Question

1. 25 mole of hydrogen reacts with 18 mole of iodine, at equilibrium 30.8 mole of HI is formed then what is the dissociation constant of HI? ​

Answer 1

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25mL of H

2

and 18mL of I

2

vapours were heated in a sealed glass tube at 465

o

C and at equilibrium 30.8mL of HI is formed. Calculate the percentage degree of dissociation of HI at 465

o

C.

Medium

Solution

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Since, the number of moles of a gas is proportional its volume (Avogadro's

Law), the volumes in mL of gases may be used instead of number of moles in the equilibrium constant expression.

Let x ml of hydrogen reacts with x ml of iodine to form 2x ml of HI.

H

2

+I

2

→2HI

Initial volume (ml) 25 18

Equilibrium volume (ml) 25−x 18−x 2x=30.8

Since, 2x=30.8 mL

So, x=15.4 mL

Vol. of H

2

at equilibrium (25−x)=(25−15.4)=9.6 mL

Vol. of I

2

at equilibrium (18−x)=(18−15.4)=2.6 mL

Let V L be the total volume.

K

c

=

[I

2

][H

2

]

[HI]

2

K

c

=

(9.6/V)(2.6/V)

(30.8/V)

2

K

c

=38.0

Consider the reverse reaction (decomposition of HI)

2HI(g)⇌H

2

(g)+I

2

(g)

K

c

′

=

K

c

1

K

c

′

=

38.0

1

K

c

′

=0.0263

Let the degree of dissociation be y.

If we start with 2 moles of HI, 0 moles of hydrogen and 0 moles of iodine, then at equilibrium, 2(1−y) moles of HI, y moles of hydrogen and y moles of iodine will be present.

K

c

′

=

[HI]

2

[H

2

][I

2

K

c

′

=

(2(1−y)/V)

2

(y/V)×(y/V)

0.0263=

4(1−y)

2

y

2

0.105=

(1−y)

2

y

2

0.3244=

(1−y)

y

1−y=

0.3244

y

1−y=3.08y

1=4.08y

y=24.5

The percentage degree of dissociation of HI at 465

o

C is 24.5 %.

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