Question

What is the percentage dissociation if 1 mole of X2Y is introduced into 1 L vessel at 100 K? Kc for the dissociation of X2Y = 8Ã10^-6. 2X2Y(g) -> 2X2(g) + Y2 (g)

Answer 1

Answer: -

2.5%

Explanation: -

2X₂Y ⇄ 2X₂ + Y₂

I 1 0 0

C -x +x +$\frac{x}{2}$

E 1-x +x +$\frac{x}{2}$

For the equation the expression of Kc is

Kc = $\frac{X2(g)²Y2(g)}{X2Y²}$

8x10⁻⁶ = ( x )² ($\frac{x}{2}$ / (1-x)²

On solving x = 2.5 x 10⁻²

Thus percentage dissociation = $\frac{2.5x10-2}{1}$ x 100

= 2.5

Answer 2

Answer: 0.025

Explanation: