Question

if 2/9 of 1 mol of HI dissociates the equilibrium constant of disintegration of acid at same temperature will be

Answer 1

Answer: The value of equilibrium constant of disintegration of acid at same temperature will be

Explanation:

                       $HI\rightleftharpoons H_2+I_2$

initial conc.        1       0      0                  

At eq'm      $(1-\frac{2}{9})$      $\frac{1}{9}$        $\frac{1}{9}$  

$\text{concentration}=\frac{\text{number of moles}}{\text{given volume in litres}}$  

Then, $[HI]=\frac{(1-\frac{2}{9})}{V}mol/L,[H_2]=[I_2]=\frac{\frac{1}{9}}{V}mol/L$,where V is= volume in litres

According to reaction $K_c$ will be given as:  

$K_c=\frac{[H_{2}]^{\frac{1}{2}}{[I_2}]^{\frac{1}{2}}}{[HI]^1}=\frac{(\frac{1}{9V})^{\frac{1}{2}}\times (\frac{1}{9V})^{\frac{1}{2}}}{(1-\frac{2}{9V})}$

Substituiting all the terms, on solving we get:

$K_c=\frac{1}{7}=0.1428$

Answer 2

Answer: 1/49

Explanation: see pics