Question

the number of moles of H2, I2, HI are 1, 2, 3 moles respectively. total pressure of the reaction mixture is 60 atom. calculate the partial pressure of the H2, I2, HI in the mixture

Answer 1

Solution
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One mole of H
2
​
, 2 moles of I
2
​
and 3 moles of HI are injected in a one litre flask.
x moles of H
2
​
will react with x moles of I
2
​
to form 2x moles of
HI. Total number of moles of HI present at equilibrium will be 3+2x
1−x moles of H
2
​
and 2−x moles
of I
2
​
will remain at equilibrium.

The equilibrium constant K
c
​
=
[H
2
​
][I
2
​
]
[HI]
2

​

45.9=
[
1 L
1-x mol
​
][
1 L
1-x mol
​
]
[
1 L
3+2x mol
​
]
2

​

45.9=
[1−x][2−x]
[3+2x]
2

​

45.9(1[2−x]−x[2−x])=3[3+2x]+2x[3+2x]
45.9(2−x−2x+x
2
)=9+6x+6x+4x
2

91.8−137.7x+45.9x
2
=9+12x+4x
2

41.9x
2
−149.7x+82.8=0

This is quadratic equation with solution
x=
2a
−b±
b
2
−4ac
​

​

x=
2(41.9)
−(−149.7)±
(−149.7)
2
−4(41.9)(82.8)
​

​

x=
83.8
149.7±92.4
​

x=
83.8
149.7±92.4
​

x=2.88 or x=0.684

The value x=2.88 is discarded as it will lead to negative
value of number of moles.
Hence, x=0.684

The equilibrium concentrations are
[HI]=3+2x=3+2(0.684)=4.368 mol/L
[H
2
​
]=1−x=1−0.684=0.316 mol/L
[I
2
​
]=2−x=2−0.684=1.316 mol/L