Given that for the reaction. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = -92kJ. What is the enthalpy of formation of ammonia from its elements?
-
-46KJ mol -1
- -184KJmol -1
- -92KJmol -1
- 184KJmol -1
- +46KJmol -1
The correct answer is: A
Explanation
In the reaction the enthalpy of formation 2 mole of NH3 is - 92kJ. Therefore the standard \(\Delta H\) of formation = \(\frac{-92kJ}{2} = -46KJmol^{-1}\).