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Standard XII

Chemistry

Covalent Bond vs Ionic Bond

Calculate the...

Question

Calculate the lattice energy of a salt $M X_{(s)}$ from the data given below:
Heat of formation of $M X (Δ H) = - 550 kJ/mol$
Heat of sublimation of $M (S) = 80 kJ/mol$
Heat of dissociation of $X_{2} (D) = 155 kJ/mol$
Ionization energy of $M (I) = 347 kJ/mol$
Electron affinity of $X (E) = - 343 kJ/mol$

- 838 kJ/mol

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- 938.5 kJ/mol

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- 711.5 kJ/mol

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- 638.5 kJ/mol

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Solution

The correct option is C $- 711.5 kJ/mol$

$Δ H_{f} = S . E + I . E + \frac{D E}{2} - E A - L E$
$- 550 = 80 + 347 + \frac{155}{2} - 343 - L . E$
L.E = -711.5 kJ/mol
Hence, the option (c) is correct.

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Similar questions

Q. Calculate the lattice energy of a salt

M X_{(s)}

from the data given below:
Heat of formation of

M X (Δ H) = - 550 kJ/mol

Heat of sublimation of

M (S) = 80 kJ/mol

Heat of dissociation of

X_{2} (D) = 155 kJ/mol

Ionization energy of

M (I) = 347 kJ/mol

Electron affinity of

X (E) = - 343 kJ/mol

Q. Calculate the lattice energy of a salt

M X_{(s)}

from the date given below:
Heat of formation of

M X

(Δ H)

= - 550

kJ/mol
Heat of sublimation of

M (S) = 80

kJ/mol
Heat of dissociation of

X_{2}

(D) = 155

kJ/mol
Ionization energy of

M (I E) = 347

kJ/mol
Electron affinity of

X (E A) = - 343

kJ/mol

Q. Use the following data to calculate second electron affinity of oxygen for the process

O^{-} (g) + e^{-} (g) \to O^{2 -} (g)

.
Heat of sublimation of

M g (s) = 147.7 k J / m o l

Ionisation energy of

M g (g)

to form

M g^{2 +} (g) = 2189 k J / m o l

Bond dissociation energy for

O_{2}

= 498.4 k J / m o l

First electron affinity of

O (g) = - 141 k J / m o l

Heat formation of

M g O (s) = - 601.7 k J / m o l

lattice energy of

M g O = - 3791 k J / m o l

Q. Given sublimation and ionization energy of Na are 108 kJ/mol and 496 kJ/mol respectively and bond dissociation energy required to make one mole chlorine atoms and its electron affinity energy are 122 kJ/mol and -349 kJ/mol.If

△ H_{f}^{0}

of NaCl(s) is -411 kJ/mol.What is its approximate lattice formation enthalpy?

Q. Use the following data to calculate the lattice energy of caesium oxide.
Enthalpy of formation of caesium oxide =

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

Enthalpy of sublimation of

C s = + 78 k J m o l^{- 1}

First ionization energy of

C s = + 375 k J m o l^{- 1}

Enthalpy of dissociation of

O_{2} (g) = 494 k J m o l^{- 1} o f O_{2} m o l e c u l e s

First electron affinity of

O = - 141 k J m o l^{- 1} o f O a t o m s

Second electron affinity of

O = + 845 k J m o l^{- 1} o f O^{1 -} i o n s

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Covalent Bond vs Ionic Bond

Standard XII Chemistry

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Calculate the lattice energy of a salt MX(s) from the data given below: Heat of formation of MX(ΔH)=−550 kJ/mol Heat of sublimation of M(S)=80 kJ/mol Heat of dissociation of X2(D)=155 kJ/mol Ionization energy of M(I)=347 kJ/mol Electron affinity of X(E)=−343 kJ/mol

The correct option is C −711.5 kJ/mol ΔHf=S.E+I.E+DE2−EA−LE −550=80+347+1552−343−L.E L.E = -711.5 kJ/mol Hence, the option (c) is correct.

The correct option is C $- 711.5 kJ/mol$

$Δ H_{f} = S . E + I . E + \frac{D E}{2} - E A - L E$
$- 550 = 80 + 347 + \frac{155}{2} - 343 - L . E$
L.E = -711.5 kJ/mol
Hence, the option (c) is correct.