Question

The standard enthalpy of formation of $N{H}_3$ is $-46kJ/mol$. If the enthalpy of formation of $H,$ from its atoms is $-436kJmo{l}^{-1}\&$ that of $N_2$ is $-712kJmo{l}^{-1},$ the average bond enthalpy of $N-H$ bond in $N{H}_3$ 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)

$\frac{1}{2}{N}_2(g)+\frac{3}{2}{H}_2(g)\to N{H}_3(g);{\Delta }_{f}H=46kJ/mol$

$2H(g)\to H_2(g);\Delta H=-436kJmo{l}^{-1}$

$2N(g)\to N_2(g);\Delta H=-712kJmo{l}^{-1}$

${\Delta }_{r}H=\Sigma B\cdot E_R-\Sigma B\cdot E_P$

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

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

$-46=(356+654)-v3x$

$-46-1010=-3x$

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

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