Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
The Gibbs free energy for the reaction will be -3.7771 [tex]\rm \times\;10^2[/tex] kJ/mol.
[tex]\rm \Delta[/tex]G is the Gibbs free energy that can be used to calculate the reversible work.
Gibbs free energy for a reaction :
[tex]\rm \Delta G\;=\;n\;\Delta G_P\;-\;n\;\Delta G_R[/tex]
[tex]\rm \Delta G_P[/tex] = Gibbs free energy of Product
[tex]\rm \Delta G_R[/tex] = Gibbs free energy of Reactant
[tex]\Delta[/tex]G = n[tex]\rm \Delta G_M_g^2^+[/tex] + n[tex]\rm \Delta G_A_l[/tex] - n[tex]\rm \Delta G_M_g[/tex] + n[tex]\rm \Delta G_A_l^3^+[/tex]
[tex]\Delta[/tex]G = 3 [tex]\times[/tex] -456.35 kJ/mol + 2 [tex]\times[/tex] 0 - 3 [tex]\times[/tex] 0 + 2 [tex]\times[/tex] -495.67 kJ/mol
[tex]\Delta[/tex]G = -377.71 kJ/mol
[tex]\Delta[/tex]G = -3.7771 [tex]\times\;10^2[/tex] kJ/mol
The Gibbs free energy for the reaction will be -3.7771 [tex]\rm \times\;10^2[/tex] kJ/mol.
For more information about [tex]\Delta[/tex]G, refer the link:
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