Question

30.
For reaction
A + 2B = 2C
Initally 2 mole of A and B are taken in 10 lit flask
at equilibrium 1 mole of C is formed then calculate
Kc for given reaction :-
(1) 6.66 (2) 3.33 (3) 2.22 (4) 1.11​

Answer 1

Explanation:

Solution

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Answer 2

The value of the equilibrium constant is 6.6

Explanation:

Initial moles of A = 2 mole

Initial moles of B = 2 mole

Moles of C at equilibrium= 1 mole

Volume of container = 10 L

Initial concentration of B=  $\frac{moles}{volume}=\frac{2moles}{10L}=0.2M$  

Initial concentration of B= $\frac{moles}{volume}=\frac{2moles}{10L}=0.2M$  

equilibrium concentration of C =$\frac{moles}{volume}=\frac{1moles}{10L}=0.1M$ [/tex]

The given balanced equilibrium reaction is,

                            $A+2B\rightleftharpoons 2C$

Initial conc.         0.2 M     0.2 M         0  

At eqm. conc.    (0.2-x) M   (0.2-2x) M   (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[C]^2}{[A]^1[B]^2}$

$K_c=\frac{(2x)^2}{(0.2-x)^1\times (0.2-2x)^2}$

we are given : 2x= 0.1 M

x= 0.05 M

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

$K_c=\frac{(0.1)^2}{(0.2-0.05)^1\times (0.2-0.1)^2}$

$K_c=6.66$

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