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← JEE Advanced 2020 · Paper 2
ChemistryNumerical+4Question 35 of 54

JEE Advanced 2020 · Paper 2, Chemistry question 17

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.

At 298 K: ΔfH0(SnOX2(s))=−581.0 kJ mol−1\Delta_fH^0(\ce{SnO2}(s)) = -581.0\ \mathrm{kJ\,mol^{-1}}, ΔfH0(COX2(g))=−394.0 kJ mol−1\Delta_fH^0(\ce{CO2}(g)) = -394.0\ \mathrm{kJ\,mol^{-1}},

S0(SnOX2(s))=56.0 J K−1mol−1S^0(\ce{SnO2}(s)) = 56.0\ \mathrm{J\,K^{-1}mol^{-1}}, S0(Sn(s))=52.0 J K−1mol−1S^0(\ce{Sn}(s)) = 52.0\ \mathrm{J\,K^{-1}mol^{-1}},

S0(C(s))=6.0 J K−1mol−1S^0(\ce{C}(s)) = 6.0\ \mathrm{J\,K^{-1}mol^{-1}}, S0(COX2(g))=210.0 J K−1mol−1S^0(\ce{CO2}(g)) = 210.0\ \mathrm{J\,K^{-1}mol^{-1}}.

Assume that the enthalpies and the entropies are temperature independent.

Answer and worked solution

Official answer: 935

The reaction. SnOX2(s)+C(s)→Sn(s)+COX2(g)\ce{SnO2(s) + C(s) -> Sn(s) + CO2(g)}.

Changes.
ΔH∘=−394.0−(−581.0)=+187.0 kJ mol−1,\Delta H^\circ=-394.0-(-581.0)=+187.0\ \mathrm{kJ\,mol^{-1}},
ΔS∘=(52.0+210.0)−(56.0+6.0)=+200.0 J K−1mol−1.\Delta S^\circ=(52.0+210.0)-(56.0+6.0)=+200.0\ \mathrm{J\,K^{-1}mol^{-1}}.

Minimum temperature. The reduction becomes feasible when ΔG∘=ΔH∘−TΔS∘≤0\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\le0:
T=187 000200=935 K.T=\frac{187\,000}{200}=935\text{ K}.
Answer: 935

Topics:Extraction and thermodynamics (Ellingham)Gibbs energy and spontaneity
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