Question

In an equilibrium, A + B ---> C + D

A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After equilibrium has established, concentration of C was thrice the equilibrium concentration of B. Calculate Kc.

Answer 1

Let the initial concentration of B be '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

Hence required value of Kc is 1.8

Hope This Helps You!

Answer 2

REFER THE ATTACHMENT

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