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  • Confused! kindly explain, The standard enthalpy of formation of NH3 is - 46 kJ mol-1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol-1 and that of N2 is –712 kJ mol-1 , the average bond enthalpy of N - H bond in NH3 is

The standard enthalpy of formation of NH3 is - 46 kJ mol-1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol-1 and that of N2 is –712 kJ mol-1 , the average bond enthalpy of N - H bond in NH3 is

  • Option 1)

    – 1102 kJ mol-1

  • Option 2)

    – 964 kJ mol-1

  • Option 3)

    + 352 kJ mol-1

  • Option 4)

    + 1056 kJ mol-1

 

Answers (1)

best_answer

As we have learnt,

 

Bond dissociation enthalpy -

It is the average of enthalpy required to dissociate the said bond present in different gaseous compound in to free atoms in gaseous state.

- wherein

N_{2}+Bond\, Energy\rightarrow 2N

 

 

For the reaction,

\\*NH_3_{(g)} \rightarrow \frac{1}{2}N_2_{(g)} + \frac{3}{2}H_2_{(g)} \\*\Delta H^o = 3\Delta H_{N-H} + \frac{1}{2}\Delta H_{N_2} + \frac{3}{2}\Delta H_{H_2} \\*46 = 3\Delta H_{N-H} + \frac{1}{2}(-712) + \frac{3}{2}(-436) \\*\Rightarrow \Delta H_{N-H} = \frac{1}{}8 (1056) = + 352\; kJmol^{-1}


Option 1)

– 1102 kJ mol-1

Option 2)

– 964 kJ mol-1

Option 3)

+ 352 kJ mol-1

Option 4)

+ 1056 kJ mol-1

Posted by

SudhirSol

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