Chapter 9, Problem 9.71QA

Interpretation Introduction

To calculate:

The value of $∆H_{rxn}^0$ for the conversion of Α-spodumene into Β-spodumene.

Answer to Problem 9.71QA

Solution:

The value of $∆H_{rxn}^0$ for the conversion of Α-spodumene into Β-spodumene is $28.0 kJ/mol.$

Explanation of Solution

1) Concept:

To calculate the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction, we can use Hess’s law. We can algebraically manipulate the two equations for which $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ values are known so that we can calculate the $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction. Then, by applying Hess’s law, we can calculate the unknown value.

2) Given:

i)

$L{i}_2{O}_{\left(s\right)}+2 A{l}_{\left(s\right)}+4 Si{O}_{2\left(s\right)}+\frac{3}{2} O_{2\left(g\right)} \to 2{\rm A} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0=-1870.6 kJ$

ii) $L{i}_2{O}_{\left(s\right)}+2 A{l}_{\left(s\right)}+4 Si{O}_{2\left(s\right)}+\frac{3}{2} O_{2\left(g\right)} \to 2{\rm B} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0=-1814.6 kJ$

3) Calculations:

Conversion reaction of Α-spodumene into Β-spodumene looks like this:

${\rm A} LiAlS{i}_2{O}_{6\left(s\right)}\to {\rm B} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0= ?$

Given equations are

$A. L{i}_2{O}_{\left(s\right)}+2 A{l}_{\left(s\right)}+4 Si{O}_{2\left(s\right)}+\frac{3}{2} O_{2\left(g\right)} \to 2{\rm A} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0= -1870.6 kJ$

$B. L{i}_2{O}_{\left(s\right)}+2 A{l}_{\left(s\right)}+4 Si{O}_{2\left(s\right)}+\frac{3}{2} O_{2\left(g\right)} \to 2{\rm B} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0= -1814.6 kJ$

Now combine equation A and B to get the desired equation. We have to reverse the equation A so that we can get Α-spodumene on the reactant side of the overall reaction.

$2{\rm A} LiAlS{i}_2{O}_{6\left(s\right)}\to L{i}_2{O}_{\left(s\right)}+2A{l}_{\left(s\right)}+4 Si{O}_{2\left(s\right)}+\frac{3}{2} O_{2\left(g\right)} ∆H_{rxn}^0= +1870.6 kJ$

$L{i}_2{O}_{\left(s\right)}+ 2A{l}_{\left(s\right)}+4Si{O}_{2\left(s\right)}+\frac{3}{2} O_{2\left(g\right)} \to 2{\rm B} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0=-1814.6 kJ$

$2{\rm A} LiAlS{i}_2{O}_{6\left(s\right)}\to 2{\rm B} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0= 1870.6 kJ-1814.6 kJ=56.0 kJ$

Now divide the overall reaction by 2.

${\rm A} LiAlS{i}_2{O}_{6\left(s\right)}\to {\rm B} LiAlS{i}_2{O}_{6\left(s\right)} ∆H_{rxn}^0= 28.0 kJ/mol$

$∆H_{rxn}^0$ for the conversion of Α-spodumene into Β-spodumene is $28.0 kJ/mol.$

Conclusion:

We have calculated enthalpy change of the conversion of Α-spodumene into Β-spodumene by using standard heat of formation of Α-spodumene and Β-spodumene and Hess’s law.