Question

one mole of a compound AB react with one mole of compound CD according to the equation AB +CD gives AD+CB when equilibrium been established it was found that 3/4 moles each of reactants AB and CD had been converted to AD and CB there is no change in volume the equilibrium constant for the reaction is​

Answer 1

Equilibrium constant $(K_{c})$  is 9.

Explanation:

$AB+CD\rightleftharpoons AD+CB$

$r_{1}\alpha [AB][CD]$

$r_{2}\alpha [AD][CB]$

$r_{1}$ and $r_{2}$ are rate of reaction of forward and backward reaction respectively.

After removing proportionality sign,

$r_{1}= K_{1}[AB][CD]$

$r_{2}=K_{2} [AD][CB]$

At equilibrium    $r_{1}=r_{2}$

$K_{1}[AB][CD]=K_{2} [AD][CB]$

$K_{c}=\frac{K_{1}}{K_{2}}=\frac{[AD][CB]}{[AB][CD]}$      ......(i)

where $K_{1}$ and $K_{2}$ are rate constant of forward and backward reaction and $K_{c}$ is the equilibrium constant of reaction.

To solve this question by equation (i)

                                             $AB+CD\rightleftharpoons AD+CB$

Moles at initial                       1         1           0       0

Moles at equilibrium       $(1-\frac{3}{4})$    $(1-\frac{3}{4})$    $\frac{3}{4}$        $\frac{3}{4}$

                                           $=\frac{1}{4}$         $=\frac{1}{4}$

Put the values in equation (i)

                   $K_{c}=\frac{[AD][CB]}{[AB][CD]}$

                   $K_{c}=\frac{\frac{3}{4}\times \frac{3}{4}}{\frac{1}{4}\times \frac{1}{4}}$

                    $K_{c}=9$

So the Equilibrium constant $(K_{c})$ for this reaction is 9.

Answer 2

Answer:

9:1... steps in attachment