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 $
