Chemical Thermodynamics Question 448
Question: The $ H-H $ bond energy is 430 kJ $ mol $ and $ Cl-Cl $ bond energy is $ 240,kJ,mo{{l}^{-1}}.\Delta H $ for $ HCl $ is $ -90,kJ $ . The $ H-Cl $ bond energy is about
Options:
A) $ 180,kJ,mo{{l}^{-1}} $
B) $ 360,kJ,mo{{l}^{-1}} $
C) $ 213,kJ,mo{{l}^{-1}} $
D) $ 425,kJ,mo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \frac{1}{2}H_2+\frac{1}{2}Cl_2\to HCl,,\Delta H=-90,KJ $
$ \therefore \Delta H=\frac{1}{2}{E_{H-H}}+\frac{1}{2}{E_{Cl-Cl}} $
or $ -90=\frac{1}{2}\times 430+\frac{1}{2}\times 240-E_{HCl} $
$ \therefore {E_{H-Cl}}=425,kJ,mo{{l}^{-1}} $ .
