At a certain temperature, Kc is 7.98 x 10 -2 for the equilibrium: 2 A (g)--> B (g) + C (g). What are the concentrations of B(g) and C (g) in equilibrium with 0.0566 moles/liter of A(g) ?

Question

1.60 x 10 ^-2 M

3.20 x 10^-3 M

1.28 x 10^-4 M

None of the above

2.56 x 10^-4 M

Answer 1

Probably best to set up an ICE table.

2A(g) <==> B(g) + C(g)

0.0566.........0.........0..........Initial

-2x..............+x.......+x.........Change

0.0566-x...........x.........x.........Equilibrium

Kc = [B][C] / [A]²

7.98x10^-2 = (x)(x) / (0.0566-x)²

7.98x10^-2 = x² / x² - 0.113x + 0.0032

Solve for x using the quadratic equation. This will give you the concentrations of B and C @ equilibrium