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)

For oxidation of iron, 4Fe_((s)) + 3O_(2(g)) to 2Fe_(2) O_(3(s)) entr

1.	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)
Answer» Solution :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.