The resting of iron occurs as:
`4Fe(s) +3O_(2)(g) rarr 2Fe_(2)O_(3)(s)`
The entalpy of formation of `Fe_(2)O_(3)(s) is -824.0 kJ mol^(-1)` and entropy change for the reaction is `+550 J K^(-1) mol^(-1)`. Calculate `Deta_(surr)S` and predict whether resuting of iron is spontaneous or not at `298 K`.
Given `Delta_(sys) H =- 553.0 J K^(-1) mol^(-1)`

Explanation

For the reaction:
`4Fe(s) +3O_(2)(g) rarr 2Fe_(3)O_(2)(s)`
`Delta_(r)H = 2Delta_(f)H^(Theta) (Fe_(2)O_(3)) = 2 xx -824.0 =- 1648.0 kJ`
The release of enegry by the system to surroundings will increase the entropy of surroundings as:
`Delta_(surr)S =+(Delta_(sys)H)/(T) = (+1648.0)/(298) = 5553.0 J K^(-1) mol^(-1)`
`Delta_(total)S = Delta_(sys)S + Delta_(surr)S`
`=- 553.0 _ 5530.2 = 4977.2 J K^(-1) mol^(-1)`
SInce there is increase in entropy `(Delta_("total")S =+ve)`, resting of iron is spontaneous process.