For the reaction below, Kc = 9.2 × 10⁻⁵. Note Kc is sometimes called K. What is the equilibrium concentration of D if the reaction begins with 0.62 M A? A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)

Answer 1

A(aq) + 2B(s) <==> C(s) + 2D(aq)

Kc = 9.2x10^-5 = [D]² / [A] (note: Kc does NOT include solids, so B and C do not appear)

It may be easier to set up an ICE table to see what happens during reaction:

A(aq) + 2B(s) <==> C(s) + 2D(aq)

0.62M..................................0 M.............Initial

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

0.62-x...................................2x.............Equilibrium

So, at equilibrium [A] = 0.62-x and [D] = 2x

Plugging these into the expression for Kc, and solving for x, we have ...

9.2x10^-5 = (2x)² / 0.62-x

4x² = 5.7x10^-5 - 9.2x10^-5x

4x² + 9.2x10^-5x - 5.7x10^-5 = 0

x = 0.00376 M

Plug this into the value for D in the ICE table...

[D] = 2x = 2 x 0.00376 = 7.5x10^-3 M

(be sure to check the math)