Given the following data Determine at what temperature the following reaction is spontaneous ?FeO(s) + C (Graphite) →Fe(s) + CO (g)

∆G=∆H-T∆S ∆H=∑∆HP- ∑∆HR ∆H=-110.5+266.3=155.8 kJ ∆S=∑SP- ∑SR ∆S=197.6+27.28-57.5+5.74=161.64 J/mol KFor reaction to be spontaneous, ∆G < 0∆H -T∆S< 0   T > ∆H∆S=155800161.64=964 K