Chemical Thermodynamics Question 264
The heat of atomization of $ PH_3(g) $ is $ 228\ kcal\ mol^{-1} $ and that of $ P_2H_4(g) $ is $ 335\ kcal\ mol^{-1} $ . The energy of the P-P bond is
Options:
A) $ 102\ \text{kcal}\ \text{mol}^{-1}$
B) $ 51kcalmo{{l}^{-1}} $
C) $ 26\ \text{kcal}\ \text{mol}^{-1}$
D) $ 204\ \text{kcal}\ \text{mol}^{-1}$
Show Answer
Answer:
Correct Answer: B
Solution:
- Bond dissociation energy of $ PH_3( g )=228\ \text{kcal}\ \text{mol}^{-1} $
$ P-H $ bond energy $ =\frac{228}{4}=57kcalmo{{l}^{-1}} $ Bond energy of $ 4(P-H)+(P-P) $
$ =355\ \text{kcal}\ \text{mol}^{-1} $ or $ 4\times 76+( P-P )=355\ \text{kcal}\ \text{mol}^{-1} $
$ =P-P $ bond energy $ =51\ \text{kcal}\ \text{mol}^{-1}$
BETA
