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 `Delta_(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

We know that `Delta S_("surrounding")=(q_("surrounding"))/(T)`
Now, `q_("surrounding")` = Heat gained by surrounding = Heat lost by system, i.e., heat lost in the surrounding during rusting of iron `= Delta_(r )H`.
`4Fe(s)+3O_(2)(g)to 2Fe_(2)O_(3)` , For this reaction,
`Delta_(r )H=2xx Delta_(f)H(Fe_(2)O_(3))-[3xxDelta_(f)H(O_(2))+4xx Delta_(f)H(Fe)]`
`Delta_(r )H=2xx-824.2=-1648.4 kJ mol^(-1) " " (Delta_(f)H(O_(2))=0 , Delta_(r )H(Fe)=0)`
It means heat gained by surrounding, i.e., `q_("surrounding")=1648.4` kJ/mol
Therefore, `Delta S_("surrounding")=(168.4xx10^(3)J mol^(-1))/(298 K)=5531 JK^(-1)mol^(-1)`
Now, `Delta S_("system")=-549 JK^(-1)mol^(-1)` (given)
`Delta S_("Total")=-549 + 5531=+4982 kJ K^(-1)mol^(-1)`
Since there is increase in entropy, i.e., `Delta S_("Total")=+ve`, rusting of iron is spontaneous process.