Calculate equilibrium constant (in multiples of `10^(-80)`) when sodium reduces Aluminium oxide to aluminium at 298 K. `Delta G^(@)_(f)` of `Na_(2)O_(3(s))` at `298 K = -377 KJ "mole"^(-1)` and `Delta G^(@)_(f)` of `Al_(2)O_(3)` at `298K = -1582 KJ "mole"^(-1)`)

Answer

9

Explanation

The given reaction is
`6Na + Al_(2)O_(3) hArr 3Na_(2) O + 2Al`
So `Delta G` of this reaction
`3(-377) - (-1582) = 451kJ`
We know that `Delta G = -2.303 RT log k`
log `= 79.04 rArr k = 9 xx 10^(-80)`