NaCl s- Na - g - Cl - g The enthalpy of above reaction is known as-A- Enthalpy of hydrationB- Lattice enthalpyC- Enthalpy of solutionD- Enthalpy of dissociation

The correct option is B Lattice enthalpyNaCl(s) ⟶ Na^+ (g) + Cl^ (g) The energy required to dissociate nbsp;one mole of an ionic compound into gaseous ions ...

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Byju's Answer

Standard XII

Chemistry

Reactivity with Halogens and Hydrogen (Group 2)

NaCl s⟶ Na + ...

Question

$N a C l_{(s)} ⟶ N a^{+} (g) + C l^{-} (g)$
The enthalpy of above reaction is known as:

Enthalpy of hydration

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Lattice enthalpy

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Enthalpy of solution

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Enthalpy of dissociation

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Solution

The correct option is B Lattice enthalpy
$N a C l (s) ⟶ N a^{+} (g) + C l^{-} (g)$
The energy required to dissociate one mole of an ionic compound into gaseous ions is defined as lattice enthalpy.

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Similar questions

Q. The change in enthalpy for hydration of ( i ) the chloride ion ; ( ii ) the iodide ion are :
Given :
* enthalpy change of solution of NaCl(s)

= - 2

kJ / mol.
* enthalpy change of solution of NaI(s)

= + 2

kJ / mol.
* enthalpy change of hydration of

N a^{+} (g) = - 390

kJ / mol.
* lattice enthalpy of NaCl

= - 772

kJ / mol.
* lattice enthalpy of NaI

= - 699

kJ / mol.

Q. The lattice enthalpy of solid

N a C l

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

and enthalpy of solution is

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

. If the hydration enthalpy of

{N a}^{+}

and

{C l}^{-}

ions are in the ratio of

3 : 2.5

, what is the enthalpy of hydration of chloride ion?

Q. Given:

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}

ions is :

Q. Lattice energy of NaCl is

- 186

kcal/mol. If solution enthalpies of Na

+

and Cl

^{-}

are

- 97

kcal/mol and

- 85

kcal/mol respectively. The enthalpy NaCl solution in kcal will be:

Q. If enthalpy of a reaction is

1482 k J / m o l

and bond dissociation enthalpy of

(C - H)

bond is

415 k J / m o l

than bond dissociation enthalpy of

(C - C l)

bond is:

C H_{2} C l_{2 (g)} \to C_{(g)} + 2 H_{(g)} + 2 C l_{(g)}

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NaCl(s)⟶Na+(g)+Cl−(g) The enthalpy of above reaction is known as:

The correct option is B Lattice enthalpyNaCl(s)⟶Na+(g)+Cl−(g) The energy required to dissociate one mole of an ionic compound into gaseous ions is defined as lattice enthalpy.

$N a C l_{(s)} ⟶ N a^{+} (g) + C l^{-} (g)$
The enthalpy of above reaction is known as:

The correct option is B Lattice enthalpy
$N a C l (s) ⟶ N a^{+} (g) + C l^{-} (g)$
The energy required to dissociate one mole of an ionic compound into gaseous ions is defined as lattice enthalpy.