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))

Calculate equilibrium constant (in multiples of 10^(-80)) when sodium

1.	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» Solution :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)`