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Question

Following method of extracting Zn is based on thermodynamics
A : 2ZnS+3O22ZnO+2SO2
B : ZnO+CZn+CO

If Gof (standard free energies of formation, in kJmol1 ) of ZnS=205.4, ZnO=318.0, SO2=300.4 and of CO=137.3
Free energy changes of the above reaction A and B (respectively) will be :

A
826.4 kJ,+180.7 kJ
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B
+826.4 kJ,180.7 kJ
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C
826.4 kJ,180.7 kJ
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D
+826.4 kJ,+180.7 kJ
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Solution

The correct option is A 826.4 kJ,+180.7 kJ
For the reaction A :

2ZnS+3O22ZnO+2SO2
ΔG0rxn=2ΔG0ZnO+2ΔG0SO22ΔG0ZnS3ΔG0O2
ΔG0rxn=2×(318.0)+2×(300.4)2×(205.4)3×0
ΔG0rxn=826kJ/mol
For the reaction B :

ZnO+CZn+CO
ΔG0rxn=ΔG0Zn+ΔG0COΔG0ZnOΔG0C
ΔG0rxn=0+(137.3)(318.0)0
ΔG0rxn=+180.7kJ/mol

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