CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

Use the follo...

Question

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 $M g O = - 3791 k J / m o l$

A

601.7 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

744.4 kJ/mol

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

1346.1 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

147.7 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $744.4 kJ/mol$
Given :
$△_{s u b} H_{M g} = 147.7 k J m o l^{- 1}$
$△_{d i s s} H_{O_{2}} = 498.4 k J m o l^{- 1}$
but we require bond dissociation energy for
$\frac{1}{2} O_{2} (g) \to O (g)$
so $= 498.4 / 2 = 249.2 k J / m o l$

$△_{I E} H_{M g} = 2189 k J m o l^{- 1}$

First electron affinity $△_{E A} H_{O} = - 141 k J m o l^{- 1}$

$△_{f} H_{M g O} = - 601.7 k J m o l^{- 1}$
$△_{l} H_{M g O} = - 3791 k J m o l^{-}$

to find second electron gain enthalpy of $O$ i.e:
$O^{-} (g) + e^{-} (g) \to O^{2 -} (g)$ $Δ H = x$

By Born Haber Cycle (Based on Hess's Law) :
$- 601.7 = - 3791 + 147.7 + 249.2 + 2189 - 141 + x$
$x = 744.4 k J / m o l$

Suggest Corrections

thumbs-up

2

Similar questions

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)

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. Calculate the electron affinity of oxygen atom in gaining two electrons to give the oxide ion

O^{2 -}

from the following data.
Standard heat of formation of

M g O_{(s)} = - 600 k J m o l^{- 1}

Lattice energy of

M g O_{(s)} = - 3860 k J m o l^{- 1}

lonization energy of

M g_{(g)}

M g^{2 +}

k J m o l^{- 1}

Dissociation energy of

O_{2 (g)}

k J m o l^{- 1}

Heat of sublimation of

M g_{(s)}

k J m o l^{- 1}

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 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^{-}

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

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon