FeO(s)+C(graphite)→Fe(s)+CO(g) ΔH°reaction=ΔH°f(product)−ΔH°f(reactants) =[ΔH°f(Fe)+ΔH°f(CO)]−[ΔH°f(FeO)−ΔH°f(C)] =[0+(−110.5)]−[−266.3−0] =156kJ−1mol−1 ΔS°reaction=ΔS°product−ΔS°reactant =[ΔS°(Fe)+ΔS°(CO)]−[ΔS°(FeO)−ΔS°(O)] =[27.28+197.6]−[57.49+5.79] =161JK−1mol−1 According to Gibb's equation, ΔG°=ΔH°−TΔS° The reaction becomes spontaneous when ΔG° is atleast zero or negative. 0=ΔH°−TΔS° TΔS°=ΔH° ⇒T=
ΔH°
ΔS°
=
156kJmol−1
161JK−1mol−1
=
156000mol−1
161JK−1mol−1
=964K The temperature at which reaction becomes spontaneous is 964 K.