Question

The standard enthalpy of formation of $NH _{3}$ is $-46 \,kJ / mol$. If the enthalpy of formation of $H ,$ from its atoms is $-436\, kJ \,mol ^{-1} \&$ that of $N _{2}$ is $-712 \,kJ \,mol ^{-1},$ the average bond enthalpy of $N - H$ bond in $NH _{3}$ is:

• A. 964 kJ mol $^{-1}$
• B. 352 kJ mol $^{-1}$
• C. -1102 kJ mol $^{-1}$
• D. +1056 kJ mol $^{-1}$

Answer 1

Answer: (B)

$\frac{1}{2} N _{2}(g)+\frac{3}{2} H _{2}(g) \rightarrow NH _{3}(g) ; \Delta_{ f } H =46\, kJ / mol$

$2H (g) \rightarrow H _{2}(g) ; \Delta H =-436 \,kJ \,mol ^{-1}$

$2N (g) \rightarrow N _{2}(g) ; \Delta H =-712\, kJ\, mol ^{-1}$

$\Delta_{r} H=\Sigma B \cdot E_{R}-\Sigma B \cdot E_{P}$

$-46=\left(\frac{1}{2} \times 712+\frac{3}{2} \times 436\right)-(3 x)$

[where $x$ is bond energy of $N-H$ bond]

$-46=(356+654)-v 3 x$

$-46-1010=-3 x$

$\frac{-1056}{-3}=x$

$\therefore x=352$ kJmol $^{-1}$