Problem 73 - Chemistry: An Atoms-Focused Approach

Question

The mineral spodumene \(\left(\operatorname{LiAlSi}_{2} \mathrm{O}_{6}\right)\) exists in two crystalline forms called \(\alpha\) and \(\beta .\) Use Hess's law and the following information to calculate \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the conversion of \(\alpha\) -spodumene into \(\beta\) -spodumene: $$\begin{array}{r}\mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow 2 \alpha-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-1870.6 \mathrm{kJ} \end{array}$$ $$\begin{array}{r}\mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow 2 \beta-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-1814.6 \mathrm{kJ} \end{array}$$

Step-by-Step Solution

Answer

The enthalpy change for the conversion of α-spodumene to β-spodumene is ΔH_conversion = 56 kJ.

Step 1: Write the given reactions and their enthalpy changes

We have the following reactions and their corresponding enthalpy changes: 1. $$\mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow{2\alpha-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s)}$$ $$\Delta H_{\mathrm{rxn}}^{\circ}=-1870.6 \mathrm{kJ}$$ 2. $$\mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow{2\beta-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s)}$$ $$\Delta H_{\mathrm{rxn}}^{\circ}=-1814.6 \mathrm{kJ}$$

Step 2: Reverse the first reaction and change its enthalpy sign

To find the enthalpy change for the conversion of \(\alpha\)-spodumene to \(\beta\)-spodumene, we need to reverse the first reaction, which will change the sign of the enthalpy change: $$2\alpha-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \rightarrow \mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g)$$ $$\Delta H_{\mathrm{rxn}}^{\circ}=1870.6 \mathrm{kJ}$$

Step 3: Add the reversed first reaction and the second reaction

Now, add the reversed first reaction and the second reaction to obtain the reaction we want, i.e., the conversion of \(\alpha\)-spodumene to \(\beta\)-spodumene: $$\begin{array}{r} 2\alpha-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \rightarrow \mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \\\ \mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow 2\beta-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \end{array}$$ $$2\alpha-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \rightarrow 2\beta-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s)$$

Step 4: Calculate the enthalpy change for the conversion of \(\alpha\)-spodumene to \(\beta\)-spodumene

Now, we can use Hess's Law to add the enthalpy changes of the reversed first reaction and the second reaction to get the enthalpy change for the conversion of \(\alpha\)-spodumene to \(\beta\)-spodumene: $$\Delta H_{\mathrm{conversion}}^{\circ} = 1870.6\,\mathrm{kJ} + (-1814.6\,\mathrm{kJ}) = 56\,\mathrm{kJ}$$ And so, the enthalpy change for the conversion of \(\alpha\)-spodumene to \(\beta\)-spodumene is \(\Delta H_{\mathrm{conversion}}^{\circ}=56\,\mathrm{kJ}\).

Key Concepts

Enthalpy Change

Enthalpy change is a key concept in thermodynamics. When dealing with chemical reactions, enthalpy change represents the heat absorbed or released, usually under constant pressure. It is an integral part of Hess's Law, which allows us to determine the enthalpy changes of reactions that haven't been directly measured. Here's how it works: Hess’s Law states that if you can express a reaction as the sum of multiple steps, the total enthalpy change for the reaction is the sum of the enthalpy changes for each step.
In our exercise, this concept is applied to find the enthalpy change, \(\Delta H_{\mathrm{rxn}}^{\circ}\), for the conversion of \(\alpha\)-spodumene to \(\beta\)-spodumene. By reversing the reaction for \(\alpha\)-spodumene, we change the sign of its enthalpy change. This step reflects an essential principle in thermodynamics: when a reaction is reversed, the sign of \(\Delta H\) changes. By then combining the reversed and forward reactions, we can calculate the overall energy change needed for the transformation from one crystalline form to another. The calculation for this exercise results in \(\Delta H_{\mathrm{conversion}}^{\circ} = 56\, \mathrm{kJ}\).

Crystalline Forms

Crystalline forms refer to different structural configurations of the same compound. These forms, often called polymorphs, have distinct physical properties due to variations in the arrangement of atoms or molecules within the crystal lattice. In chemistry, crystalline structures are significant because they affect the material's properties, behavior, and stability.
In this exercise, we examine two polymorphic forms of spodumene: \(\alpha\) and \(\beta\). Both are composed of the same chemical formula \(\text{LiAlSi}_2 \text{O}_6\), but their atomic arrangement, and thus their crystalline structure, differs. This structural variance impacts how they interact with energy changes, such as enthalpy.

• \(\alpha\)-spodumene: Often more stable at low temperatures.
• \(\beta\)-spodumene: Typically stable at higher temperatures or when subjected to different pressures.

The conversion between these forms requires an understanding of their thermodynamic characteristics, allowing us to perform reactions and compute energy changes using principles such as Hess’s Law.

Spodumene Conversion

Spodumene conversion describes the transformation process where one crystalline form of spodumene changes into another. This conversion, in our context, is from \(\alpha\)-spodumene to \(\beta\)-spodumene, and understanding this process involves dissolution of the crystal lattice and reformation into a new structure.
During the conversion, it’s important to consider the energy implications, particularly the enthalpy change associated with rearranging the crystal lattice. The specific reactions given imply breaking and forming chemical bonds, resulting in energy exchange. According to the data, converting \(\alpha\) to \(\beta\) releases energy, making it a process favorable under certain conditions.

• Initial Reaction: Formation of \(\alpha\)-spodumene.
• Second Reaction: Formation of \(\beta\)-spodumene.

The key application of Hess's Law here allows us to calculate the net energy involved in this transformation through simplified reaction steps. By conducting this methodical calculation, the transformation’s enthalpy change is accurately given as \(\Delta H_{\mathrm{conversion}}^{\circ} = 56\, \mathrm{kJ}\), indicating the energy requisite for such crystalline adjustments.