The enthalpy of atomization of PH3(g) is +954 kJ/mol and that of P2H4 is +1.488 MJ/mol. The bond ene — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy of atomization of PH3(g) is +954 kJ/mol and that of P2H4 is +1.488 MJ/mol. The bond energy of the P-P bond is
Answer: C
💡 Solution & Explanation
BE(P-H) = 954/3 = 318 kJ/mol. P2H4 -> 2P + 4H: 1488 = BE(P-P) + 4*318 = BE(P-P) + 1272. BE(P-P) = 213 kJ/mol.
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