Question

When 1.00 mol of A and 0.800 mol of B are placed in a 2.00 L container and allowed to come to equilibrium, the resulting mixture is found to be 0.20M in D. What is the value of K at equilibrium?

Answer 1

You can't solve this without knowing the equation.
A + 2B => D or
2A + B = 2D or
etc.

Answer 2

2A (aq) + B (aq) <-> C (g) + D (g)

Answer 3

(A) = 1 mol/2L = 0.500M
(B) = 0.800/2 = 0.400 M

,,,,,,,,,,,,,,,,,,,,,,2A (aq) + B (aq) <-> C (g) + D (g)
I.....................0.5.........0.4...............0.........0
C,,,,,,,,,,,,,,,,,,,,-2x.........-x...............+x........+x
E.................0.5-2x......0.4-x.............x..........x

The problems tells you that x = 0.20M.
Write the K expression for the reaction, plug in the values for A, B, C, D, and solve for K. Post your work if you get stuck.

Answer 4

What is the k expression?

Answer 5

Keq = (concn of right side)^coefficients/(concn left side)^coeff or
Keq = (C)(D)/(A)^2(B)