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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

`+ 23 kJ`

B

`+ 92 kJ`

C

`+ 46 kJ`

D

`+ 184 kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta_(r) H = Sigma a_(i) Delta_(f) H_("products") - Sigma b_(i) Delta_(f)H_("reactants")`
`= Delta_(f) H_(N_(2)) + 3 Delta_(f) H_(H_(2)) - 2 Delta_(f) H_(NH_(3))`
`= 0 + 0 - 2 (- 46) = + 92 kJ//mol`
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