What is the equilibrium concentration of D if the reaction begins with 0.48 M A?

Question

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 0.48 M A?

 A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)

Answer 1

For this question, it is first important to understand what K stands for. Kc is the equilibrium constant at which the reactants and products are in equilibrium, meaning there is no net change in the reactants and product concentration. For every reaction, you can write a Kc expression. This is typically written as products to the power of their respective coefficient divided by the reactants to the power of their respective coefficient. Within the Kc expression, solids are not considered only reactants or products with aq written next to them are considered. For this particular reaction, the Kc expression is Kc=[D]^2 / [A]^1 or Kc=[D]^2 / [A].

Now, we know what the value of Kc is but we don't know what D is nor do we know what A is because remember that the Kc expression accounts for products and reactants in their equilibrium condition and the concentration that we are given for A is not its equilibrium concentration. This is where the ICE table comes in:

ICE stands to initial, change, equilibrium

Now, we do have the initial concentration of A but what about D? Now, if you think about it initially, before the reaction has started, we have 0 product that means D initially is 0. Now, in terms of change if you think about the reaction, your reactant are what become your product. So, the change expression for reactant will always have a - sign , while the change expression for the products will always have a + sign. Please look at the ICE table that I have made below for this problem and then I will explain in further details

A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)

Initial 0.48 0

Change -x +2x

Equilibrium 0.48-x 2x

Note that for D's change expression, I wrote 2 before the X and for A I only wrote X. This is because of the coefficient. For A, the coefficient is 1 so I just wrote X but for D the coefficient is 2 so I wrote 2X. So, for the change expression always consider the coefficient.

Now, that we have the equilibrium expression by combining initial with change, we can insert this into the Kc expression:

Kc= [2x]^2 / [0.48-x]

or 9.2 × 10⁻⁵ = [2x]^2 / [0.48-x]

Typically, x is negligible for the subtraction part( there is alot more that can happen here but that is a whole other discussion). So, the expression just becomes 9.2 × 10⁻⁵ = [2x]^2 / [0.48]

Now, if you do the math and solve for x, you get 0.003323

If you put this x into the equilibrium expression for D, you get 2 * 0.003323 = 0.006645 M and that is the final answer.

It is easier to explain all these minute details live and understand them but I hope this written version helps!