Bond energies of `N-=N, H-H` and `N-H` bonds are 945,463 & 391 kJ `mol^(-1)` respectively, the enthalpy of the following reactions is : `N_(2)(g)+3H_(

Question

Bond energies of `N-=N, H-H` and `N-H` bonds are 945,463 & 391 kJ `mol^(-1)` respectively, the enthalpy of the following reactions is : `N_(2)(g)+3H_(

Bond energies of `N-=N, H-H` and `N-H` bonds are 945,463 & 391 kJ `mol^(-1)` respectively, the enthalpy of the following reactions is :
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - `+93 KJ`
`DeltaH = 6xxDeltaH_(N-H) - DeltaH_(N-N) - 3DeltaH_(H-H) = 93 KJ`

5 views Answered 2023-02-05 15:29