1. Tardigrade
  2. Question
  3. Chemistry
  4. The heat of atomisation of PH3(g) and P2H4(g) are 954 kJ mol-1 and 1485 kJ mol-1 respectively .The P-P bond energy in kJ mol-1 is

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Q. 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

KEAMKEAM 2009Thermodynamics

Solution:

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 $=1485-1272=213\, kJ\, mol ^{-1}$