The following molecular scenes depict the aqueous reaction

Step 1 of 3

Here, we are going to calculate the KC for the given reaction.

Given that

The reaction is as follows:

\(\text{2D}\rightleftharpoons E\)     —- (1)

Total No. of red spheres (D)  in A vessel = 3

Total No. of blue spheres (E)  in A vessel = 3

Concentration of each ball = 0.0100 mol

Therefore, the concentration of D is

\([D]=3\ \text{spheres}\times\frac{0.01\ \text{mol}}{1.0\ \text{sphere}}\times\frac{1}{1.0\ \text{Liter}}\)

         \(= 0.03\ \text{mol/L} = 0.03\ \text{M}\)

Similarly, the concentration of E is

\([E] = 3\ \text{spheres}\times \frac{0.01\ \text{mol}}{1.0\ \text{sphere}}\times\frac{1}{1.0\ \text{Liter}}\)

         \(= 0.03\ \text{mol/L} = 0.03\ \text{M}\)

We know that, the equilibrium constant is defined as

\(K_c=\frac{[E]}{[D]^2}\)     —- (2)

\(\Rightarrow K_c=\frac{(0.03)}{(0.03)^2}\)

\(\Rightarrow K_c=33.33\)

Therefore, the \(K_c\) when the reaction in scene A is at equilibrium, is 33.33.