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What is the P-P bond energy in kJ/mol, if the heat of atomisation of $PH_3(g)$ and $P_2H_4(g)$ are 954 kJ/mol and 1485 kJ/mol, respectively?


(a) 213
(b) 426
(c) 1262
(d) 107

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Answer: 213
Atomisation energy of $PH_3 = 3\times BE(P-H) = 954$ kJ/mol
$\therefore BE(P-H) = \frac{954 kJ/mol}{3} = 318$ kJ/mol
Atomisation energy of $P_2H_4 = BE(P-H)\times 4 + BE(P-P) = 1485$ kJ/mol
$\therefore BE(P-P)$ = Atomisation energy of $P_2H_4 -4 \times BE(P-H) = 1485-4\times 318 = 1485 -1272 = 213$ kJ/mol
answered Feb 27, 2014 by mosymeow_1
 

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