The standard enthalpy of formation of NH₃(g) is -46 ~kJ mol⁻¹ . If the enthalpy of formation of H(g) is 218 ~kJ mol⁻¹ and the average N-H bond enthalpy in ammonia is 391 ~kJ mol⁻¹ , the standard enthalpy of formation of N(g) is __________ kJ mol⁻¹ .

Step-by-step solution

The atomization of ammonia can be represented as: NH₃(g) N(g) + 3 H(g) The enthalpy of atomization ( _ atom H ) is the sum of the bond enthalpies of all bonds in the molecule. Since ammonia has three N-H bonds: _ atom H = 3 BE(N-H) = 3 391 = 1173 ~kJ mol⁻¹ Using Hess's Law, the enthalpy of atomization is also given by: _ atom H = _ f H ^ ( N(g) ) + 3 _ f H ^ ( H(g) ) - _ f H ^ ( NH₃(g) ) Substituting the given values: 1173 = _ f H ^ ( N(g) ) + 3(218) - (-46) 1173 = _ f H ^ ( N(g) ) + 654 + 46 1173 = _ f H ^ ( N(g)