`Fe_(2)O_(2)(s)+(3)/(2)C(s)to(3)/(2)CO_(2)(g)+2Fe(s)` `DeltaH^(@)=+234.12KJ ` `C(s)+O_(2)(g)toCO_(2)(g) DeltaH^(@)=-393.5KJ` Use these equations and `DeltaH^(@)` value to calculate `DeltaH^(@)` for this reaction : `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)`A. `-1648.7kJ`B. `-1255.3 kJ`C. `-1021.2kJ`D. `-129.4kJ`

`Fe_(2)O_(2)(s)+(3)/(2)C(s)to(3)/(2)CO_(2)(g)+2Fe(s)` `DeltaH^(@)=+234

1.	`Fe_(2)O_(2)(s)+(3)/(2)C(s)to(3)/(2)CO_(2)(g)+2Fe(s)` `DeltaH^(@)=+234.12KJ ` `C(s)+O_(2)(g)toCO_(2)(g) DeltaH^(@)=-393.5KJ` Use these equations and `DeltaH^(@)` value to calculate `DeltaH^(@)` for this reaction : `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)`A. `-1648.7kJ`B. `-1255.3 kJ`C. `-1021.2kJ`D. `-129.4kJ`
Answer» Correct Answer - A `2Fe(s)+(3)/(2)CO_(2)(g)toFe_(2)O_(3)(s)+(3)/(2)C(s)" "(DeltaH^(@)=-234.1kJ)` . . .(i) `C(s)+O_(2)(g)toCO_(2)(g)" "(DeltaH^(@)=-393.5kJ)` . . . (ii) Multiplying eq. (i) by 2 and eq. (ii) by 3 and on adding both equations, we get: `4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)," "DeltaH^(@)=(-234.1xx2)+(-3xx393.5)` `=-1648.7kJ`

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