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 $$298 K: \triangle_fH^0(SnO_2(s)) = -581.0$$ KJ mol$$^{−1}$$, $$\triangle_fH^0 (CO_2(g)) = -394.0$$ KJ mol$$^{−1}$$, $$S^0(SnO_2(s)) = 56.0 J K^{-1} mol^{-1}$$, $$S^0(Sn(s)) = 52.0 J K^{-1} mol^{-1}$$ , $$S^0(C(s)) = 6.0 J K^{-1} mol^{-1}$$, $$S^0(CO_2(g)) = 210.0 J K^{-1} mol^{-1}$$.
Assume that the enthalpies and the entropies are temperature independent.

The reduction to be studied is

$$SnO_2(s) + C(s) \rightarrow Sn(s) + CO_2(g) \qquad -(1)$$

The standard Gibbs free-energy change is obtained from

$$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \qquad -(2)$$

Step 1 : Standard enthalpy change, $$\Delta H^{\circ}$$
For reaction $$-(1)$$, use the given standard enthalpies of formation (in kJ mol$$^{-1}$$):

$$\Delta H^{\circ} = \Bigl[\Delta_fH^{\circ}(CO_2) + \Delta_fH^{\circ}(Sn)\Bigr] - \Bigl[\Delta_fH^{\circ}(SnO_2) + \Delta_fH^{\circ}(C)\Bigr]$$

Elements in their standard state have $$\Delta_fH^{\circ}=0$$, hence

$$\Delta H^{\circ} = \bigl[-394.0 + 0\bigr] - \bigl[-581.0 + 0\bigr]$$

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

Step 2 : Standard entropy change, $$\Delta S^{\circ}$$
Given entropies (in J K$$^{-1}$$ mol$$^{-1}$$):

$$\Delta S^{\circ} = \bigl[S^{\circ}(CO_2) + S^{\circ}(Sn)\bigr] - \bigl[S^{\circ}(SnO_2) + S^{\circ}(C)\bigr]$$

$$\Delta S^{\circ} = (210.0 + 52.0) - (56.0 + 6.0) = 262.0 - 62.0 = 200.0 \text{ J K}^{-1}\text{ mol}^{-1}$$

Step 3 : Temperature for spontaneity
The reduction becomes feasible when $$\Delta G^{\circ} \le 0$$. Setting $$\Delta G^{\circ}=0$$ in $$-(2)$$ gives the minimum temperature, $$T_{\text{min}}$$:

$$0 = \Delta H^{\circ} - T_{\text{min}}\Delta S^{\circ}$$

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

Convert $$\Delta H^{\circ}$$ to joules to match units:
$$\Delta H^{\circ} = 187.0 \text{ kJ mol}^{-1} = 187\,000 \text{ J mol}^{-1}$$

$$T_{\text{min}} = \frac{187\,000}{200} = 935 \text{ K}$$

Minimum temperature ≈ $$935 \text{ K}$$.