Chemistry · Thermodynamics & Thermochemistry

JEE Advanced 2020 — Paper 2 — Question 32

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(SnO2( s))=−581.0 kJ mol−1,ΔfH0(CO2(g))=−394.0 kJ mol−1298 \mathrm{~K} ; \Delta_{f} H^{0}\left(\mathrm{SnO}_{2}(\mathrm{~s})\right)=-581.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{f} H^{0}\left(\mathrm{CO}_{2}(g)\right)=-394.0 \mathrm{~kJ} \mathrm{~mol}^{-1},

S0(SnO2(s))=56.0 J K−1 mol−1,S0(Sn(s))=52.0 J K−1 mol−1S^{0}\left(\mathrm{SnO}_{2}(s)\right)=56.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, S^{0}(\mathrm{Sn}(\mathrm{s}))=52.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1},

S0(C(s))=6.0 J K−1 mol−1,S0(CO2( g))=210.0 J K−1 mol−1S^{0}(\mathrm{C}(s))=6.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, S^{0}\left(\mathrm{CO}_{2}(\mathrm{~g})\right)=210.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1},

Assume that the enthalpies and the entropies are temperature independent.

Answer: 935

Numerical answer — enter this value.

Step-by-step solution

SnO2( s)+C(s)⟶Sn(s)+CO2( g)\mathrm{SnO}_{2}(\mathrm{~s})+\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{Sn}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})

ΔHr0=−394.0+581=187kJmol−1\Delta H_{r}^{0}=-394.0+581=187 \mathrm{kJmol}^{-1}

ΔS0=210+52−6−56=200Jmol−1 K−1\Delta \mathrm{S}^{0}=210+52-6-56=200 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}

Equilibrium temperature (T)eq=187×1000200=935 K(\mathrm{T})_{\mathrm{eq}}=\frac{187 \times 1000}{200}=935 \mathrm{~K}

So, T>935 K\mathrm{T}>935 \mathrm{~K} (for spontaneity)

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Gibbs Free Energy and the Third Law of Thermodynamics
Tin is obtained from cassiterite by reduction with coke. Use the data… | JEE Advanced 2020 PYQ with Solution · DhiX AI