Question

Bond dissociation enthalpies of $H_{2_{(}g)}$ and $N_{2_{(}g)}$ are 436.0 $KJ$ $mo{l}^{-1}$ and 941.8 $KJ$ $mo{l}^{-1}$, respectively and enthalpy of formation of $N{H}_{3_{(}g)}$ is $-46KJ$ $mo{l}^{-1}$.

What is the enthalpy of atomization of $N{H}_{3_{(}g)}$ and the average bond enthalpy of $N-H$ bond respectively (in $KJ$ $mo{l}^{-1}$)?

• A. 1170.9, 390.3
• B. 117, 300
• C. 300, 200
• D. 2000, 1975

Answer 1

Answer: (A)

1170.9, 390.3

$N_{2\left(g\right)}+3H_{2\left(g\right)}\to 2N{H}_{3\left(g\right)};\Delta H=-2X46kJmo{l}^{-1}$

$\Delta H=\sum (BE{)}_{reactants}-\sum (BE{)}_{products}$

$=(941.8+3\times 436)-\left(6x\right)=-2\times 46$

(Here $x=$ BE of $N-H$ bonds)

$x=$ $390.3kJmo{l}^{-1}$

$N{H}_3\to N+3\left(H\right)$

Heat of atomisation $=3\times 390.3=1170.9$ $kJmo{l}^{-1}$

Hence, the correct option is $A$