Question

1 mole of A (g) is heated to 300° C in closed
one litre vessel till the following equilibrium is
reached A(g) B(g). The equilibrium
constant for the reaction at 300°C is 4. What
is the concentration of B (in mol. lit) at
equilibrium?
10.2 2 0.6 30.8 40.1​

Answer 1

Hope the above solution helps :)

Plz follow me :)

Answer 2

Answer : The concentration of B at equilibrium is 0.8 M

Solution :  Given,

Moles of A = 1 mole

Volume of solution = 1 L

First we have to calculate the concentration of A.

$\text{Concentration of }A=\frac{\text{Moles of }A}{\text{Volume of solution}}=\frac{1mole}{1L}=1M$

Now we have to calculate the concentration of B at equilibrium.

The given equilibrium reaction is,

                             $A\rightleftharpoons B$

Initially                 1        0

At equilibrium    (1-x)    x

The expression of $K_c$ will be,

$K_c=\frac{[B]}{[A]}$

$4=\frac{x}{(1-x)}$

By solving the term x, we get:

$x=0.8$

Thus, the concentration of B at equilibrium = x = 0.8 M