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Standard XII

Chemistry

Lattice Energy

Calculate the...

Question

Calculate the electron affinity of fluorine atom using the following data (only magnitude in nearest integer in kj/mol).
All the values are in $k J$ ${m o l}^{- 1}$ at $25^{o} C$ , $Δ H_{d i s s} (F_{2}) = 160$ ; $Δ H_{r}^{o}$ ( $N a F (s)) = - 571; I . E$ $[N a (g)] = 494, Δ H_{v a p} [N a (s)] = 101$ . Lattice energy of $N a F (s) = - 894$ .

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Solution

From Born-Hyber cycle in the diagram,
$Δ H_{f}^{o} = Δ H_{v a p, N a (s)}^{o} + \frac{1}{2} Δ H_{d i s, F}^{o} + I . E_{N a} + E . A_{F} + L . E_{N a F}$
On simplifying
$E . A_{F} = Δ H_{f}^{o} - Δ H_{v a p, N a (s)}^{o} - \frac{1}{2} Δ H_{d i s, F}^{o} - I . E_{N a} - L . E_{N a F}$
$\Rightarrow E . A_{F} = - 571 - 101 - \frac{1}{2} \times 160 - 494 + 894 = - 352 k J / m o l$

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Q. Calculate the electron gain enthalpy of fluorine atom using the following data.

All the values are in

k J m o l^{- 1}

25^{o} C

Δ H_{d i s s} (F_{2}) = 160, Δ_{f} H (N a F (s)) = - 571

, I.E.

[N a (g)] = 494, Δ H_{v a p} [N a (s)] = 101

. Lattice enthalpy of

N a F (s) = 894

Also, make Born - Haber's cycle.

Q. Use the following data to calculate the lattice energy of caesium oxide.
Enthalpy of formation of caesium oxide =

- 233 k J m o l^{- 1}

Enthalpy of sublimation of

C s = + 78 k J m o l^{- 1}

First ionization energy of

C s = + 375 k J m o l^{- 1}

Enthalpy of dissociation of

O_{2} (g) = 494 k J m o l^{- 1} o f O_{2} m o l e c u l e s

First electron affinity of

O = - 141 k J m o l^{- 1} o f O a t o m s

Second electron affinity of

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Q. Use the following data to calculate the lattice energy of caesium oxide.
Enthalpy of formation of caesium oxide

= - 233 kJ/mol

Enthalpy of sublimation of

C s = + 78 kJ/mol

First ionization energy of

C s = + 375 kJ/mol

Enthalpy of dissociation of

O_{2} (g) = 494 kJ/mol

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Q. Calculate the electron affinity of oxygen atom in gaining two electrons to give the oxide ion

O^{2 -}

from the following data.
Standard heat of formation of

M g O_{(s)} = - 600 k J m o l^{- 1}

Lattice energy of

M g O_{(s)} = - 3860 k J m o l^{- 1}

lonization energy of

M g_{(g)}

to give

M g^{2 +}

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Enthalpy of formation of cesium oxide =

- 233 k J / m o l

Enthalpy of sublimation of

C s = + 78 k J / m o l

First ionization energy of

C s = + 375 k J / m o l

Enthalpy of dissociation of

O_{2} (g) = 494 k J / m o l

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First electron affinity of

O = - 141 k J / m o l

of O atoms
Second electron affinity of

O = + 845 k J / m o l

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Calculate the electron affinity of fluorine atom using the following data (only magnitude in nearest integer in kj/mol). All the values are in kJ mol−1 at 25oC, ΔHdiss(F2)=160; ΔHor (NaF(s))=−571; I.E [Na(g)]=494, ΔHvap[Na(s)]=101. Lattice energy of NaF(s)=−894.

From Born-Hyber cycle in the diagram,ΔHof=ΔHovap,Na(s)+12ΔHodis,F+I.ENa+E.AF+L.ENaFOn simplifying E.AF=ΔHof−ΔHovap,Na(s)−12ΔHodis,F−I.ENa−L.ENaF⇒E.AF=−571−101−12×160−494+894=−352kJ/mol