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

Options

A. 46
B. `-46`
C. 92
D. `-92`

Answer

C

Explanation

`(1)/(2) N_(2)(g) + (3)/(2) H_(2) (g) to NH_(3) (g)`
`Delta H_(r) = - 46` kJ/mol
` 2NH_(3) (g) to N_(2) (g) + 3H_(2)(g)`
`DeltaH_(r) = -2 DeltaH_(r)`
= `-2(-46)`
= 92 kJ