1.00 mol of A and 2.00 mol of B are placed in a 5 Liter container. after equilibrium has been established, .50 mol of D is present in the container. What is the equilibrium constant, Kc?

A + 2B <---> 2C + D

Kc = (C)^2(D)/(A)(B)^2

Initial:
A = 1.00/5.00 = ??
B = 2.00/5.00 = ??
C = 0
D = 0

equilibrium:
D = 0.50/5.00 from the problem
C = 2*D
A = initial-D
B = initial-2D

Substitute into Kc and solve.

To find the equilibrium constant, Kc, we need to set up an expression using the concentrations of the species at equilibrium. The equilibrium constant expression is given by:

Kc = [C]^2 [D] / [A][B]^2

In this case, we know the initial number of moles of A, B, and the final number of moles of D. However, we don't know the equilibrium concentrations of C and D. To find these values, we need to use the given information about the initial moles and the stoichiometry of the reaction.

Let's find the change in the number of moles for each species:

For A: Initial moles = 1.00 mol, Change = -x, Equilibrium moles = 1.00 - x
For B: Initial moles = 2.00 mol, Change = -2x, Equilibrium moles = 2.00 - 2x
For C: Initial moles = 0 mol, Change = +2x, Equilibrium moles = 2x
For D: Initial moles = 0 mol, Change = +0.50 mol, Equilibrium moles = 0.50

Since two moles of C are formed for every one mole of D, we can say that 2x = 0.50, and therefore x = 0.25.

Now, we can substitute these equilibrium concentrations into the expression for Kc:

Kc = [(2x)^2 (0.50)] / [(1.00 - x)(2.00 - 2x)^2]

Plugging in the values, we have:

Kc = [(2(0.25))^2 (0.50)] / [(1.00 - 0.25)(2.00 - 2(0.25))^2]

Kc = [(0.50)^2 (0.50)] / [(0.75)(1.50)^2]

Kc = (0.25)(0.25)(0.50) / (0.75)(1.50)(1.50)

Kc = 0.03125 / 1.6875

Kc ≈ 0.0185

Therefore, the equilibrium constant, Kc, is approximately 0.0185.