Question #543c1

1 Answer
Jul 25, 2016

The given reversible gaseous reaction :

2X_2Y(g)" "rightleftharpoons" "2X_2(g)" "+" "Y_2(g)2X2Y(g) 2X2(g) + Y2(g)

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

color(red)(C)" "-alpha" "mol" " alpha" "mol" " alpha/2" " "molC α mol α mol α2 mol

color(red)(E)" "1-alpha" "mol" " alpha" "mol" " alpha/2" " "molE 1α mol α mol α2 mol

Where alphaα is the degree of dissociation of X_2Y(g)X2Y(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[X2Y(g)]=(1α)M

[X_2(g)]=alphaM[X2(g)]=αM

[Y_2(g)]=alpha/2M[Y2(g)]=α2M

Now concentration equilibrium constant

K_c=([X_2(g)]^2[Y_2(g)])/[X_2Y(g)]^2Kc=[X2(g)]2[Y2(g)][X2Y(g)]2

=>8*10^-6=((alpha)^2*alpha/2)/(1-alpha)^28106=(α)2α2(1α)2

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

=>alpha^3/2=8*10^-6α32=8106

=>alpha=16^(1/3)*10^-2~~2.5*10^-2->α=16131022.5102 No. of moles dissociated per mole of reactant

So the percent dissociation =2.5*10^-2*100=2.5%=2.5102100=2.5%