Calculate the nbsp- enthalpy of nbsp- formation of MgF-2 from thefollowing data- Enthalpy of sublimation for Mg-s- - Mg-g- - 146-4 -kJmol-1 nbsp- Enthalpy of dissociation of F-2-g- - 2F -g- - 155-8-kJmol-1 Ionisation energy of Mg-g- - Mg-2-g- - 2186 -kJmol-1 Electron gain enthalpy of 2F - 2F-1- 2-322-6 - -645-2 -kJmol-1 Lattice enthalpy of MgF-2 - -2922-5 -kJmol-1 nbsp-1079-5- kJmol-12200- kJmol-11099-5- kJmol-12200- 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^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1099.5 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2200 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2200 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $M g F_{2} = - 1079.5 k J m o l^{- 1}$

Suggest Corrections

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

M g^{+ 2}

is +2170

k J m o l^{- 1}

Dissociation energy of

O_{2 (g)}

is +494

k J m o l^{- 1}

Heat of sublimation of

M g_{(s)}

is +150

k J m o l^{- 1}

Join BYJU'S Learning Program

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program