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Standard XII

Chemistry

Enthalpy

Calculate the...

Question

Calculate the lattice formation enthalpy in ${kJ mol}^{- 1}$ of $L i F$ given that
Enthalpy of sublimation of lithium = $155.3 {kJ mol}^{- 1}$
Dissociation enthalpy of half mole of $F_{2} = 75.3 kJ$
Ionization enthalpy of lithium $= 520 {kJ mol}^{- 1}$
Electron gain enthalpy of 1 mol of $F (g) = - 333 kJ Δ_{f} H_{overall} = - 594 {kJ mol}^{- 1}$

- 1011.6 {kJ mol}^{- 1}

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- 1200.4 {kJ mol}^{- 1}

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- 984.6 {kJ mol}^{- 1}

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- 1488.6 {kJ mol}^{- 1}

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Solution

The correct option is A $- 1011.6 {kJ mol}^{- 1}$
Given,
$L i (s) \to L i (g); Δ H_{Sub}^{o} = + 155.3 {kJ mol}^{- 1} L i (g) \to L i^{+}; I E = 520 {kJ mol}^{- 1} \frac{1}{2} F_{2} (g) \to F (g); Δ H_{F - F}^{o} = 75.3 {kJ mol}^{- 1} F (g) + e^{- 1} \to F^{1 -}; Δ H_{e g} = - 141 {kJ mol}^{- 1} Δ_{f} H_{overall} = - 594 {kJ mol}^{- 1} F o r L i F; Δ H_{f} = Δ H_{Sub}^{o} + \frac{1}{2} Δ H_{F - F} + Δ H_{i} + Δ H_{e g} + Δ H_{Lattice} - 594.1 = 155.2 + 75.3 + 520 - 333 + Δ H_{Lattice} Δ H_{Lattice} = - 1011.6 {kJ mol}^{- 1}$
Hence, the lattice formation enthalpy of $L i F$ is given by $- 1011.6 {kJ mol}^{- 1}$ .

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Similar questions

Q. Calculate the lattice enthalpy in

k J m o l^{- 1}

L i F

given that
Sublimation of lithium is

155.3 k J m o l^{- 1}

Dissociation of half mole of

F_{2} = 75.3 k J

Ionization enthalpy of lithium

= 520 k J m o l^{- 1}

Electron gain enthalpy of 1 mol of

F (g) = - 333 k J Δ_{f} H_{o v e r a l l} = - 594 k J m o l^{- 1}

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

First electron affinity of

O = - 141 kJ/mol

Second electron affinity of

O = + 845 kJ/mol

Q. Calculate the lattice enthalpy of

C a C l_{2}

, given that :
Enthalpy of sublimation for

C a (s) \to C a (g) = 121 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} (g) \to 2 C l (g) = 242.8 k J m o l^{- 1}

Ionisation energy of

C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}

Electron gain enthalpy of

2 C l \to 2 C l^{- 1} = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}

Enthalpy of formation of

C a C l_{2} = - 795 k J m o l^{- 1}

Q. Calculate the lattice enthalpy(nearest integer value) in

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

L i F

, given that enthalpy of,

(i) sublimation of lithium is

155.3 k J

{m o l}^{- 1}

;

(ii) dissociation of

1 / 2

mole of

F_{2}

75.3 k J

;

(iii) ionization enthalpy of lithium is

520 k J

{m o l}^{- 1}

;

(iv) electron gain enthalpy of 1 mole of

F (g)

- 333 k J

;

(v)

Δ_{f} H

overall is

- 594 k J

{m o l}^{- 1}

Use the following data to calculate $Δ_{l a t t i c} H^{⊖}$ for NaBr. $Δ_{s u b} H^{⊖}$ for sodium metal = 108.4 kJ $m o l^{- 1}$ , ionisation enthalpy of sodium = 496 kJ $m o l^{- 1}$ , electron gain enthalpy of bromine = -325 kJ $m o l^{- 1}$ , bond dissociation enthalpy of bromine = 192 kJ $m o l^{- 1}$ , $Δ_{f} H^{⊖}$ for NaBr(s) = -360. 1 kJ $m o l^{- 1}$

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Enthalpy

Standard XII Chemistry

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