1.

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

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`


Discussion

No Comment Found

Related InterviewSolutions