Question

In the mining of antimony, the sulphide ore Sb₂S₃ is refined by reducing it to its elemental form by reaction with Fe to form FeS. The reaction is:

Sb₂S₃ (s) + 3 Fe (s) -> 2 Sb(s) + 3 FeS(s)

Relevant thermodynamic data are

	deltaformH (kJ/mol)	S naught (J/K x mole)
Sb2S3	-174.91	182.0
FeS(s)	-101.67	60.33
Fe(s)		27.31
Sb(s)		45.70

What is the entropy change at 298 K in the surroundings during the refining process given above? (Answer is 437 J/ K x mole I just need the steps to get there!)

Answer 1

The first thing to do is to calculate the Enthalpy Change (products - Reactants)

Enthalpy products =3*(FeS) = 3*-101.67 = -305.01 KJ

Enthalpy reactants = -174.91 = ENthalpy Sb2S3

Enthalpy (products - reactants ) = -305.01 - (-174.91) = -130.1 KJ

Enthalpy for pure compounds is equal to zero

ENtropy is surroundings is the negative ratio of the enthalpy to the temperature

S is for entropy

S (surroundings) = -(-130.1)/298 = 0.436577 KJ/K mole

multiply by 1000 to change KJ to joules

S surroundings = 436.6 J / K mole

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