Bond dissociation enthalpies of H2(g) and N2(g) are 436.0 kJmol−1 an – InterviewSolution

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1.	Bond dissociation enthalpies of H2(g) and N2(g) are 436.0 kJmol−1 and 941.3 kJmol−1 respectively. Enthalpy of formation of NH3(g) is −46.0 kJmol−1. What is the enthalpy of atomization ofNH3.
Answer» Bond dissociation enthalpies of $H_{2} (g) a n d N_{2} (g)$ are $436.0 k J m o l^{- 1}$ and $941.3 k J m o l^{- 1}$ respectively. Enthalpy of formation of $N H_{3}$ (g) is $- 46.0 k J m o l^{- 1}$ . What is the enthalpy of atomization of $N H_{3}$ .

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