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Entropy

Generally alk...

Question

Generally alkali metals form their oxides by an average increase of $6 k J m o l^{- 1} K^{- 1}$ in their entropies.If at $25^{o} C, △ G_{f}^{o} o f A l_{2} O_{3} (s) = - 1582 k J m o l^{- 1}; △ H_{f}^{o} o f L i_{2} O (s) = 1244 k J / m o l a n d △ H_{f}^{o}$ of $N a_{2} O (s) = 1411 k J / m o l .$ The aluminium oxide can be reduced to Al metal by which metal.

A

N a

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B

L i

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C

both

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D

none

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Solution

The correct option is B $L i$
For $L i_{2} O,$
$2 L i (s) + \frac{1}{2} O_{2} (g) \to L i_{2} O (s)$
$△ G_{f}^{o} (L i_{2} O) = △ H_{f}^{o} - T . △ S^{o}$
$= 1244 - (298 \times 6)$
$= - 544 k J m o l^{- 1}$

For $N a_{2} O,$
$2 N a (s) + \frac{1}{2} O_{2} (g) \to N a_{2} O (s)$
$△ G_{f}^{o} (N a_{2} O) = △ H_{f}^{o} - T . △ S^{o}$
$= 1411 - (298 \times 6)$
$= - 377 k J m o l^{- 1}$

Now for reaction
$A l_{2} O_{3} (s) + 6 N a (s) \to 3 N a_{2} O (s) + 2 A l (s)$
$△ G^{o} = 3 \times △ G_{f}^{o} (N a_{2} O) - △ G_{f}^{o} (A l_{2} O_{3})$
$= 3 \times (- 377) - (- 1582)$
$= + 451 k J m o l^{- 1}$
Similarly, when $L i$ is used
$△ G^{o} = 3 \times △ G_{f}^{o} (L i_{2} O) - △ G_{f}^{o} (A l_{2} O_{3})$
$= 3 \times (- 544) - (- 1582)$
$= - 1632 + 1582 = - 50 k J m o l^{- 1}$
$△ G^{o}$ is negative when Li is used for the reduction, so the reaction is feasible.

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Similar questions

Q. Generally alkali metals form their oxides by an average increase of

6 k J m o l^{- 1} K^{- 1}

in their entropies.If at

25^{o} C, △ G_{f}^{o} o f A l_{2} O_{3} (s) = - 1582 k J m o l^{- 1}; △ H_{f}^{o} o f L i_{2} O (s) = 1244 k J / m o l a n d △ H_{f}^{o}

N a_{2} O (s) = 1411 k J / m o l .

The aluminium oxide can be reduced to Al metal by which metal.

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. What is the heat of formation of liquid methyl alcohol in

k J m o l^{- 1},

use the following data.
Heat of vaporization of liquid methyl alcohol =

38 k J m o l^{- 1}

.
Heat of formation of gaseous atoms from the elements in their standard states:

H = 218 k J m o l^{- 1}, C = 715 k J m o l^{- 1}, O = 249 k J m o l^{- 1}

Average bond energies :

C - H = 415 k J m o l^{- 1} C - O = 356 k J m o l^{- 1} O - H = 463 k J m o l^{- 1}

Q. What is the heat of formation of liquid methyl alcohol in

k J m o l^{- 1},

use the following data.
Heat of vaporization of liquid methyl alcohol =

38 k J m o l^{- 1}

.
Heat of formation of gaseous atoms from the elements in their standard states:

H = 218 k J m o l^{- 1}, C = 715 k J m o l^{- 1}, O = 249 k J m o l^{- 1}

Average bond energies :

C - H = 415 k J m o l^{- 1} C - O = 356 k J m o l^{- 1} O - H = 463 k J m o l^{- 1}

Q. Compute the heat of formation of liquid methyl alcohol (in kJ mol

^{- 1}

) using the following data. The heat of vaporisation of liquid methyl alcohol is 38 kJ mol

^{- 1}

The heat of formation of gaseous atoms from the elements in their standard states:

H =

^{- 1}

C = 715

^{- 1}

O = 249

^{- 1}

Average bond energies:

C - H = 415

^{- 1}

C - O = 356

^{- 1}

O - H = 463

^{- 1}

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