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}} $



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