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Standard XII

Chemistry

Hess' Law

Calculate the...

Question

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

719 k J m o l^{- 1}

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

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

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832 k J m o l^{- 1}

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Solution

The correct option is B $- 719 k J m o l^{- 1}$
$K (s) \to K (g); Δ H_{s u b} = 89 k J m o l^{- 1}$

$\frac{1}{2} C l_{2} (g) \to C l (g); \frac{B D E}{2} = \frac{244}{2} = 122 k J m o l^{- 1}$

$K (g) \to K^{+} (g); I E = 425 k J m o l^{- 1}$

$C l (g \to C l^{-} (g); E_{A} = - 355 k J m o l^{- 1}$

$K (s) + \frac{1}{2} C l_{2} (g) \to K C l (s); Δ H_{f}^{0} = - 438 k J m o l^{- 1}$

According to Born-Haber cycle
$H_{f}^{0} = Δ H_{s u b} + \frac{B D E}{2} + I E + E A + U$
$- 438 = 89 + 122 + 425 - 355 + U$
$U = - 719 k J m o l^{- 1}$

hence, lattice energy of $K C l$ = $- 719 k J m o l^{- 1}$

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Similar questions

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. Calculate the enthalpy of formation of

M g F_{2}

from the following 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. Calculate the lattice enthalpy in

k J m o l^{- 1}

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}

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Hess' Law

Standard XII Chemistry

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Calculate the lattice enthalpy of KCl from the following data at standard states- Enthalpy of sublimation of K=89 kJmol−1 Enthalpy of dissociation of Cl2=244 kJmol−1 Ionisation energy of K=425 kJmol−1 Electron gain enthalpy of Cl=−355 kJmol−1 Enthalpy of formation of KCl=−438 kJmol−1