(\right[1\left(154.4\right)]-[1\left(129.3\right)\left]\right)J/K\left(7\right)\end{array}$

Therefore,$\Delta S^{°}=25.1J/K.\left(8\right)$

Finally, find T at equilibrium. $\Delta G^{°}=0$

$\begin{array}{c}\Delta G^{°}=\Delta H^{°}-T\Delta S^{°}\\ 0=\Delta H^{°}-T\Delta S^{°}\\ T\Delta S^{°}=\Delta H^{°}\\ T=\frac{\Delta H^{°}}{\Delta S^{°}}\\ =\frac{28kJ\times \frac{1000J}{1kJ}}{25.1J/K}\\ T=1115.5K.\end{array}$

Therefore, the T at which$\alpha \to \beta$ conversion is feasible is at 1115.5K.