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 0.78 M A? A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)

Answer 1

Hi Halie,

For starters, we need an equilibrium expression for this reaction. Be mindful that solids and liquids are NOT included in equilibrium expressions.

Kc = [D]² / [A]

We need an ICE table to determine the concentration of B at equilibrium when [A]= 0.78 M

A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)

I 0.78M -- -- 0

C -x -- -- +2x

E 0.78-x -- -- 2x

Now we can plug the values in row "E" into the equilibrium expression.

Kc = [D]² / [A]

9.2 x10^-5 = (2x)² / (0.78-x)

(0.78)(9.2 x10^-5) = 4x²

x= 0.0042

(Note: the ratio of x to the original concentration is less than 5% so we can neglect the (-x) part.)

At equilibrium, the concentration of D = 2x (see ICE table.)

So, 2(0.0042) = 0.0085M