The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Options

A. `+46.0 kJ`
B. `+92.0 kJ`
C. `-23 kJ`
D. `-92 kJ`

Answer

B

Explanation

The given reaction involve decomposition of `2` mol of ammonia `DeltaH_(f)(NH_(3))=-46.0`, `DeltaH_("decomposition")"of" NH_(3)=+46`
`:. DeltaH "of given reaction"=2xx46=92 kJ`