What is the standard enthalpy of rection (in kJ) when two moles of `Fe_(2)O_(3)`(s) reacts with `H_(2)` gas to give Fe metal?
`DeltaH_(f)^(@)` of `Fe_(2)O_(3)(s)andH_(2)O(l)` are -824.2 and `-285.83kJmol^(-1)` respectively.

Options

A. `-66.58`
B. `-33.3`
C. `-538.37`
D. `-1110.03`

Answer

A

Explanation

When ` Fe_(2)O_(3)(s)` react with ` H_(2) (g)` , it will give Fe and `H_(2)O` as follows : ` Fe_(2) O_(3) + 3H_(2) to 2Fe_(2) + 2H_(2)O `
when 2 moles of ` Fe_(2)O_(3)` are used , the reaction becomes as follows :
` [Fe_(2)O_(3) + 3H_(2) to 2 Fe + 3H_(2)O]`
Given, ` Delta H^(@)""_(f) " for " Fe_(2)O_(3) (s) = - "824.2 kJm mol"^(-1)`
and ` Delta H^(@)""_(f) " for " H_(2)O (l) = - 285.83 " kJ mol"^(-1)`
` because ` Standard enthalpy of reaction
` (DeltaH^(@)""_(R)) = sum Delta H^(@) ""_(f) " (Product)" - sum Delta H^(@) ""_(f) ` (Reacant)
and `{:( H^(@)""_(f) " for " H_(2) = 0 ),(H^(@)""_(f) " for Fe = 0 "):}]` (as are present in free state)
Thus . ` Delta H^(@) ""_(R) = [6xx - 285.83] - [2xx(-824.2)] `
` DeltaH^(@) ""_(R) = (-1714.98) + (1648.4) `
` Delta H^(@) ""_(R) = - 66.58 Kj//mol `
Hence , option (a) is the correct answer .