The enthalpy of formation of ammonia is `- 46.2 mol^(-1)`. The enthalpy change for the reaction
`2NH_(3) rarr N_(2) + 3 H_(2)` is

Options

A. `42 kJ`
B. `64 kJ`
C. `80 kJ`
D. `92 kJ`

Answer

D

Explanation

The enthalpy of formation of ammonia describes the following thermochemical equation:
`(1)/(2) N_(2) (g) + (3)/(2) H_(2) (g) rarr 2 NH_(3)`
To get the enthalpy change for the given reaction, we need to reverse it and then multiply it by 2. Thus,
`Delta H = - (2) (46.2) = 92.4 kJ`