# Question 543c1

Jul 25, 2016

The given reversible gaseous reaction :

$2 {X}_{2} Y \left(g\right) \text{ "rightleftharpoons" "2X_2(g)" "+" } {Y}_{2} \left(g\right)$

$\textcolor{red}{I} \text{ "1" "mol" " 0" "mol" " 0" " } m o l$

$\textcolor{red}{C} \text{ "-alpha" "mol" " alpha" "mol" " alpha/2" " } m o l$

$\textcolor{red}{E} \text{ "1-alpha" "mol" " alpha" "mol" " alpha/2" " } m o l$

Where $\alpha$ is the degree of dissociation of ${X}_{2} Y \left(g\right)$

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

$\left[{X}_{2} Y \left(g\right)\right] = \left(1 - \alpha\right) M$

$\left[{X}_{2} \left(g\right)\right] = \alpha M$

$\left[{Y}_{2} \left(g\right)\right] = \frac{\alpha}{2} M$

Now concentration equilibrium constant

${K}_{c} = \frac{{\left[{X}_{2} \left(g\right)\right]}^{2} \left[{Y}_{2} \left(g\right)\right]}{{X}_{2} Y \left(g\right)} ^ 2$

$\implies 8 \cdot {10}^{-} 6 = \frac{{\left(\alpha\right)}^{2} \cdot \frac{\alpha}{2}}{1 - \alpha} ^ 2$

Neglecting $\textcolor{red}{\alpha}$ comparing with 1

$\implies {\alpha}^{3} / 2 = 8 \cdot {10}^{-} 6$

$\implies \alpha = {16}^{\frac{1}{3}} \cdot {10}^{-} 2 \approx 2.5 \cdot {10}^{-} 2 \to$ No. of moles dissociated per mole of reactant

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