Question

A(g) + 3B(g ) is equilibrium to 4(C). Initial concentration of A is equal to that of B. The equilibrium concentration of A and C are equal.Then Kc is equal to

Answer 1

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Answer 2

Answer: The value of $K_c$ is 8.

Explanation:

Let initial concentration of  $A$ = x

Initial concentration of $B$ = x

The given balanced equilibrium reaction is,

       $A(g)+3B(g)\rightleftharpoons 4C(g)$

Initial conc.         x        x     0

At eqm. conc.    (x-y) M   (x-3y) M   (4y) M

[A] at eqm = x-y

[C] at eqm= 4y

Given : x - y = 4y

x = 5y

Thus [A] at eqm = x- y = 5y -y = 4y

[B] at eqm = x - 3y = 5y - 3y = 2y

[C] at eqm = 4y

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[C]^4}{[B]^3[A]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(4y)^4}{(2y)^3\times (4y)}$

$K_c=8$

Thus the value of the equilibrium constant is 8.