Question #543c1

1 Answer
P dilip_k
Jul 25, 2016

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%#

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