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Calculate equilibrium constant (in natural number multiple of $10^{- 80}$ ) when sodium reduces Aluminium oxide to aluminium at 298K. $Δ G_{f}^{0}$ of $N a_{2} O_{(s)}$ at $298 K = - 377 K J m o l e^{- 1}$ and $Δ G_{f}^{0} o f A l_{2} O_{3}$ at $298 K = - 1582 K J m o l e^{- 1}$

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Detailed Solution

The given reaction is
$6 N a + A l_{2} O_{3} ⇌ 3 N a_{2} O + 2 A l$
So $Δ G$ of this reaction
$3 (- 377) - (- 1582) = 451 k J$
We know that $Δ G = - 2.303$ RT log k
$\log k = 79.04 \Rightarrow k = 9 \times 10^{- 80}$

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