What is the equilibrium constant for the reversible reaction in aqueous medium below given that the respective concentrations of A, B, C, and D are 0.0117 M, 0.00440 M, 0.00550 M, and 0.00780 M

Question

What is the equilibrium constant for the reversible reaction in aqueous medium below given that the respective concentrations of A, B, C, and D are 0.0117 M, 0.00440 M, 0.00550 M, and 0.00780 M

Answer 1

What is the reversible reaction? Is it A + B <==> C + D ? If so, the proceed as follows:

Keq = [C][D]/[A][B] = (0.0055)(0.0078)/(0.117090.0044) = 0.833