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:

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

And the volume concentration of HI at eqm. =

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.

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