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Covalent Bond vs Ionic Bond

Use the follo...

Question

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 $O_{2} (g) = 494 k J / m o l$ of $O_{2}$ molecules
First electron affinity of $O = - 141 k J / m o l$ of O atoms
Second electron affinity of $O = + 845 k J / m o l$ of $O^{-}$ ions

- 2090 k J / m o l

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2090 k J / m o l

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- 2290 k J / m o l

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- 2390 k J / m o l

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Solution

The correct option is A $- 2090 k J / m o l$
Given,
$2 C s (s) \to 2 C s (g); Δ_{S u b} H^{o} = (+ 78 \times 2) k J m o l^{- 1}$
$2 C s (g) \to 2 C s^{+}; I E = (2 \times + 375) k J m o l^{- 1}$
$\frac{1}{2} O_{2} (g) \to O (g); Δ_{O - O} H^{o} = (494 \times \frac{1}{2}) k J m o l^{- 1}$
$O + e^{- 1} \to O^{1 -}; E_{A 1} = - 141 k J m o l^{- 1}$
$O^{1 -} + e^{- 1} \to O^{2 -}; E_{A 2} = + 845 k J m o l^{- 1}$
$2 C s^{+} (g) + O_{2} (g) \to C s_{2} O (s); Δ H^{0} = - 233 k J m o l^{- 1}$
Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber Cycle .
Let us now calculate the lattice enthalpy of $C s_{2} O$ (s) by following steps given below :
$Δ_{f} H^{0} = Δ_{S u b} H^{o} + I E + Δ_{O - O} H^{o} + E_{A 1} + E_{A 2} + Δ_{l a t t i c e} H^{o}$
$- 233 = 2 (78) + 2 (375) + \frac{494}{2} - 141 + 845 + Δ_{l a t t i c e} H^{o}$
$Δ_{l a t t i c e} H^{o} = - 2090 k J m o l^{- 1}$
the lattice energy of cesium oxide is $- 2090 k J m o l^{- 1}$

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Similar questions

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 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 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 kJ/mol

Ionisation energy of

M g (g)

to form

M g^{2 +} (g) = 2189 kJ/mol

Bond dissociation energy for

O_{2}

= 498.4 kJ/mol

First electron affinity of

O (g) = - 141 kJ/mol

Heat formation of

M g O (s) = - 601.7 kJ/mol

lattice energy of

M g O = - 3791 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. 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

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Covalent Bond vs Ionic Bond

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