Question

Given are the bond enthalpies: ${ϵ}_{N\equiv N}=945{\text{kJ mol}}^{-1},{ϵ}_{H-H}=436{\text{kJ mol}}^{-1}$ and ${ϵ}_{(N-H)}=391{\text{kJ mol}}^{-1}$. The enthalpy change of the reaction
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is

• A. $-93{\text{kJ mol}}^{-1}$
• B. $-89{\text{kJ mol}}^{-1}$
• C. $-105{\text{kJ mol}}^{-1}$
• D. $105{\text{kJ mol}}^{-1}$

Answer 1

The correct option is A $-93{\text{kJ mol}}^{-1}$

Given are the bond enthalpies: ${ϵ}_{N\equiv N}=945{\text{kJ mol}}^{-1},{ϵ}_{H-H}=436{\text{kJ mol}}^{-1}$ and ${ϵ}_{(N-H)}=391{\text{kJ mol}}^{-1}$.
Hence, $\Delta H={ϵ}_{(N\equiv N)}+3{ϵ}_{(H-H)}-6{ϵ}_{(N-H)}$
$=(945+3\times 436-6\times 391){\text{kJ mol}}^{-1}=-93{\text{kJ mol}}^{-1}$