1. Write the reduction reaction

Cassiterite is $SnO_2$. Reduction by coke gives:

$SnO_2(s) + C(s) \rightarrow Sn(s) + CO_2(g)$

We need the minimum temperature at which this reaction becomes feasible, i.e. when

$\Delta G^\circ = 0$

So,

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

At threshold temperature,

$T_{\min} = \frac{\Delta H^\circ}{\Delta S^\circ}$

provided $\Delta H^\circ > 0$ and $\Delta S^\circ > 0$.

2. Calculate $\Delta H^\circ$ of the reaction

Using standard enthalpies of formation:

$\Delta H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$

Given:

• $\Delta_f H^\circ[SnO_2(s)] = -581.0\,kJ\,mol^{-1}$
• $\Delta_f H^\circ[CO_2(g)] = -394.0\,kJ\,mol^{-1}$
• For elements in standard state: $\Delta_f H^\circ[Sn(s)] = 0, \quad \Delta_f H^\circ[C(s)] = 0$

Therefore,

$\Delta H^\circ = [0 + (-394.0)] - [(-581.0) + 0]$

$\Delta H^\circ = -394.0 + 581.0 = 187.0\,kJ\,mol^{-1}$

3. Calculate $\Delta S^\circ$ of the reaction

$\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$

Given:

• $S^\circ[SnO_2(s)] = 56.0\,J\,K^{-1}\,mol^{-1}$
• $S^\circ[Sn(s)] = 52.0\,J\,K^{-1}\,mol^{-1}$
• $S^\circ[C(s)] = 6.0\,J\,K^{-1}\,mol^{-1}$
• $S^\circ[CO_2(g)] = 210.0\,J\,K^{-1}\,mol^{-1}$

So,

$\Delta S^\circ = [52.0 + 210.0] - [56.0 + 6.0]$

$\Delta S^\circ = 262.0 - 62.0 = 200.0\,J\,K^{-1}\,mol^{-1}$

4. Find the minimum temperature

Convert enthalpy to joules:

$\Delta H^\circ = 187.0\,kJ\,mol^{-1} = 187000\,J\,mol^{-1}$

Now,

$T_{\min} = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{187000}{200.0}$

$T_{\min} = 935\,K$

5. Final answer

The minimum temperature for reduction is:

$\boxed{935}$

This matches the stored correct answer.