one mole of HI is introduced into a vessel held at constant temperature . when equilibrium is reached it is found that 0.1mole of i2 have been formed . find kc
Answers
Answer:
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
Explanation:
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