Consider the reversible reaction in aqueous medium shown below. at equilibrium, the concentrations of a, b, c, and d are 0.0053 m, 0.0085 m, 0.0040 m, and 0.0035 m, respectively. what is the equilibrium constant for the reaction? a+3b⇌3c+d

The equilibrium constant for the reaction is 0.0689

Explanation:

Step 1: Data given

The concentrations of A = 0.0053 M

The concentration of B = 0.0085 M

The concentration of C = 0.0040 M

The concentration of D = 0.0035 M

Step 2: The balanced equation:

A+3B⇌3C+D

Step 3: Calculate the equilibrium constant

K = [C]³[D] / [A][B]³

K = (0.0040)³(0.0035) / (0.0053)(0.0085)³

K = 0.0689

The equilibrium constant for the reaction is 0.0689