Question

What is percentage dissociation if 1mole of x2y is introduce in 1 litre at 1000k kc for the dissociation of x2y is 1÷ 1000?

Answer 1

Answer: -

The dissociation reaction is not given in the question.

The probable dissociation for X₂Y is

      X₂Y        =    2 X + Y

I       1 mol/L         0       0

C        -c               2c      c

E        (1-c)             2c      c

The expression for Kc is

Kc = $\frac{[X]2[Y]}{[x2y]}$

  10-3  = $\frac{[2c]2[c]}{1-c}$

On solving

c = 0.063 mol / L

Percentage dissociation = $\frac{0.063}{1}$ x 100

                                          = 6.3 %