Chemical Thermodynamics Question 402
Question: The molar entropies of HI (g) and I (g) at 298 K are 206.5, 114.6, and $ 180.7Jmo{{l}^{-1}}{{K}^{-1}} $ respectively. Using the $ \Delta G{}^\circ $ given Below, calculate the bond energy of HI. $ HI( g )\xrightarrow{{}}H( g )+I( g );\Delta G{}^\circ =271.8kJ $
Options:
A) $ 282.4kJmo{{l}^{-1}} $
B) $ 298.3kJmo{{l}^{-1}} $
C) $ 290.1kJmo{{l}^{-1}} $
D) $ 315.4kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \Delta {{S}^{{}^\circ }}=-206.5+114.6+180.7=88.8 $
$ \Delta {{G}^{{}^\circ }}=\Delta {{H}^{{}^\circ }}-T\Delta {{S}^{{}^\circ }} $
$ \Delta {{H}^{{}^\circ }}=271.8+298\times 88.8\times {{10}^{-3}} $
$ \Delta {{H}^{{}^\circ }}=298.3kJmo{{l}^{-1}} $
