The standard enthalpy of formation of NH_3 is -46.0 kJ mol^-1 . If the enthalpy of formation of H_2 from its atoms is -436 kJ mol^-1 and that of N_2 i - Chemistry MCQ AIEEE 2026 | ExamTest.live

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The standard enthalpy of formation of NH_{3} is - 46.0 kJ mol^{- 1} . If the enthalpy of formation of H_{2} from its atoms is - 436 kJ mol^{- 1} and that of N_{2} is - 712 kJ mol^{- 1}, the average bond enthalpy of N - H bond in NH_{3} is:

Solution & Step-by-step Explanation

The reaction for the formation of NH_{3} is: \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to NH_{3} (g) Δ H_{f} = - 46.0 kJ mol^{- 1} Alternatively, considering the balanced equation: N_{2} + 3 H_{2} \to 2 NH_{3} Δ H = 2 \times (- 46) = - 92 kJ The enthalpy of formation of molecules from atoms is the negative of bond enthalpy.Bond enthalpy of N \equiv N = 712 kJ mol^{- 1} Bond enthalpy of H - H = 436 kJ mol^{- 1} Let the bond enthalpy of N - H be x . Δ H = \sum B.E. (Reactants) - \sum B.E. (Products) - 92 = [(B.E. of N_{2}) + 3 (B.E. of H_{2})] - [2 \times 3 \times (B.E. of N - H)] - 92 = [712 + 3 (436)] - 6 x - 92 = [712 + 1308] - 6 x - 92 = 2020 - 6 x 6 x = 2112 x = 352 kJ mol^{- 1}

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The standard enthalpy of formation of NH_{3} is - 46.0 kJ mol^{- 1} . If the enthalpy of formation of H_{2} from its atoms is - 436 kJ mol^{- 1} and that of N_{2} is - 712 kJ mol^{- 1}, the average bond enthalpy of N - H bond in NH_{3} is:

- 964 kJ mol^{- 1}

+ 352 kJ mol^{- 1}

+ 1056 kJ mol^{- 1}

- 1102 kJ mol^{- 1}

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