Problem: “Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is Fe2 O3(s) + 2Al(s) ⟶ Al2 O3(s) + 2Fe(s). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.

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FREE Expert Solution

Fe2O3(s) + 2 Al(s) ⟶ Al2O3(s) + 2 Fe(s)



ΔG°rxn=ΔH°rxn-TΔS°rxn



Step 1: We first need to convert ΔH˚ from kJ to J:


ΔH°rxn=-851.8 kJmol×103 J1 kJ

ΔH°rxn851800 J/mol

**heat is released to the surroundings


Step 2: We can use the following equation to solve for ΔS˚rxn:


ΔS°rxn=S°f, prod-S°f, react


Note that we need to multiply each S˚ by the stoichiometric coefficient since S˚ is in J/mol • K.


ΔS°rxn=1 mol Al2O350.9 Jmol·K+(2 mol Fe)(27.3 Jmol·K)           -1 mol Fe2O387.4 Jmol·K+2 mol Al28.3 Jmol·K

ΔS°rxn = –10.2 J/mol • K



Step 3: Now that we have ΔH˚rxn and ΔS˚rxn, we can now solve for ΔG˚rxn

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Problem Details

“Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is Fe2 O3(s) + 2Al(s) ⟶ Al2 O3(s) + 2Fe(s). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.

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