Question

What is the percentage dissociation if 1 mole of X2Y is introduced into 1 L vessel at 1000 K? Kc for the dissociation of X2Y = 1*10-6.
2 X2Y(g) ßà 2 X2 (g) + Y2 (g)

Answer 1

Given that,

                           $X_{2}Y    ⇆     2X_{2}   +    Y_{2}$

Initial moles                1                         0                  0

Equili. moles            (1-α)                     2α                 α

$K_{c} = \frac{4\alpha ^{3} }{(1-\alpha )}  = 10^{-6}$

As   α very less than 1 therefore (1-α) = 1

$4\alpha ^{3} = 10^{-6}$

$\alpha ^{3} = 0.25*10^{-6}$

$\alpha = 0.63*10^{-2} = 0.0063$

Conclusion:

percentage of dissociation = 0.63 %

Answer 2

Answer: Percent dissociation is 0.6%

Explanation:

moles of $X_2Y$ = 1 mole

Volume of solution = 1 L

Initial concentration of $X_2Y=\frac{moles}{\text {volume in Liters}}=\frac{1mole}{1L}=1M$[/tex]

$2X_2Y(g)\rightarrow 2X_2(g)+Y_2(g)$

Initial      1M                   0             0

Eqm  $1-2\alpha$     $2\alpha$     $1\alpha$  

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[X_2]^2\times [Y_2]}{[X_2Y]^2}$

Now put all the given values in this expression, we get :

$1\times 10^{-6}=\frac{(2\alpha)^2\times (\alpha)}{(1-2\alpha)^2}$

By solving we get :

$\alpha=0.006$

Thus the percent dissociation is $0.006\times 100=0.6\%$

Thus percent dissociation is 0.6%