Question

Enthalpy of formation of $2$ mol of $NH_3 (g)$ is $- 90\, kJ,$ and $\Delta H_{H-H}$ and$\Delta H_{N-H}$are respectively $435 \,kJ\, mol^{-1}$ and $390\, kJ \,mol^{-1}$. The value of $\Delta H_{N-N}$ is

• A. $-472.5\, kJ$
• B. $-945\, kJ\, mol^{-1}$
• C. $472.5\, kJ$
• D. $945\, kJ\, mol^{-1}$

Answer 1

Answer: (D)

$N_{2}\left(g\right)+3H_{2}\left(g\right)\to2NH_{3}\left(g\right);\Delta H=-90kJ$
Now $-90=\Delta H_{N=N}+3\Delta H_{H-H}+6\Delta H_{N-H}$
or $\Delta H_{N=N}=6\Delta H_{N-H}-3\Delta H_{H-H}-90 $
$= 6 \times 390 -3 \times 435-90$
$= 945 kJ$