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Standard XII

Chemistry

Examples of Ionic Compounds

Caesium chlor...

Question

Caesium chlorides is formed according to the following equation $C s (s) + 0.5 C l_{2} (g) \to C s C k (s)$ . The enthalpy of sublimation of $C s$ , enthalpy of dissociation of chlorine, ionization energy of $C s$ and electron affinity of chlorine are $81.2, 243.0, 375.7$ and $- 348.3 k J m o l^{- 1}$ . The energy change involved in the formation of $C s C l$ is $- 388.6 K J m o l^{- 1}$ . Calculate the lattice energy of $C s C l$ .

618.7 k J m o l^{- 1}

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1237.4 k J m o l^{- 1}

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- 1237.4 k J m o l^{- 1}

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

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Solution

The correct option is A $618.7 k J m o l^{- 1}$
$△ H_{f} = △ H_{s u b} + I . E + \frac{1}{2} △ H_{d i s s} + E . A + μ$
$\Rightarrow - 388.6 = 81.2 + 375.7 + \frac{243}{2} + (- 348.3) + μ$
$\Rightarrow μ = 618.7 k J m o l^{- 1}$

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Similar questions

Q. Cesium chlorides is formed according to the following equation

C s (s) + 0.5 C l_{2} (g) \to C s C l (s)

. The enthalpy of sublimation of Cs, the enthalpy of dissociation of chlorine, ionization energy of

C s

and electron affinity of cholrine are

81.2, 243.0, 375.7

and

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

. The energy involved in the formation of

C s C l

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

. Calculate the lattice energy of

C s C l

Q. Caesium chloride is formed according to the following equation:

C s (s) + 0.5 {C l}_{2} (g) ⟶ C s C l (s)

The enthalpy of sublimation of

C s

, enthalpy of dissociation of chlorine, ionization energy of

C s

and electron affinity of chlorine are

81.2, 243.0, 375.7

and

- 348.3 k J

{o l}^{- 1}

. The energy change involved in the formation of

C s C l

388.6 k J . {m o l}^{- 1}

. Calculate the lattice energy of

C s C l

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

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

= - 233 kJ/mol

Enthalpy of sublimation of

C s = + 78 kJ/mol

First ionization energy of

C s = + 375 kJ/mol

Enthalpy of dissociation of

O_{2} (g) = 494 kJ/mol

First electron affinity of

O = - 141 kJ/mol

Second electron affinity of

O = + 845 kJ/mol

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

- 233 k J / m o l

Enthalpy of sublimation of

C s = + 78 k J / m o l

First ionization energy of

C s = + 375 k J / m o l

Enthalpy of dissociation of

O_{2} (g) = 494 k J / m o l

O_{2}

molecules
First electron affinity of

O = - 141 k J / m o l

of O atoms
Second electron affinity of

O = + 845 k J / m o l

O^{-}

ions

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The correct option is A 618.7 kJ mol−1△Hf=△Hsub+I.E+12△Hdiss+E.A+μ⇒−388.6=81.2+375.7+2432+(−348.3)+μ⇒μ=618.7kJmol−1

The correct option is A $618.7 k J m o l^{- 1}$
$△ H_{f} = △ H_{s u b} + I . E + \frac{1}{2} △ H_{d i s s} + E . A + μ$
$\Rightarrow - 388.6 = 81.2 + 375.7 + \frac{243}{2} + (- 348.3) + μ$
$\Rightarrow μ = 618.7 k J m o l^{- 1}$