Equilibrium and using ICE tables

Question

For the reaction below, Kc = 4.60 × 10⁻⁶. Note Kc is sometimes called K. What is the equilibrium concentration of B if the reaction begins with 0.700 M A? A (g) ⇌ 3 B (g)

Answer 1

First, write out your reaction:

A (g) ⇌ 3 B (g); Kc = 4.60 x 10^-6

I want to first point out that the equilibrium constant (Kc) is quite small, which indicates that the forward reaction is not super favorable. In a container of this substance, there would be more A (reactants) than B (products).

Let's set up an ICE table to document all of our values.

Rxn: A (g) ⇌ 3 B (g); Kc = 4.60 x 10^-6 Notes:

I (initial): 0.700 0 Always 0 to start on the products side

C (concentration): -x. +3x. Notice the coefficients**

E (equilibrium): 0.700-x. 0+3x Sum the values

Kc is the equilibrium constant, which is why we use the values from the equilibrium (E) column. We are given out Kc as Kc = 4.60 x 10^-6, so let's plug that in.

4.60 x 10^-6 = (3x)^3/(.700-x)

These numbers are set up in the format of (products)^coefficient/(reactants)^coefficient. This is used in calculations for K constants and Q (reaction quotient) constants.

For this problem, let's assume the 5% rule to simplify our math, as our Kc is very small:

4.60 x 10^-6 = (3x)^3/(.700)

Solve: x≈4.92×10−3

Lastly, do not forget to plug this back in! What we have just calculated is the change in concentrations to reach equilibrium; it is not the actual concentration of B (what your question is asking).

Once we plug it into the concentration of B (3x), our final answer is [B]≈1.48×10−2 M