the inital concentration of A is double the the inital concentration of B At equilibrium the concentration of B was found to be one third of the concentration of C the value of equilibrium constant is
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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
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