# Question #543c1

##### 1 Answer

**The given reversible gaseous reaction :**

#2X_2Y(g)" "rightleftharpoons" "2X_2(g)" "+" "Y_2(g)#

Where

Volume of the vessel being 1 L

the concentrations of the reactants and products **at equilibrium** will be as follows

#[X_2Y(g)]=(1-alpha)M#

#[X_2(g)]=alphaM#

#[Y_2(g)]=alpha/2M#

Now concentration equilibrium constant

#K_c=([X_2(g)]^2[Y_2(g)])/[X_2Y(g)]^2#

#=>8*10^-6=((alpha)^2*alpha/2)/(1-alpha)^2#

Neglecting#color(red)(alpha)# comparing with 1

#=>alpha^3/2=8*10^-6#

**No. of moles dissociated per mole of reactant**

So the percent dissociation