Chemical Thermodynamics Question 400

Question: The heat of atomization of $ PH_3(g) $ is $ 228kcalmo{{l}^{-1}} $ and that of $ P_2H_4(g) $ is $ 335kcalmo{{l}^{-1}} $ . The energy of the P-P bond is

Options:

A) $ 102kcalmo{{l}^{-1}} $

B) $ 51kcalmo{{l}^{-1}} $

C) $ 26kcalmo{{l}^{-1}} $

D) $ 204kcalmo{{l}^{-1}} $

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Answer:

Correct Answer: B

Solution:

  • Bond dissociation energy of $ PH_3( g )=228kcalmo{{l}^{-1}} $

$ P-H $ bond energy $ =\frac{228}{3}=76kcalmo{{l}^{-1}} $ Bond energy of $ 4(P-H)+(P-P) $

$ =355kcalmo{{l}^{-1}} $ or $ 4\times 76+( P-P )=355kcalmo{{l}^{-1}} $

$ =P-P $ bond energy $ =51kcalmo{{l}^{-1}} $



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