Q.

The heat of atomisation of PH3(g) is 228 K.Cal mol–1 and that of P2H4(g) is 355 K.Cal mol–1 The energy of the P–P bond is (in K.Cal);

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answer is 2.

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Detailed Solution

PH3(g) → P(g) + 3H(g) ΔH=228 Kcal3B.E(P-H) = 228BEP-H = 228/3355=304+BEp-pBEP-P = 335-304           = 51 Kcal
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The heat of atomisation of PH3(g) is 228 K.Cal mol–1 and that of P2H4(g) is 355 K.Cal mol–1 The energy of the P–P bond is (in K.Cal);