Question

The molecular scenes above depict the aqueous reaction 2 d ⇌ e where d is red and e is blue, and each sphere represents 0.0100 mol. the volume in scene a is 1.00 l, and in scenes b and c, it is 0.500 l. (a) if the reaction in scene a is at equilibrium, calculate kc.

Answer 1

The value of Kc for the reaction at equilibrium in scene a is 1.000.

Step-by-step explanation:

To calculate Kc for the given reaction, we first need to determine the concentrations of the reactant and product. In scene a, the volume is 1.00 L, and each sphere represents 0.0100 mol. Since there are 2 spheres of d and 1 sphere of e, we have [d] = [e] = 0.0100 mol/L. Therefore, the concentration of d or e in scene a is 0.0100 M.

Now, the mathematical expression for Kc is Kc = [e]^2 / [d]^2. Substituting the concentration values we obtained earlier, we get Kc = (0.0100 M)^2 / (0.0100 M)^2 = 1.000. Therefore, the value of Kc for the reaction at equilibrium in scene a is 1.000.