A mixture containing 8.07 mol of hydrogen and 9.08mol of iodine was heated at `448^(@)C` till equilibrium was attained when 13.38 mol of hydrogen iodide was obtained. Calculate the percentage dissociation of hydrogen iodide at `448^(@)C`.

Options

A. `13.2%`
B. `19.8%`
C. `18.9%`
D. `21.4%`

Answer

D

Explanation

The reaction of formation of HI is as follows: `{:(H _(3) ,+, I _(2), hArr, 2HI), (8.07,,9.08,,0), (8.07-x,, 9.08-x,,2x):}`
Therefore, from the above data x can be calculated as : `2x = 13.38 impliesx = ( 13.38)/(2) = 6.69`
The equilibrium constant can be calculated as:
`K ._(C) = ([ HI]^(2))/( [H_(2)] [I_(2)])= ((2x ) ^(2))/((9.08 -x) ( 8.07 -x))= ((13. 38)^2)/((9.08 - 6.69) xx (8.07- 6.69) ) = 54.279`
To calculate the dissociation constatn of HI, consider the following reaction,
`2HI hArr H _(2) + I _(2)`
`{:(1, 0, 0), ( 1-2x, x, x):}`
From the above data, the dissociation constatn is given by `K _(C) =(x xx x)/( (1- 2x ) ^(2)) = (x ^(2))/((1- 2x) ^(2))`
But `K _(C) = (1)/(K ._(C)) = (1)/( 54. 279) = 0.01842`
.x. can be calculated as `((x)/(1 - 2x )) ^(2) = (1)/( 54.279) implies (x)/(1-2x) = sqrt ((1)/(54.279))= (1)/( 7.3674) or 6.3674x=1 -2x.`
Therefore, `x = (1)/( 9.3674 ) = 106.75 xx 10 ^(-3) mol`
`alpha = 2x = 2 xx 106. 75 xx 10 ^(-3) = 213 . 5 xx 10 ^(-3)` Percentage dissociation `= 21. 35 %`