1.

For oxidation of iron. `4Fe(s)+3O_(2)(g)rarr2Fe_(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^(Ө)` for this reaction is `-1648xx10^(3)J"mol"^(-1)`)

Answer» One decides the spontaneity of a reaction by considering
`DeltaS_("total")(DeltaS_("sys")+DeltaS_(surr")).` For calculating `DeltaS_("surr")` have to consider the heat absorbed by the surroundings which is equal to `-Delta_(r)H^(Ө)` . At temperature T, entropy change of the surroundings is
`DeltaS_("surr")=-(Delta_(r)H^(Ө))/(T)` (at constant pressure)
`=-((-1648xx19^(3)J "mol"^(1)))/(298K)`
`5530 JK^(-1)"mol"^(-1)`
Thus, total entropy change for this reaction
`Delta_(r)S_("total")=5530JK^(-1)"mol"^(-1)+`
`" " (-549.4JK^(-1)"mol"^(-1))`
`=4980.6 JK^(-1)"mol"^(-1)`
This shows that the above reaction is spontaneous.


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