JEE Challenger
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ChemHardAdv 2020 • Paper 2

Minimum Temperature for Cassiterite Reduction by Coke

Tin is extracted from cassiterite ore via reduction with coke. Determine the threshold temperature (in K\text{K}) above which the reduction of cassiterite by coke becomes spontaneous. Use the thermodynamic data at 298 K298\text{ K}: ΔfH0(SnO2(s))=−581.0 kJ mol−1\Delta_f H^0(\text{SnO}_2(s)) = -581.0\text{ kJ mol}^{-1}, ΔfH0(CO2(g))=−394.0 kJ mol−1\Delta_f H^0(\text{CO}_2(g)) = -394.0\text{ kJ mol}^{-1}, S0(SnO2(s))=56.0 J K−1mol−1S^0(\text{SnO}_2(s)) = 56.0\text{ J K}^{-1}\text{mol}^{-1}, S0(Sn(s))=52.0 J K−1mol−1S^0(\text{Sn}(s)) = 52.0\text{ J K}^{-1}\text{mol}^{-1}, S0(C(s))=6.0 J K−1mol−1S^0(\text{C}(s)) = 6.0\text{ J K}^{-1}\text{mol}^{-1}, S0(CO2(g))=210.0 J K−1mol−1S^0(\text{CO}_2(g)) = 210.0\text{ J K}^{-1}\text{mol}^{-1}. Assume enthalpy and entropy values remain constant over temperature.

Official Numerical Answer935

Topics & Concepts

ThermodynamicsEnthalpyEntropy and Second LawGibbs Free Energy

Step-by-Step Solution

To find the minimum temperature above which the reduction of cassiterite (SnO2\text{SnO}_2) by coke (C\text{C}) becomes spontaneous, we consider the overall chemical reaction:

SnO2(s)+C(s)⟶Sn(s)+CO2(g)\text{SnO}_2(s) + \text{C}(s) \longrightarrow \text{Sn}(s) + \text{CO}_2(g)

Step 1: Calculate the standard enthalpy of reaction (ΔrH0\Delta_r H^0)

The standard enthalpy change of the reaction is given by: ΔrH0=[ΔfH0(Sn(s))+ΔfH0(CO2(g))]−[ΔfH0(SnO2(s))+ΔfH0(C(s))]\Delta_r H^0 = \left[ \Delta_f H^0(\text{Sn}(s)) + \Delta_f H^0(\text{CO}_2(g)) \right] - \left[ \Delta_f H^0(\text{SnO}_2(s)) + \Delta_f H^0(\text{C}(s)) \right]

Since the standard enthalpy of formation of elements in their standard state is zero (ΔfH0(Sn(s))=0\Delta_f H^0(\text{Sn}(s)) = 0 and ΔfH0(C(s))=0\Delta_f H^0(\text{C}(s)) = 0):

ΔrH0=0+(−394.0 kJ mol−1)−[−581.0 kJ mol−1+0]\Delta_r H^0 = 0 + (-394.0\text{ kJ mol}^{-1}) - \left[ -581.0\text{ kJ mol}^{-1} + 0 \right] ΔrH0=−394.0+581.0=+187.0 kJ mol−1=187000 J mol−1\Delta_r H^0 = -394.0 + 581.0 = +187.0\text{ kJ mol}^{-1} = 187000\text{ J mol}^{-1}

Step 2: Calculate the standard entropy of reaction (ΔrS0\Delta_r S^0)

The standard entropy change of the reaction is given by: ΔrS0=[S0(Sn(s))+S0(CO2(g))]−[S0(SnO2(s))+S0(C(s))]\Delta_r S^0 = \left[ S^0(\text{Sn}(s)) + S^0(\text{CO}_2(g)) \right] - \left[ S^0(\text{SnO}_2(s)) + S^0(\text{C}(s)) \right]

Substitute the given absolute entropy values: ΔrS0=(52.0+210.0)−(56.0+6.0)\Delta_r S^0 = (52.0 + 210.0) - (56.0 + 6.0) ΔrS0=262.0−62.0=200.0 J K−1mol−1\Delta_r S^0 = 262.0 - 62.0 = 200.0\text{ J K}^{-1}\text{mol}^{-1}

Step 3: Determine the minimum temperature for spontaneity

For a reaction to be spontaneous, the change in Gibbs free energy (ΔrG0\Delta_r G^0) must be negative (ΔrG0<0\Delta_r G^0 < 0): ΔrG0=ΔrH0−TΔrS0<0\Delta_r G^0 = \Delta_r H^0 - T \Delta_r S^0 < 0

At the minimum threshold temperature (TminT_{\text{min}}), the system is at equilibrium where ΔrG0=0\Delta_r G^0 = 0: ΔrH0−TminΔrS0=0\Delta_r H^0 - T_{\text{min}} \Delta_r S^0 = 0 Tmin=ΔrH0ΔrS0T_{\text{min}} = \frac{\Delta_r H^0}{\Delta_r S^0}

Substituting the values of ΔrH0\Delta_r H^0 and ΔrS0\Delta_r S^0: Tmin=187000 J mol−1200.0 J K−1mol−1=935 KT_{\text{min}} = \frac{187000\text{ J mol}^{-1}}{200.0\text{ J K}^{-1}\text{mol}^{-1}} = 935\text{ K}

Thus, the minimum temperature at which the reduction of cassiterite by coke will take place is 935 K935\text{ K}.

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