f the enthalpy of sublimation of Li is 155 kJ enthalpy of dissociation of is , ionization enthalpy of Li is electron gain enthalpy of F is standard enthalpy of formation of LiF is The magnitude of lattice enthalpy of Li

Question

If the enthalpy of sublimation of Li is 155 kJ ${mol}^{- 1},$ enthalpy of dissociation of $F_{2}$ is ${150 kJ mol}^{- 1}$ , ionization enthalpy of Li is ${520 kJ mol}^{- 1},$ electron gain enthalpy of F is ${−313 kJ mol}^{- 1},$ standard enthalpy of formation of LiF is ${−594 kJ mol}^{- 1} .$ The magnitude of lattice enthalpy of LiF is __________ ${kJ mol}^{- 1} .$ (Nearest Integer) [2026]

Smart Achievers · Accepted Answer

(1031)

$- 594 = 155 + 520 + \frac{150}{2} - 313 + (L.E.)$

$L.E. = - 1031 {kJ mol}^{- 1}$

Solution

Ans.
(1031)

$- 594 = 155 + 520 + \frac{150}{2} - 313 + (L.E.)$

$L.E. = - 1031 {kJ mol}^{- 1}$

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Solution

Ans. (1031) -594=155+520+1502-313+(L.E.) L.E.=-1031 kJ mol-1

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Ans.
(1031)

$- 594 = 155 + 520 + \frac{150}{2} - 313 + (L.E.)$

$L.E. = - 1031 {kJ mol}^{- 1}$