Question

The heat of atomisation of $PH_3(g)$ and $P_2H_4(g)$ are $954 kJ\, mol^{-1}$ and $1485\, kJ\, mol^{-1}$ respectively .The P-P bond energy in $kJ\, mol^{-1}$ is

• A. 213
• B. 426
• C. 318
• D. 1272

Answer 1

Answer: (A)

213

Answer 2

In $PH _{3( g )}$, energy required to break $3 P - H$ bonds $=954 \,kJ\, mol ^{-1}$ $\therefore$ Energy required to break $1 P - H$ bond $=\frac{954}{3}=318 \,kJ\, mol ^{-1}$ In $P _{2} H _{4(g)}$, energy of $1 P - P$ bond $+4 P - H$ bonds $=1485 \,kJ\, mol ^{-1}$ $\because$ Energy of $1 P - H$ bond $=318 kJ mol ^{-1}$ $\therefore$ Energy of $4 P - H$ bonds $=318 \times 4=1272 \,kJ\, mol ^{-1}$ Thus, the $P - P$ bond energy $=1485-1272=213\, kJ\, mol ^{-1}$