Calculate the enthalpy of formation of MgF 2 from thefollowing dataEnthalpy of sublimation for Mg s - Mg g -146-4 kJ mol -1Enthalpy of dissociation of F 2 g - 2 F g -155-8 kJ mol -1Ionisation energy of Mg g - Mg 2- g -2186 kJmol -1Electron gain enthalpy of 2 F - 2 F -1-2 -322-6-645-2 kJ mol -1Lattice enthalpy of MgF 2-2922-5 kJmol -1A- -1079-5 kJ mol -1B- 2200 kJmol -1C- 1099-5 kJ mol -1D- 2200 kJmol -1

The correct option is A 1079.5 kJmol^ 1Mg(s) → Mg(g); Δ H_sub = 146.4 kJmol^ 1 F_2(g) → 2F(g); BDE = 155.8 kJmol^ 1 Mg(g) → Mg^2+(g) ; IE = 2186 kJmol^ ...

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Standard XII

Chemistry

Heat of Formation

Calculate the...

Question

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

- 1079.5 k J m o l^{- 1}

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Solution

The correct option is A $- 1079.5 k J m o l^{- 1}$
$M g (s) \to M g (g); Δ H_{s u b} = 146.4 k J m o l^{- 1}$

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

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

$2 F (g) + 2 e^{-} \to 2 F^{-} (g); E_{A} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}$

$M g (s) + F_{2} (g) \to M g F_{2} (s); Δ_{f} H = - 2922.5 k J m o l^{- 1}$

According to Born-Haber cycle
$Δ_{f} H = Δ H_{s u b} + B D E + I E + E_{A} + U$
$Δ_{f} H = 146.4 + 155.8 + 2186 - 645.2 - 2922.5$
$Δ_{f} H = - 1079.5 k J m o l^{- 1}$
hence, enthalpy of formation of $M g F_{2} = - 1079.5 k J m o l^{- 1}$

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Similar questions

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 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 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 electron affinity of oxygen atom in gaining two electrons to give the oxide ion

O^{- 2}

from the following data.
Standard heat of formation of

M g O_{(s)} = - 600 k J m o l^{- 1}

Lattice energy of

M g O_{(s)} = - 3860 k J m o l^{- 1}

lonization energy of

M g_{(g)}

to give

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is +2170

k J m o l^{- 1}

Dissociation energy of

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Heat of sublimation of

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Heat of Formation

Standard XII Chemistry

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