L)\rightleftharpoons H(G),\] \[{{\Delta }_{R}}{{S}^{{}^\circ }}=174.4-77.4=97J/K-mol\] \[\THEREFORE \text{ }\Delta {{G}^{{}^\circ }}=\Delta {{H}^{{}^\circ }}-T\Delta {{S}^{{}^\circ }}=0\] \[T=\frac{\Delta {{H}^{{}^\circ }}}{\Delta {{S}^{{}^\circ }}}\] \[=\frac{60.8\times 1000}{97}=626.8K\]

"> L)\rightleftharpoons H(G),\] \[{{\Delta }_{R}}{{S}^{{}^\circ }}=174.4-77.4=97J/K-mol\] \[\THEREFORE \text{ }\Delta {{G}^{{}^\circ }}=\Delta {{H}^{{}^\circ }}-T\Delta {{S}^{{}^\circ }}=0\] \[T=\frac{\Delta {{H}^{{}^\circ }}}{\Delta {{S}^{{}^\circ }}}\] \[=\frac{60.8\times 1000}{97}=626.8K\]

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

NEET Chemistry in NEET . 9 months ago

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[b] \[Hg(L)\rightleftharpoons H(G),\] \[{{\Delta }_{R}}{{S}^{{}^\circ }}=174.4-77.4=97J/K-mol\] \[\THEREFORE \text{ }\Delta {{G}^{{}^\circ }}=\Delta {{H}^{{}^\circ }}-T\Delta {{S}^{{}^\circ }}=0\] \[T=\frac{\Delta {{H}^{{}^\circ }}}{\Delta {{S}^{{}^\circ }}}\] \[=\frac{60.8\times 1000}{97}=626.8K\]

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Posted on 21 Aug 2024, this text provides information on NEET related to Chemistry in NEET. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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