20g of HI is heated at 327 C in a bulb of 1 l capacity. Calculate the volume percentage of H2 I2 and HI at equilibrium. Given that the mass law constant for the equation 2HI--->H2 + I2 is 0.0559 at 327 C when concentration are expressed in moles/l

Dear Student,

The reaction of HI as follows:
                                        $$\begin{gathered} 2HI \stackrel{}{\to }{H}_2 + I_2 \\ at t = 0 2 0 0 \\ at eqm 2-2x x x \\ K = \frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2} \\ K = \frac{x \times x}{(2-2x{)}^2} \\ rooting both side: \\ \sqrt{K} = \frac{x}{2-2x} \\ Putting the values of K \\ \sqrt{0.0559} =\frac{x}{2-2x} \\ 0.2364 =\frac{x}{2-2x} \\ 0.2364(2-2x) = x \\ 0.4728 - 0.4728x = x \\ 1.4728x = 0.4728 \\ x = 0.32 mol \end{gathered}$$

So, when 2 mol of HI is heated 0.32 mol of Hydrogen gas and iodine is produced.
2 mol of HI = 128 g of HI 
128 g of HI forms = 22.4 lit of H₂ x 0.32 
1 g of HI forms = 7.168/128 
20 g of HI forms 
                             $$\begin{gathered} = \frac{7.168}{128}\times 20 \\ = 1.12 L of H_2 \& I_2 \end{gathered}$$

And the volume concentration of HI at eqm. = 
                              $$\begin{gathered} \frac{20}{128} mol of HI = 1 lit \\ 0.156 mol of HI contains = 1 lit \\ 1 mol contains of HI contains = 1/0.156 \\ Remaining concentration of HI at eqm = 0.156 - (0.156 \times 0.32) \\ = 0.156 - 0.05 \\ = 0.106 \\ So, the volume of HI\hspace{0.17em}at eqm = \frac{0.106}{0.156} = 0.68 L \end{gathered}$$

As the volume of bulb is fixed at eqm so, 
The HI and H₂ and I₂  are in the ratio = 0.68 : 2.24 
So, if the total volume cant exceed 1L , hence the HI will be at eqm = 0.68/ 2.24 = 0.30 L 
And the volume of H₂ at eqm = 0.35 L 
and the volume of I₂ at eqm = 0.35 L 
So, the volume percentage of HI at eqm = 30 % 
 the volume percentage of H₂ at eqm = 35 % 
& the volume percentage of I₂ at eqm = 35 % 

Regards.