Calculate the entropy change for the rusting of iron according to the reaction : 4Fe(s) + 3O_(2)(g) rarr Fe_(2)O_(2)(s) , DeltaH^(@) =- 1648 kJ mol^(-1) Given that the standard entropies of Fe,O_(2)and Fe_(2)O_(3) are 27.3,205.0 and 87.4 JK^(-1) mol^(-1) respectively.Will the reaction be spontandeous at room temperature ( 25^(@)C) ?Justify your answer with appropriate calculations.

Calculate the entropy change for the rusting of iron according to the

1.	Calculate the entropy change for the rusting of iron according to the reaction : 4Fe(s) + 3O_(2)(g) rarr Fe_(2)O_(2)(s) , DeltaH^(@) =- 1648 kJ mol^(-1) Given that the standard entropies of Fe,O_(2)and Fe_(2)O_(3) are 27.3,205.0 and 87.4 JK^(-1) mol^(-1) respectively.Will the reaction be spontandeous at room temperature ( 25^(@)C) ?Justify your answer with appropriate calculations.
Answer» SOLUTION :`Delta_(r) S^(@)= SigmaS^(@) `( PRODUCTS ) `- SigmaS^(@) ` ( Reactants) `=2S^(@) (Fe_(2)O_(3)) -[4 S^(@) ( Fe) + 3S^(@) ( O_(2))]` `= 2xx 87.4- [4 xx 27.3 + 3 xx 205.0 ]JK^(-1) mol^(-1) =- 549 JK^(-1) mol^(-1)` This is the entropy change for the reaction, i.e., system `( Delta S_("system"))` Now, `Delta_(r) G^(@) = Delta_(r) H^(@) - TDelta_(r) S^(@)= - 1648000 J mol^(_1) -298 K xx ( - 549 .4 J K^(-1) mol^(-1))` `= - 1648000 +163721 J K^(-1) mol^(_1) = - 1484279 J K^(-1) mol^(-1)` As `Delta G^(@)`is - ve , the reaction is spontaneous. Alsternatively, as the reaction is exothermic, heat given out by THEREACTION is absorbed by the surroundings at room temperature `(25^(@)C)`. Hence, entropy ofthe surroundings increases. `DeltaS_("surroundings")= (1648xx10^(3)JK^(-1) mol^(-1))/( 298 K ) = 5530 JK^(-1) mol^(_1)` `:. Delta S_("total") = DeltaS_("system") + DeltaS_("surroundings") = -549.4 + 5530 JK^(-1) mol^(-1) = + 4980. 6JK^(-1) mol^(-1)` As `Delta S_("total") ` is `+ ve` ,therefore, the reaction is spontaneous.