A+2B--->2C Kc=2.27

2c--->D Kc=0.144

Calculate the equilibrium constant for the reaction
D--->A+2B

Add eqn 1 to eqn 2 to obtain

A+2B ==> 2C
2C ==> D
------------
A + 2B ==> D so K' = k1*k2.
You want the reverse of this rxn; therefore, you want 1/K'.

To calculate the equilibrium constant for the reaction D --> A + 2B, you can use the following equation:

K_eq = (Kc1 * Kc2)^(1/2)

Where Kc1 and Kc2 are the equilibrium constants for the two reactions given:

1) A + 2B ---> 2C with Kc = 2.27
2) 2C ---> D with Kc = 0.144

So, plugging in the values into the equation:

K_eq = (2.27 * 0.144)^(1/2)

Simplifying this further:

K_eq = (0.32712)^(1/2)

Taking the square root:

K_eq = 0.571

Therefore, the equilibrium constant (K_eq) for the reaction D --> A + 2B is approximately 0.571.