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The standard enthalpy of formation of NH3 is -46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is -436 kJ/mol and that of N2 is -712 kJ/mol, the average bond enthalpy of N-H bond in NH3 is:

  • Option 1)

    -1102 kJ/mol

  • Option 2)

    -964 kJ/mol

  • Option 3)

    352 kJ/mol

  • Option 4)

    1056 kJ/mol


Answers (1)


As discussed in

Lavasier and Laplce's Law -

Enthalpy of decomposition of a compound is numerically equal to the formation of that compound and only sign changes.

- wherein

C_{(s)}+O_{2(g)}\rightarrow CO_{2}

\Delta H= 393.5 \, KJ

CO_{2(g)}\rightarrow C_{(s)}+O_{2(g)}

\Delta H= -393.5 \, KJ


 NH_{3}(g)\rightarrow \frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)

\Delta H^{\circ}=-(-46)


\Delta H^{\circ}=3\Delta H_{N-H}+\frac{1}{2}\Delta H_{N\equiv N}+\frac{3}{2}\Delta H_{H-H}

46=3.\Delta H_{N-H}+\frac{1}{2}\times (-712)+\frac{3}{2}\times (-436)

Solving we get 

\Delta H_{N-H}=352\: KJmol^{-1}

Option 1)

-1102 kJ/mol

Option is incorrect

Option 2)

-964 kJ/mol

Option is incorrect

Option 3)

352 kJ/mol

Option is correct

Option 4)

1056 kJ/mol

Option is incorrect

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