Given that for the reaction. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = -92kJ. What is the enthalpy of formation of ammonia from its elements?

A. -46KJ mol -1
B. -184KJmol -1
C. -92KJmol -1
D. 184KJmol -1
E. +46KJmol -1

Correct Answer:

Option A – -46KJ mol -1

Explanation

In the reaction, the enthalpy of formation 2 mole of NH3 is – 92kJ. Therefore the standard ΔH of formation = −92kJ/2 =−46KJmol−1.