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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)`) |
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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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