Q20.88 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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:

Step-by-Step Solution

Answer

As a result, the T value at which $\mathrm{\alpha }\to \mathrm{\beta }$ conversion is possible is 1115.5K.

Step 1: Definition of Delta

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.

Step 2: Solve for Δ S ° using the entropy constants

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

Step 3: Solve for T at the equilibrium

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.