For oxidation of iron, `4Fe_((s)) + 3O_(2(g)) to 2Fe_(2) O_(3(s))`
entropy change is `-549.4 "JK"^(-1) "mol"^(-1)` at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous ? `Delta_(r) H^( Theta )` for this reaction is `1648 xx 10^(3) "J mol"^(-1)`

Explanation

One decides the spontaneity of a reaction by considering :
`DeltaS_("total")=(DeltaS_("sys") + DeltaS_("surr")).` For calculating.
`DeltaS_("surr")` we have to consider the heat absorbed by the surroundings which is equal to `- Delta_(r) H^( Theta )`. At temperature T, entropy change of the surroundings is
`DeltaS _("surr") =-(Deltar H^( Theta ) )/( T) ("at constant pressure")`
`=- ((-1648 xx 10^(3) "J mol"^(-1) ))/(298 K)= 5530 "kJ mol"^(-1)`
Thus, total entropy change for this reaction
`Delta_("total") = 5530 "JK"^(-1) "mol"^(-1) + (-549 .4 "JK mol"^(-1) )`
`=4980.6 "JK mol"^(-1)`
This shows that the above reaction is spontaneous.