MCQ
Calculate the $N-N$ bond energy in $N_2H_4$ from given bond enthalpy data.
$\varepsilon_{N-H} = 393\, kJ/mole$
$\varepsilon_{H-H} = 436\, kJ/mole$
$\Delta H_{vap}[N_2H_4(l)] = 18\, kJ / mole$
$N_2H_4(l) + H_2(g) \to 2NH_3(g) : \Delta H = -142\, kJ/mole$
$\varepsilon_{N-H} = 393\, kJ/mole$
$\varepsilon_{H-H} = 436\, kJ/mole$
$\Delta H_{vap}[N_2H_4(l)] = 18\, kJ / mole$
$N_2H_4(l) + H_2(g) \to 2NH_3(g) : \Delta H = -142\, kJ/mole$
.......$ kJ/mole$
- A$210$
- ✓$190$
- C$180$
- D$150$
