Chemical Thermodynamics Question 321
The $ H-H $ bond energy is 430 kJ $ mol^{-1} $ and $ Cl-Cl $ bond energy is $ 240,kJ,mol^{-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}} $ .
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