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For oxidation of iron, 4Fe(s) + 3O2(g) t...

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

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta G =Delta H – T.DeltaS `
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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))

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) )

Knowledge Check

  • 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

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