Question

The equilibrium system shown below has a K = 8.3 x 10^-2. If [D] = 0.38 mol/L, what is the [A]? 2A(g) + B(s) <=====> C(s) + 3D(g) *a)0.081 mol/Lb) 8.1 mol/Lc) 0.81 mol/Ld) 0.0081 mol/Le) 0.00081 mol/L

Answer 1

Given the reaction:

2 A (g) + B (s) <=====> C (s) + 3 D (g)

Let's find the expression for the equilibrium constant of that reaction:

K = [D]³ / [A]²

In that expression we don't include B and C because they are solids, and solids aren't included.

According to the problem:

K = 8.3 x 10^(-2)

[D] = 0.38 mol/L

Replacing the given values:

K = [D]³ / [A]²

8.3 x 10^(-2) = (0.38)³ / [A]²

Then we can solve that equation for [A]:

8.3 x 10^(-2) = (0.38)³ / [A]²

[A]² = (0.38)³/ 8.3 x 10^(-2)

[A]² = 0.661

[A] = √0.661

[A] = 0.81 mol/L

Then the answer to the problem is c) 0.81 mol/L