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Examples of Ionic Compounds

Which of the ...

Question

Which of the following reaction involves the liberation of energy?

A

N a_{(s)} \to N a_{(g)}^{+}

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B

C l_{2 (g)} \to 2 C l_{(g)}

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C

N a_{(g)}^{+} + C l_{(g)}^{-} \to N a C l_{(s)}

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D

N a C l_{(s)} \to N a_{(g)}^{+} + C l_{(g)}^{-}

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Solution

The correct option is C $N a_{(g)}^{+} + C l_{(g)}^{-} \to N a C l_{(s)}$
(A) $N a C O \to N a C O^{+}$
This is equivalent to
$N a (s) - -- \to {Δ H}_{1} N a (g) - -- \to {Δ H}_{2} N a^{+} (8)$
$Δ H_{1} =$ Heat of cultivation
$Δ H_{2} =$ Ionisation energy
Thus in this case net energy is absorbed

(B) $C l_{2} (q) \to 2 C l (g)$
This reaction involves dissociation of $C l_{2}$ molecules in chlorine atoms
Thus Net energy is required (for bond breaking)

(C) $N a^{+} (g) + C l (g) \to N a C l (s)$
This represent the formation of an ionic lattice from the gaseous isolated sodium and chloride ions
Since lattice energy of formation is negative
Net energy is liberated

(D) $N a C l (s) \to N a^{+} (g) + C l^{\oplus} (g)$
Thus involves breaking the lattice crystal which simply requires energy and also lattice energy of acisociation is positive
Thus Net energy is absorbed.

$∴$ Correct Ans $(C)$

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Similar questions

N a C l (s) + a q ⟶ N a^{\oplus} (a q) + C l^{⊖}

Δ H = 3.9 k J

N a^{\oplus} (g) + C l^{⊖} (g) ⟶ N a C l (s) Δ H = - 788 k J

C l^{⊖} (g) + a q ⟶ C l^{⊖} (a q) Δ H = - 394.1 k J

Then the enthalpy of hydration of

N a^{\oplus}

Q. How much heat is liberated when one mole of gaseous

N a^{\oplus}

combines with one mole of

C l^{⊖}

ion to form solid

N a C l

Use the data given below:

N a (s) + \frac{1}{2} C l_{2} (g) ⟶ N a C l (s)

Δ H = - 98.23 k c a l

N a (s) ⟶ N a (g)

Δ H = + 25.98 k c a l

N a (g) ⟶ N a^{\oplus} + e^{-}

Δ H = + 120.0 k c a l

C l_{2} (g) ⟶ 2 C l (g)

Δ H = + 58.02 k c a l

C l^{⊖} (g) ⟶ C l (g) + e^{-}

Δ H = + 87.3 k c a l

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)

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

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

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

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

The enthalpy of formation of NaCl will be -

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

N a C l_{(s)} ⟶ N a^{+} (g) + C l^{-} (g)

The enthalpy of above reaction is known as:

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