Hello there!

Greetings!

Let us take the initial concentration of B be ss 'x'

Then , according to question, initial concentration of A will be '2x'.

At initial stage ,

A - 2x

B - x

C - 0

D - 0

Now, let 'y' be the change . Then at equilibrium

A - 2x - y
B - x - y
C - y
D - y

At equilibrium, concentration of C is thrice the concentration of B.

We can represent this as ->
y = 3(x - y)
y = 3x -3y
y = 3x/4

Plugging the values of y in 2x - y, we get equilibrium concentration of A as 5x/4.

For equilibrium concentration of B we get x/4.

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

Kc = y x y / (2x - y)(x - y)

Kc =( 3x/4 )^2 / 5x/4 x x/4

Kc = 9/5

Kc = 1.8

Thus, required value of Kc is 1.8

Thankyou