Calculate the lattice enthalpy of CaCl 2- given that -Enthalpy of sublimation for Ca s - Ca g -121 kJ mol -1Enthalpy of dissociation of Cl 2 g - 2 Cl g -242-8 kJ mol -1Ionisation energy of Ca g - Ca - g -2422 kJ mol -1Electron gain enthalpy of 2 Cl - 2 Cl -1 -2 -355 kJ mol -1-710 kJ mol -1Enthalpy of formation of CaCl 2-795 kJ mol -1A- -2670-8 kJ mol -1B- -2870-8 kJ mol -1C- -2770-8 kJ mol -1D- -2970-8 kJ mol -1

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Enthalpy

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Question

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

- 2670.8 k J m o l^{- 1}

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- 2770.8 k J m o l^{- 1}

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- 2870.8 k J m o l^{- 1}

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- 2970.8 k J m o l^{- 1}

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Solution

The correct option is C $- 2870.8 k J m o l^{- 1}$
Lattice energy can be Born Haber cycle if the enthalpies are known.

We are given,
$C a (s) \to C a (g); Δ_{s u b} H = 121 k J m o l^{- 1}$

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

$C a (g) \to C a^{+ +} (g) + 2 e^{- 1}; I E = 2422 k J m o l^{- 1}$

$2 C l (g) + 2 e^{- 1} \to 2 C l^{- 1} (g); Δ_{e g} H = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}$

$C a^{+ +} (g) + 2 C l^{- 1} (g) \to C a C l_{2} (s); Δ_{l a t t i c e} H = ? k J m o l^{- 1}$

$Δ H_{f}^{0} = - 795 k J m o l^{- 1}$

According to Born Haber cycle
$Δ H_{f}^{0} = Δ H_{s u b} + Δ_{C l - C l} H + I E + Δ_{e g} H + Δ_{l a t t i c e} H$

putting all the values,
$- 795 = 121 + 242.8 + 2422 - 710 + Δ_{l a t t i c e} H Δ_{l a t t i c e} H = - 2870.8 k J m o l^{- 1}$

the lattice enthalpy of $C a C l_{2} i s - 2870 k J m o l^{- 1}$

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Similar questions

Q. If X is the lattice enthalpy of

C a C l_{2}

, find X. Given that the enthalpy of
(i) Sublimation of

C a (s)

121 k J m o l^{- 1}

(ii) Dissociation of

C l_{2} (g)

2 C l (g)

242.8 k J m o l^{- 1}

(iii) Ionization of

C a (g)

C a^{2 +} (g)

2422 k J m o l^{- 1}

(iv) Electron gain enthalpy for

C l (g)

C l^{- 1} (g)

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

(v)

△_{f} H

overall is

- 795 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}

Q. Calculate the enthalpy of formation of

M g F_{2}

from thefollowing data-
Enthalpy of sublimation for

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

Enthalpy of dissociation of

F_{2} (g) \to 2 F (g) = 155.8 k J m o l^{- 1}

Ionisation energy of

M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

Electron gain enthalpy of

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

Lattice enthalpy of

M g F_{2} = - 2922.5 k J m o l^{- 1}

Q. If X is the lattice entahlpy of

C a {C l}_{2}

,find

X

. Given that the enthalpy of
(i) sublimation of

C a

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

(ii) dissociation of

{C l}_{2}

C l

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

;
(iii) ionization of

C a

{C a}^{2 +}

2422 k J {m o l}^{- 1}

;
(iv) electron gain enthalpy for

C l

{C l}^{-}

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

;
(iv)

Δ_{f} H

overall is

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

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Enthalpy

Standard XII Chemistry

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