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Standard XII

Chemistry

Hess' Law

If X is the l...

Question

If X is the lattice enthalpy of $C a C l_{2}$ , find X. Given that the enthalpy of
(i) Sublimation of $C a (s)$ is $121 k J m o l^{- 1}$
(ii) Dissociation of $C l_{2} (g)$ to $2 C l (g)$ is $242.8 k J m o l^{- 1}$
(iii) Ionization of $C a (g)$ to $C a^{2 +} (g)$ is $2422 k J m o l^{- 1}$
(iv) Electron gain enthalpy for $C l (g)$ to $C l^{- 1} (g)$ is $- 355 k J m o l^{- 1}$
(v) $△_{f} H$ overall is $- 795 k J m o l^{- 1}$

2870.8 kJ/mol

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-2372.6 kJ/mol

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-2870.8 kJ/mol

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2372.6 kJ/mol

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Solution

The correct option is C -2870.8 kJ/mol
Given :
$△_{s u b} H_{C a} = 121 k J m o l^{- 1}$
$△_{d i s s} H_{C l_{2}} = 242.8 k J m o l^{- 1}$
$△_{I E} H_{C a} = 2422 k J m o l^{- 1}$
$△_{E G} H_{C l} = - 355 k J m o l^{- 1}$
$△_{f} H_{C a C l_{2}} = - 795 k J m o l^{- 1}$
$△_{l} H_{C a C l_{2}} =$ Lattice energy

using Hess's law :

$△_{f} H_{C a C l_{2}} = △_{s u b} H_{C a} + △_{d i s s} H_{C l - C l} + △_{I E} H_{C a} + 2 \times △_{E G} H_{C l}$ + $△_{l} H_{C a C l_{2}}$
$- 795 = 121 + 242.8 + 2422 - (355 \times 2) + △_{l} H_{C a C l_{2}}$
Thus,
$△_{l} H_{C a C l_{2}}$ = $- 2870.8 k J / m o l$

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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. 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. 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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Hess' Law

Standard XII Chemistry

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