Chemical Thermodynamics Question 418

Question: What is the normal boiling point of mercury- Given: $ \Delta H_{f}^{{}^\circ }( Hg,l )=0;S{}^\circ ( Hg,l )=77.4J/K $ -mol $ \Delta H_{f}^{{}^\circ }( Hg,g )=60.8kJ/mol;S{}^\circ ( Hg,g )=174.4J/K-mol $

Options:

A) 624.8 K

B) 626.8 K

C) 636.8 K

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ Hg(l)\rightarrow H(g), $

$ {\Delta_{R}}{{S}^{{}^\circ }}=174.4-77.4=97J/K-mol $

$ \therefore \Delta {{G}^{{}^\circ }}=\Delta {{H}^{{}^\circ }}-T\Delta {{S}^{{}^\circ }}=0 $

$ T=\frac{\Delta {{H}^{{}^\circ }}}{\Delta {{S}^{{}^\circ }}} $

$ =\frac{60.8\times 1000}{97}=626.8K $



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