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}} $
Show Answer
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}} $
