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Calculate Delta(r)S(m)^(Theta) for the r...

Calculate `Delta_(r)S_(m)^(Theta)` for the reaction:
`4Fe(s)+3O_(2)(g) rarr 2Fe_(2)O_(3)(s)`
Given that `S_(m)^(Theta)(Fe) = 27.3J K^(-1) mol^(-1)`,
`S_(m)^(Theta)(O_(2)) = 205.0J K^(-1)mol^(-1)`and `S_(m)^(Theta)(Fe_(2)O_(3)) = 87.4 J K^(-1) mol^(-1)`.

Text Solution

Verified by Experts

`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.
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Knowledge Check

  • Calculate standard entropy change in the reaction Fe_(2)O_(3)(s)+3H_(2)(g) rarr 2Fe(s)+3H_(2)O(l) Given : S_(m_(0))(Fe_(2)O_(3).S)=87.4,S_(m)^(@)(Fe,S)=27.3 S_(m)^(@)(H_(2),g)=130.7,S_(m)^(@)(H_(2)O,l)=69.9JK^(-1)mol^(-1)

    A
    `-212.5JK^(-1)mol^(-1)`
    B
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    C
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  • in the given equation 4Fe(s)+3O_(2)(g) to 2Fe_(2)O_(3)(s) the entropy change is =-549.4 JK^(-1) mol^(-1) at 298 K (Delta_rH^(-)=-1648 xx10^(3)Jmol^(-1)) .the above reactions is

    A
    spontaneous
    B
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    C
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  • Which of the following statements is correct about the reaction given below : 4Fe(s)+3O_(2)(g) rarr 2Fe_(2)O_(3)(s)

    A
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    B
    Total mass of reactants = total mass of product , therefore, law of multiple proportions is followed
    C
    Amount of `Fe_(2)O_(3)` can be increased by taking one of the reactants (iron or oxygen) in excess.
    D
    Amount of `Fe_(2)O_(3)` produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.
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