Question

The bond enthalpy of $H_2\left(g\right)$ is $436\text{kJ/mol}$ and that of $N_2\left(g\right)$ is $941.3\text{kJ/mol}$. Calculate the average bond enthalpy of an N - H bond in ammonia.
$\Delta H_f^{\circ }\left(N{H}_3\right)=-46\text{kJ/mol}$

• A. $452.5\text{kJ/mol}$
• B. $390.2\text{kJ/mol}$
• C. $39\text{kJ/mol}$
• D. $1170.6\text{kJ/mol}$

Answer 1

The correct option is B $390.2\text{kJ/mol}$

$N_2\left(g\right)\to 2N\left(g\right);\Delta H=941.3\text{kJ/mol}\dots \dots \left(i\right)$
$H_2\left(g\right)\to 2H\left(g\right);\Delta H=436\text{kJ/mol}\dots \dots \left(ii\right)$
$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\to N{H}_3\left(g\right);\Delta H=-46\text{kJ/mol}\dots \dots \left(iii\right)$
The required equation can be obtained by taking $\frac{1}{2}$ times eqn (i), $\frac{3}{2}$ times eqn (ii) and on reversing one time eqn (iii)
$\frac{1}{2}{N}_2\left(g\right)\to N\left(g\right);\Delta H=\frac{941.3}{2}\text{kJ/mol}$
$\frac{3}{2}{H}_2\left(g\right)\to 3H\left(g\right);\Delta H=\frac{436\times 3}{2}\text{kJ/mol}$
$\underset{–}{N{H}_3\left(g\right)\to \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right);\Delta H=46\text{kJ/mol}}$
$\underset{–}{N{H}_3\left(g\right)\to N\left(g\right)+3H\left(g\right);\Delta H=1170.6\text{kJ/mol}}$
Since, there are three $N-H$ bonds in $N{H}_3$, the average bond enthalpy is obtained by dividing the above value of $\Delta H$ by 3.
$\therefore$ Average bond enthalpy of an $N-H$ bond $=\frac{1170.6}{3}=390.2\text{kJ/mol}$