The standard enthalpy of atomization of is and that of is . What is the bond dissociation enthalpy o — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For $PH_3$, atomization involves breaking $3$ identical $P-H$ bonds. $3 \times BE(P-H) = 954 \Rightarrow BE(P-H) = 318 \text{ kJ mol}^{-1}$. For $P_2H_4$, atomization involves breaking $1$ $P-P$ bond and $4$ $P-H$ bonds. $BE(P-P) + 4 \times BE(P-H) = 1485$. Substituting $BE(P-H)$ yields: $BE(P-P) + 4(318) = 1485 \Rightarrow BE(P-P) = 1485 - 1272 = 213 \text{ kJ mol}^{-1}$.