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The enthalpy of formation of ammonia is `-46.0 kJ mol^(-1)`. The enthalpy for the reaction `2N_(2)(g)+6H_(2)(g)to4NH_(3)(g)`
A. `-46.0 kJ`
B. `46.0 kJ`
C. `184.0 kJ`
D. `-184.0kJ`

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Correct Answer - D
`DeltaH_(f)` refers to the formation of one mole of `NH_(3)`.
`:. DeltaH_(f)` for `4 "moles"=4xx-46=-184 kJ`

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