Calculate lattice energy for the change, `Li^(+)(g)+Cl^(-) (g) rarr LiCl(g)` Given that `{:(DeltaH_("sublimation")" of "Li=160.67 kJ mol^(-1)",",,Delt

Question

Calculate lattice energy for the change, `Li^(+)(g)+Cl^(-) (g) rarr LiCl(g)` Given that `{:(DeltaH_("sublimation")" of "Li=160.67 kJ mol^(-1)",",,Delt

Calculate lattice energy for the change,
`Li^(+)(g)+Cl^(-) (g) rarr LiCl(g)`
Given that
`{:(DeltaH_("sublimation")" of "Li=160.67 kJ mol^(-1)",",,DeltaH_("Dissociation")" of "Cl_(2) =244.34 kJ mol^(-1)","),(DeltaH_("ionisation")" of "Li(g)=520.07 kJ mol^(-1)",",,DeltaH_(E.A)" of "Cl(g)= -365.26 kJ mol^(-1)","),(DeltaH_(f)" of "LiCl(s)= -401.66 kJ mol^(-1)",",,):}`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Considering the different changes that occur in the formation of solid lithium chloride based on the date given the lattice energy of the above can be constituted as:
`DeltaH_(f)^(@)=DeltaH_("subl.")+DeltaH_("I.E.")+1/2 DeltaH_("Diss.")+DeltaH_("E.A")+DeltaH_("lattice")`
or`" "DeltaH_("lattice")=DeltaH_(f)^(@)-DeltaH_("subl.")-DeltaH_("I.E.")-1/2 DeltaH_("Diss.")-DeltaH_("E.A")`
`=-839.31 kJ mol^(-1)`