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The enthalpy of formation of NH(3) is -4...

The enthalpy of formation of `NH_(3)` is `-46 kJ//mol` The enthalpy change for reaction :
`2NH_(3)(g) rarr N_(2)(g) + 3H_(2)(g)` is :

A

`+23kJ`

B

`+92kJ`

C

`+46kJ`

D

`+184kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
The amount of heat absorbed or evolved when 1 mole of the substance is directly obtained from its constituent element is called heat or enthalpy of formation.
The reaction for the formation of `NH_(3)` is
`1/2N_(2)+3/2H_(2)toNH_(3),DeltaH=-45kJ"mol"^(-1)`
Heat of formtioin of `NH_(3)=-46kJ"mol"^(-1)`
The enthalpy change for the given reaction
`2NH_(3(g))toN_(2(g))+3H_(2(g)),Delta H_(1)=?`
`DeltaH_(1)=-(2xxDeltaH)=-(2xx-46)=92kJ`
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