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 ; Δf H0( SnO 2(s))=-581.0 kJ mol -1, Δf H0( CO 2(g))=-394.0 kJ mol -1, S0( SnO 2(s))=56.0 J K -1 mol -1, S0( Sn (s))=52.0 J K -1 mol -1, S0( C (s))=6.0 J K -1 mol -1, S0( CO 2(g))=210.0 J K -1 mol -1, Assume that the enthalpies and the entropies are temperature independent.

Question

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 ; Δf H0( SnO 2(s))=-581.0 kJ mol -1, Δf H0( CO 2(g))=-394.0 kJ mol -1, S0( SnO 2(s))=56.0 J K -1 mol -1, S0( Sn (s))=52.0 J K -1 mol -1, S0( C (s))=6.0 J K -1 mol -1, S0( CO 2(g))=210.0 J K -1 mol -1, Assume that the enthalpies and the entropies are temperature independent. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

SnO 2( s )+ C ( s ) longrightarrow Sn ( s )+ CO 2( g ) Δ H r 0=-394.0+581=187 kJ mol -1 Δ S0=210+52-6-56=200 J mol -1 K -1 Equilibrium temperature ( T ) text eq =(187 × 1000/200)=935 K So, T >935 K (for spontaneity)

SnO2​(s)+C(s)⟶Sn(s)+CO2​(g) ΔHr0​=−394.0+581=187kJmol−1 ΔS0=210+52−6−56=200Jmol−1K−1 Equilibrium temperature (T)eq ​=200187×1000​=935K So, T>935K (for spontaneity)

$S n O_{2} (s) + C (s) ⟶ S n (s) + C O_{2} (g)$
$Δ H_{r}^{0} = - 394.0 + 581 = 187 k J m o l^{- 1}$
$Δ S^{0} = 210 + 52 - 6 - 56 = 200 J m o l^{- 1} K^{- 1}$
Equilibrium temperature $(T)_{eq} = \frac{187 \times 1000}{200} = 935 K$
So, $T > 935 K$ (for spontaneity)