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For oxidation of iron. 4Fe(s)+3O(2)(g)...

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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To determine why the oxidation of iron is spontaneous despite a negative entropy change, we can analyze the thermodynamic principles involved. ### Step-by-Step Solution: 1. **Identify the Given Data:** - The reaction is: \[ 4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s) ...
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For oxidation of iron, 4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3) (s) entrop change is -549.4 J K^(-1)mol^(-1) at 298K. In spite of negative entropy change of this reaction, why is the reaction spontaneous? ( Delta_(r ) H^(theta) for this reaction is -1648 kJ mol^(-1) )

For oxidation of iron,, 4Fe(s) + 3O_(2)(g) rarr 2Fe_(2)O_(3)(s) , entropy change is - 549 .4 JK^(-1) mol^(-1) at 298 K . Inspite of the negative entropy change of this reaction , why is the reaction spontaneous ? ( Delta H ^(@) for this reaction is - 1648 xx 10^(3) J mol^(-1))

Knowledge Check

  • For oxidation of iron, 4Fe(s) + 3O2(g) to 2Fe2O3(s) entropy change is – 549.4 JK​ ^(–1 ) ​mol^(​–1) at 298K. Delta r H^(@) for this reaction is – 1648 xx 10​^( 3)​ J mol^(​–1) Above reaction is :-

    A
    Spontaneous
    B
    Non-spontaneous
    C
    At equilibrium
    D
    Cant predict

  • 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
    non-spontaneous
    C
    both (a) and (b)
    D
    none of these

  • The entropy change for a certain non-spontaneous reaction is 150 J/K.mol. at 298 K. The minimum value of DeltaH for the reaction is _________.

    A
    `-44.7 `kJ
    B
    44.7 kJ
    C
    0.448 kJ
    D
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