Question #543c1
1 Answer
The given reversible gaseous reaction :
2X_2Y(g)" "rightleftharpoons" "2X_2(g)" "+" "Y_2(g)2X2Y(g) ⇌ 2X2(g) + Y2(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[X2Y(g)]=(1−α)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)^2⇒8⋅10−6=(α)2⋅α2(1−α)2 Neglecting
color(red)(alpha)α comparing with 1
=>alpha^3/2=8*10^-6⇒α32=8⋅10−6
So the percent dissociation