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The heat of formation of NH(3)(g) is -46...

The heat of formation of `NH_(3)(g)` is `-46 " kJ mol"^(-1)`. The `DeltaH` (in `" kJ mol"^(-1)`) of the reaction, `2NH_(3)(g)rarrN_(2)(g)+3H_(2)(g)` is

A

`+184 kJ`

B

`+23 kJ`

C

`+92 kJ`

D

`+46 kJ`

Text Solution

Verified by Experts

The correct Answer is:
C
`2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)`
`DeltaH_(r )= - (2 xx "enthalpy of formation of" NH_3) = -(2 xx -46) = 92 kJ`.
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