CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Enthalpy

Calculate the...

Question

Calculate the lattice enthalpy in $k J m o l^{- 1}$ of $L i F$ 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 $F (g) = - 333 k J Δ_{f} H_{o v e r a l l} = - 594 k J m o l^{- 1}$

A

- 1011.6 {kJ mol}^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

- 1200.4 {kJ mol}^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

- 984.6 {kJ mol}^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

- 1488.6 {kJ mol}^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 1011.6 {kJ mol}^{- 1}$
Given,
$L i (s) \to L i (g); Δ H_{S u b}^{o} = + 155.3 k J m o l^{- 1} L i (g) \to L i^{+}; I E = 520 k J m o l^{- 1} \frac{1}{2} F_{2} (g) \to F (g); Δ H_{F - F}^{o} = 75.3 k J m o l^{- 1} F (g) + e^{- 1} \to F^{1 -}; Δ H_{e g} = - 141 k J m o l^{- 1} Δ_{f} H_{o v e r a l l} = - 594 k J m o l^{- 1} F o r L i F; Δ H_{f} = Δ H_{S u b}^{o} + \frac{1}{2} Δ H_{F - F} + Δ H_{i} + Δ H_{e g} + Δ H_{L a t t i c e} - 594.1 = 155.2 + 75.3 + 520 - 333 + Δ H_{L a t t i c e} Δ H_{L a t t i c e} = - 1011.6 k J m o l^{- 1}$
Hence, the lattice enthalpy of LiF is given by $- 1011.6 k J m o l^{- 1}$ .

Suggest Corrections

thumbs-up

4

Similar questions

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

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

;

Δ_{f} H

- 594 k J

{m o l}^{- 1}

Q. Calculate the lattice formation enthalpy in

{kJ mol}^{- 1}

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}

Q. Calculate the lattice enthalpy of

K C l

from the following data at standard states-
Enthalpy of sublimation of

K = 89 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} = 244 k J m o l^{- 1}

Ionisation energy of

K = 425 k J m o l^{- 1}

Electron gain enthalpy of

C l = - 355 k J m o l^{- 1}

Enthalpy of formation of

K C l = - 438 k J m o l^{- 1}

Q. Calculate the lattice enthalpy of

K C l

from the following data at standard states-
Enthalpy of sublimation of

K = 89 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} = 244 k J m o l^{- 1}

Ionisation energy of

K = 425 k J m o l^{- 1}

Electron gain enthalpy of

C l = - 355 k J m o l^{- 1}

Enthalpy of formation of

K C l = - 438 k J m o l^{- 1}

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}

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon