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The standard enthalpy of formation of `Fe_(2)O_(3)(s)` is - 824.2 kJ `mol^(-1)` . Calculate the enthalpy change for the reaction `: 4 Fe (s) + 3O_(2) (g) rarr 2Fe_(2) O_(3) (s)` |
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Answer» Correct Answer - `-1648.4kJ` `Delta_(r)H^(@) = [ 2 xx Delta_(f) H^(@) ( Fe_(2)O_(3))] - [ Delta_(f)H^(@) ( Fe)+ 3Delta_(f) H^(@) (O_(2))]` `= [2(-824.2) ] -[0+0]` `= - 1648.4kJ` |
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