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Ways to Define Concentration

If X is the l...

Question

If X is the lattice entahlpy of $C a {C l}_{2}$ ,find $X$ . Given that the enthalpy of
(i) sublimation of $C a$ is $121 k J {m o l}^{- 1}$
(ii) dissociation of ${C l}_{2}$ to $C l$ is $242.8 k J {m o l}^{- 1}$ ;
(iii) ionization of $C a$ to ${C a}^{2 +}$ is $2422 k J {m o l}^{- 1}$ ;
(iv) electron gain enthalpy for $C l$ to ${C l}^{-}$ is $- 355 k J {m o l}^{- 1}$ ;
(iv) $Δ_{f} H$ overall is $- 795 k J {m o l}^{- 1}$ .

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Solution

Given:
$Δ_{s u b} H_{C a} = 121 k J {m o l}^{- 1}$
$Δ H_{C l - C l} = 242.8 k J {m o l}^{- 1}$
$Δ_{f} H_{{C a}^{2 +}} = 2422 k J {m o l}^{- 1}$
$Δ_{e g} H_{C l} = - 355 k J {m o l}^{- 1}$
$Δ_{f} H_{C a {C l}_{2}} = - 795 k J {m o l}^{- 1}$
$Δ_{f} H_{C a {C l}_{2}} = Δ_{s u b} H_{C a} + Δ H_{C l - C l} + Δ_{f} H_{{C a}^{2 +}} + 2 \times Δ_{e g} H_{C l} + Δ_{l} H_{C a {C l}_{2}}$
or $- 795 = 121 + 242.8 + 2422 - (355 \times 2) + Δ_{l} H_{C a {C l}_{2}}$
$∴$ $Δ_{l} H_{C a {C l}_{2}} = - 2870.8 k J$ ${m o l}^{- 1}$

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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}

△_{f} H

- 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}

△_{f} H

- 795 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 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 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}

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