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For the reaction N(2)H(4) (g) rarr N(2...

For the reaction
`N_(2)H_(4) (g) rarr N_(2)H_(2)(g) +H_(2)(g)" "Delta_(r)H^(@)=109 KJ//mol`
Calculate the bond enthalpy of `N=N`.
Given : `B.E. (N-N) =163 KJ//mol, B.E. (N-H) =391 KJ//mol, B.E. (H-H)=436 KJ//mol`

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To calculate the bond enthalpy of the N=N bond for the reaction \[ \text{N}_2\text{H}_4(g) \rightarrow \text{N}_2\text{H}_2(g) + \text{H}_2(g) \] with the given enthalpy change \( \Delta_rH^\circ = 109 \, \text{kJ/mol} \), we can use the bond enthalpy values provided: - B.E. (N-N) = 163 kJ/mol - B.E. (N-H) = 391 kJ/mol ...
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