Bond dissociation enthalpies of H2(s) and N2(g) are 436.0kJmol−1 and enthalpy of formation of NH3(g) is −46kJmol−1. What is enthalpy of atomization of NH3(g)? What is the average bond enthalpy of N−H bond?
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Solution
N2(g)+3H2(g)⟶2NH3(g);ΔH=−2×46kJ/mol ΔH=∑(BE)R−∑(BE)P=(941.8+3×436)−(6x)=−2×46 (here x=BE of N-H bond) x=390kJmol−1 NH3⟶N+3H Heat of atomization =3×390.3=1170.9kJmol−1