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Byju's Answer

Standard XII

Chemistry

Hess' Law

The lattice e...

Question

The lattice energy of NaCl(s) using the following data will be:
heat of sublimation of $N a (s) = S$
$(I E)_{1}$ of $N a (g) = I$
bond dissociation energy of $C l_{2} (g) = D$
electron affinity of $C l (g) = - E$
heat of formation of $N a C l (s) = - Q$

Lattice energy

- U = S + I + \frac{D}{2} - E - Q

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Lattice energy

- U = S - I + \frac{D}{2} - E - Q

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Lattice energy

- U = S + I + \frac{D}{2} + E - Q

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Lattice energy

- U = S - I - \frac{D}{2} + E + Q

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Solution

The correct option is A Lattice energy $- U = S + I + \frac{D}{2} - E - Q$
An estimate of the strength of the bonds in an ionic compound can be obtained by measuring the lattice energy of the compound, which is the energy given off when oppositely charged ions in the gas phase come together to form a solid.
The Lattice energy of NaCl from its elements Sodium and Chlorine in their stable forms is modeled in five steps in the diagram:
Enthalpy change of atomization enthalpy of lithium
Ionization enthalpy of lithium
Atomization enthalpy of fluorine
Electron affinity of fluorine
Lattice enthalpy
$- U = S + I + \frac{D}{2} - E - Q$

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Similar questions

Q. The correct folrmula of Q by using Born-Haber cycle for the preparation of

M g B r_{2} (s)

will be:
S

=

heat of sublimation of Mg(s)
I

=

ionisation energy of Mg(g)
D

=

dissociation energy of Br

_{2}

(g)
-E

=

electron affinity of Br(g)
-U

=

lattice energy of MgBr

_{2}

(s)
Q

=

heat of formation of MgBr

_{2}

(s)

Q. According to Born-Haber's cycle, the enthalpy of formation of ionic compounds can be determined. The formation of NaCl involves following steps –
(i)

N a (s) S - ------------ \to Energy of subtimation N a (g)

(ii)

N a (g) I - ---------- \to Ionisation potenital N a^{+} (g) + e^{-}

(iii)

C l_{2} (g) D - -------------- \to Bond dissociation energy 2 C l (g)

(iv)

C l (g) + e^{-} E - --------- \to Electron - affinity C l^{-} (g)

(v)

C l^{-} (g) + N a^{+} (g) U - ------- \to Lattice energy N a C l (s)

The enthalpy of formation of NaCl will be -

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 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. 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

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The lattice energy of NaCl(s) using the following data will be:heat of sublimation of Na(s)=S(IE)1 of Na(g)=Ibond dissociation energy of Cl2(g)=Delectron affinity of Cl(g)=−Eheat of formation of NaCl(s)=−Q

The lattice energy of NaCl(s) using the following data will be:
heat of sublimation of $N a (s) = S$
$(I E)_{1}$ of $N a (g) = I$
bond dissociation energy of $C l_{2} (g) = D$
electron affinity of $C l (g) = - E$
heat of formation of $N a C l (s) = - Q$