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What is the normal boiling point of mercury? Given : `DeltaH _(f)^(@)(Hg,l)=0,S^(@)(Hg,l)=77.4 J//K-"mol"` `DeltaH _(f)^(@)(Hg,g)= 60.8 kJ//"mol", S^(@)(Hg, g)=174.4 J//K-"mol"`A. 624.8 KB. 626.8 KC. 636.8 KD. None of these |
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Answer» Correct Answer - B `HgiffHg(g)`, `Delta_(r )S^(@)=174.4-77.4 = 97 J//K-mol` `because" "DeltaG^(@)=DeltaH^(@)-T.DeltaS^(@)=0` `" "T=(DeltaH^(@))/(DeltaS^(@))` `" "=(60.8 xx1000)/(97)=626.8K` |
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