Question

Aluminum is one of the most versatile metals. It is produced by the Hall-Heroult process, in which molten cryolite, Na_3AlF_6, is used as a solvent for the aluminum ore. Cryolite undergoes very slight decomposition with heat to produce a tiny amount of F_2, which escapes into the atmosphere above the solvent. K_c is 2 times 10^-104 at 1300 K for the reaction. Na3AlF6(/) 3 Na(/) + Al(/) + 3 F2(g) What is the concentration of F_2 over a bath of molten cryolite at this temperature? 8 times 10^-312 M 3 times 10^-35 M 8 times 10^74 M 6 times 10^-74 M 5 times 10^102 M

Answer 1

Na3AlF6 (l) <------------------------> 3 Na(l) + Al (l) + 3 F2 (g)

x 0 0 0 ----------------> initial

-y +y +y +y -------------------> changed

x-y y y y -------------------> equilibrium

Kc = [F2]^3

Kc = x^3

2 x 10^-104 = x^3

x = 3 x 10^-35 M

concnetration of F2 = 3 x 10^-35 M