Generally alkali metals form their oxides by an average increase of 6 kJ mol -1 K -1 in their entropies-If at 25- C - G f 0 of Al 2 O 3 s - -1582 kJ mol -1 - H f o of Li 2 O s -1244 kJ - mol and H f o of N a 2 O s -1411 kJ - mol- The aluminium oxide can be reduced to Almetal by which metal-A- NaB- noneC- LiD- both

The correct option is C LiFor Li_2O, 2Li(s) +12O_2(g)→ Li_2O(s) G^o_f (Li_2O) = H^o_f T . S^o = 1244 (298 × 6) = 544 kJ mol^ 1 For Na_2O, 2Na(s) + 12O ...

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Standard XII

Chemistry

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.

N a

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L i

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both

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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. The oxides of metals which are least reactive can be reduced to their respective metals by simple heating.

Q. The oxides of active metals like potassium, sodium, aluminium, calcium and magnesium do not get reduced by hydrogen into their corresponding metals.

Q. Lithium forms

L i_{2} O

while other alkali metals form peroxides and super oxides because:

Q. Alkali metals cannot be extracted by reduction of their oxides and other compounds because:

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Entropy

Standard XII Chemistry

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Generally alkali metals form their oxides by an average increase of 6 kJ mol−1K−1 in their entropies.If at 25oC,△Gof of Al2O3(s)=−1582 kJ mol−1; △Hof of Li2O(s)=1244 kJ/mol and △Hof of Na2O(s)=1411 kJ/mol. The aluminium oxide can be reduced to Al metal by which metal.