Question

Question #543c1

Answer 1

The given reversible gaseous reaction :

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

`color(red)(I)" "1" "mol" " 0" "mol" " 0" " "mol`

`color(red)(C)" "-alpha" "mol" " alpha" "mol" " alpha/2" " "mol`

`color(red)(E)" "1-alpha" "mol" " alpha" "mol" " alpha/2" " "mol`

Where `alpha` is the degree of dissociation of `X_2Y(g)`

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`

`=>alpha=16^(1/3)*10^-2~~2.5*10^-2->` No. of moles dissociated per mole of reactant

So the percent dissociation `=2.5*10^-2*100=2.5%`