Question #543c1

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Question #543c1

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Answer 1 · 2016-07-25T10:30:18

The given reversible gaseous reaction :

$2 X_{2} Y (g) ⇌ 2 X_{2} (g) + Y_{2} (g)$ $2X_2Y(g)" "rightleftharpoons" "2X_2(g)" "+" "Y_2(g)$

$I 1 m o l 0 m o l 0 m o l$ $color(red)(I)" "1" "mol" " 0" "mol" " 0" " "mol$

$C - α m o l α m o l \frac{α}{2} m o l$ $color(red)(C)" "-alpha" "mol" " alpha" "mol" " alpha/2" " "mol$

$E 1 - α m o l α m o l \frac{α}{2} m o l$ $color(red)(E)" "1-alpha" "mol" " alpha" "mol" " alpha/2" " "mol$

Where $α$ $alpha$ is the degree of dissociation of $X_{2} Y (g)$ $X_2Y(g)$

Volume of the vessel being 1 L
the concentrations of the reactants and products at equilibrium will be as follows

$[X_{2} Y (g)] = (1 - α) M$ $[X_2Y(g)]=(1-alpha)M$

$[X_{2} (g)] = α M$ $[X_2(g)]=alphaM$

$[Y_{2} (g)] = \frac{α}{2} M$ $[Y_2(g)]=alpha/2M$

Now concentration equilibrium constant

$K_{c} = \frac{{[X_{2} (g)]}_{2}^{2} [Y_{2} (g)]}{{[X_{2} Y (g)]}_{2}^{2}}$ $K_c=([X_2(g)]^2[Y_2(g)])/[X_2Y(g)]^2$

$\Rightarrow 8 \cdot 10^{- 6} = \frac{{(α)}^{2} \cdot \frac{α}{2}}{{(1 - α)}^{2}}$ $=>8*10^-6=((alpha)^2*alpha/2)/(1-alpha)^2$

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

$\Rightarrow \frac{α^{3}}{2} = 8 \cdot 10^{- 6}$ $=>alpha^3/2=8*10^-6$

$\Rightarrow α = 16^{\frac{1}{3}} \cdot 10^{- 2} \approx 2.5 \cdot 10^{- 2} \to$ $=>alpha=16^(1/3)*10^-2~~2.5*10^-2->$ No. of moles dissociated per mole of reactant

So the percent dissociation $= 2.5 \cdot 10^{- 2} \cdot 100 = 2.5 %$ $=2.5*10^-2*100=2.5%$

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Question #543c1

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