Question

Given the following bond dissociation enthalpies $\left(\text{kJ mo}{\text{l}}^{\text{-1}}\right)$ . $P\equiv P-490,P-P-209,$ $N\equiv N-946,$ $N-N-160$ Compare the enthalpy changes for the process (i) $2P_2(g)\to P_4(g)$ (ii) $2N_2(g)\to N_4(g)$ Choose the correct option.

• A. $P_4$ is feasible but ${\text{N}}_{\text{4}}$ is not feasible
• B. $P_4$ is not feasible but ${\text{N}}_{\text{4}}$ is feasible
• C. Both ${\text{P}}_4$ and ${\text{N}}_{\text{4}}$ are feasible
• D. Both $P_4$ and ${\text{N}}_{\text{4}}$ are not feasible

Answer 1

Answer: (A)

The process involves replacement of $\text{2}\left(\text{X}\equiv \text{X}\right)$ units by 6 X?X bonds in the $X_4$ tetrahedron. Energy is released when bonds are formed and thus species formed is stable. (i) $2P^2(g)\to P_4(g)$
$\Delta H_t=-6\times BE(P-P)+2BE(P\equiv P)$ (BE =Bonds energy ) $=-6\times 209+2\times 490=-2\text{74 kJ mo}{\text{l}}^{\text{-}}$ Formation of $P_4$ is exothermic thus is feasible. (ii) $2N_2(g)\to N_4(g)$ $\Delta H_t=-6BE(N-N)+2BE(N\equiv N)$ $=-6\times 160+2\times 946=+9\text{32 kJ mo}{\text{l}}^{\text{-1}}$ Formation of $N_4$ is endothermic, thus is not feasible.