Data given for the following reaction is as follows:
FeO$$_{(s)}$$ + C$$_{(graphite)}$$ $$\rightarrow$$ Fe$$_{(s)}$$ + CO$$_{(g)}$$
Substance                $$\Delta_f H°$$(kJ mol$$^{-1}$$)                               $$\Delta S°$$(J mol$$^{-1}$$ K$$^{-1}$$)
FeO$$(s)$$                                          -266.3                                        57.49
C$$(graphite)$$                                  0                                              5.74
Fe$$(s)$$                                              0                                               27.28
CO$$(g)$$                                           -110.5                                         197.6
The minimum temperature in K at which the reaction becomes spontaneous is _________. (Integer answer)

First, we recall the thermodynamic criterion for spontaneity at constant pressure and temperature:

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

A reaction becomes spontaneous when $$\Delta G^\circ<0$$. The borderline (minimum) temperature at which spontaneity just starts is obtained by setting $$\Delta G^\circ=0$$, giving

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

Hence we need to calculate the standard enthalpy change $$\Delta H^\circ$$ and the standard entropy change $$\Delta S^\circ$$ for the reaction

$$\text{FeO}(s)+\text{C (graphite)} \;\longrightarrow\; \text{Fe}(s)+\text{CO}(g)$$

We start with the enthalpies of formation. The formula for the reaction enthalpy is

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

Substituting the given data (all values in kJ mol−1):

$$\Delta_f H^\circ(\text{Fe})=0,\quad \Delta_f H^\circ(\text{CO})=-110.5,$$ $$\Delta_f H^\circ(\text{FeO})=-266.3,\quad \Delta_f H^\circ(\text{C})=0$$

We have

$$\Delta H^\circ =\bigl[0+(-110.5)\bigr]-\bigl[(-266.3)+0\bigr]$$ $$=\;(-110.5)-(-266.3)$$ $$=\;-110.5+266.3$$ $$=\;155.8\ \text{kJ mol}^{-1}$$

Next, for the entropy change we use the analogous formula

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

Substituting the given standard entropies (all values in J mol−1 K−1):

$$S^\circ(\text{Fe})=27.28,\quad S^\circ(\text{CO})=197.6,$$ $$S^\circ(\text{FeO})=57.49,\quad S^\circ(\text{C})=5.74$$

We obtain

$$\Delta S^\circ =\bigl[27.28+197.6\bigr]-\bigl[57.49+5.74\bigr]$$ $$=\;224.88-63.23$$ $$=\;161.65\ \text{J mol}^{-1}\,\text{K}^{-1}$$

To keep units consistent with $$\Delta H^\circ$$ (kJ), we convert $$\Delta S^\circ$$ from joules to kilojoules:

$$\Delta S^\circ = \dfrac{161.65}{1000}=0.16165\ \text{kJ mol}^{-1}\,\text{K}^{-1}$$

Now we substitute the values into the temperature expression:

$$T_{\text{min}} =\dfrac{\Delta H^\circ}{\Delta S^\circ} =\dfrac{155.8\ \text{kJ mol}^{-1}}{0.16165\ \text{kJ mol}^{-1}\,\text{K}^{-1}}$$ $$=963.96\ \text{K}\;\approx\;964\ \text{K}$$

So, the answer is $$964$$.