InterviewSolution
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InterviewSolution
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Find the electron affinity of chlorine from the following data. Enthalpy of formation of `LiCI` is `-97.5 kcal mol^(-1)`, lattice energy of `LiCI =- 197.7 kcal mol^(-1)`. Dissociation energy of chlirine is `57.6 kcal mol^(-1)`, sublimation enthalpy of lithium `=+ 38.3 kcal mol^(-1)`, ionisation energy of lithium `=123.8 kcal mol^(-1)`. |
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Answer» `CI +e^(-) rarr CI^(Theta) DeltaH =?` Given: `Li(s) +(1)/(2)CI_(2)(g) rarr LiCI(s), DeltaH_(1) =- 97.5 kcal mol^(-1)` `Li^(o+) +CI^(Theta) (g)rarr LiCI(s), DeltaH_(2) =- 197.7 kcal mol^(-1)` `CI(g) rarr 2CI(g), DeltaH_(3) = 57.6 kcal mol^(-1)` `Li(s) rarr Li(g), DeltaH_(4) = 38.3 kcal mol^(-1)` `Li(g) rarr Li^(o+)(g) +e^(-), DeltaH_(5) = 123.8 kcal mol^(-1)` Rewritting equations: `LI(s) +(1)/(2)CI(2)(g) rarr LiCI(s), DeltaH_(1) = - 97.5 kcal mol^(-1)` `LICI(s) rarr Li^(o+)(g) +CI^(Theta)(g), DeltaH_(2) = 197.7 kcal mol^(-1)` `(1)/(2)[2CI_(2)(g) rarr CI_(2)(g)], DeltaH_(3) =(1)/(2)xx-57.6 kcal mol^(-1)` `Li(g) rarr Li(s), DeltaH_(4) =- 38.3 kcal mol^(-1)` `Li^(o+)(g) +e^(-) rarr Li(g), DeltaH_(5) =- 123.8 kcal mol^(-1)` `{:ulbar(CI(g)+e^(-)CI^(Theta)(g)):}` `DeltaH = - 97.5 +197.7 -(1)/(5) xx 57.6 - 38.3 - 123.8` `=- 90.7 kcal mol^(-1)` or Operating: `DeltaH = DeltaH_(1) - DeltaH_(2) - (1)/(2)DeltaH_(3)-DeltaH_(4)-DeltaH_(5)` `=97.5 -(-197.7) -(1)/(2)(57.6)-38.3 -123.8` `=- 90.7 kcal mol^(-1)` |
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