Question

43.
The reaction A+B C + D is studied in a one
litre Vessel at 250°C. The initial concentration of
A was 3n and of B was n. After equilibrium was
attained then equilibrium concentration of C was
found to be equal to equilibrium concentration of
B. What is the concentration of D at equilibrium :-
(2)(3n -
(1)
(3)(n+2)
(4)n

Answer 1

The concentration of D at equilibrium is $\frac{n}{2}$

Explanation:

Initial concentration of A= 3n

Initial concentration of B= n

The given balanced equilibrium reaction is    

                    $A+B\rightleftharpoons C+D$

Initial conc            3n       n        0     0  

At eqm. conc.    (3n-x) M   (n-x) M   (x) M     (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[C]\times [D]}{[A]\times [B]}$

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

$K_c=\frac{x\times x}{(3n-x)\times (n-x)}$

We are given :  at eqm $[C]=[B]$

$x=(n-x)$

$x=\frac{n}{2}$

Thus concentration of D at equilibrium , $[D]=x=\frac{n}{2}$

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