Question

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 1

`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.