Calculate the $N-N$ bond energy in $N_2H_4$ from the given bond enthalpy data.
$\varepsilon_{N-H} = 393 \ kJ/mol$
$\varepsilon_{H-H} = 436 \ kJ/mol$
$\Delta H_{vap}[N_2H_{4(l)}] = 18 \ kJ/mol$
$N_2H_{4(l)} + H_{2(g)} \to 2NH_{3(g)} : \Delta H = -142 \ kJ/mol$
....... $kJ/mol$

  • A
    $210$
  • B
    $190$
  • C
    $180$
  • D
    $150$

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