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Given (i) 2Fe(2)O(3)(s) rarr 4Fe(s)+3...

Given
(i) `2Fe_(2)O_(3)(s) rarr 4Fe(s)+3O_(2)(g)`,
`DeltaG^(@)=+1487.0 kj mol^(-1)`
(ii) `2CI(g)+O_(2) rarr 2CO_(2)(g)`,
`Delta_(r )G^(@)=-514.4 kj mol^(-1)`
Free energy change `Delta_(r)G^(@)` for the reaction
`2Fe_(2)O_(3)(s)+6CO(g) rarr 4Fe(s)+6cO_(2)(g)` will be :

A

`-1 1 2 .4 K J mol^(-1)`

B

`-5 6 . 2 K J mol^(-1)`

C

`-1 6 8 . 2 K J mol^(-1)`

D

`- 2 0 8 .0 K J mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta G_("reaction") = Delta G_(1) + 3Delta G_(2) = 1487 + 3 xx (-514.4) = -56.2 kJ mol^(-1)`
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