Question

Aluminum is one of the most versatile metals. It is produced by the Hall-Heroult process, in which molten cryolite, Na3AlF6, is used as a solvent for the aluminum ore. Cryolite undergoes very slight decomposition with heat to produce a tiny amount of F2, which escapes into the atmosphere above the solvent. Kc is 2 10-104 at 1300 K for the reaction. Na3AlF6(l) 3 Na(l) Al(l) 3 F2(g) What is the concentration of F2 over a bath of molten cryolite at this temperature

Answer 1

Answer:

$3.0*10^{-35}$

Explanation:

The equation for this process can be given as:

$Na_3AlF_6 \rightleftharpoons 3Na_{(l)} + Al_{(l)}+3F_2_{(g)}$

Also given that:

$K_c = 2.0*10^{-104}$

If we construct an ICE Table; we have:

                      $Na_3AlF_6$    $\rightleftharpoons$     $3Na_{(l)}$    $+$   $Al_{(l)}$   $+$    $3F_2_{(g)}$

Initial               -                           -                  -              0  

Change           -                           -                  -             +x

Equilibrium      -                           -                  -             x

In equilibrium constant, liquids are not involved; as such:

$K_c = [x]^3$

$2.0*10^{-104}=x^3$

$x =\sqrt[3]{2.0*10^{-104}}$

$x=2.7*10^{-35}$

$x$ ≅ $3.0*10^{-35}$