Question

when one mole of HI is heated, in a closed vessel of 2 litres capacity, at equilibrium, 0.25 mole of H2 is formed. determine the equilibrium constant of the reaction.​

Answer 1

Answer:

In the reaction, H

2

(g)+I

2

(g)⇌2HI(g), in a 2 litre flask 0.4 mole of each H

2

and I

2

are taken. At equilibrium 0.5 mol of HI are formed. The value of equilibrium constant K

c

will be 11.1.

0.5 moles of HI will be obtained from 0.25 moles of H

2

and 0.25 moles of I

2

.

0.4−0.25=0.15 moles of H

2

and 0.4−0.25=0.15 moles of I

2

will remain.

The equilibrium constant

K

c

=

[H

2

][I

2

]

[HI]

2

K

c

=

(

2 L

0.15 mol

)×(

2 L

0.15 mol

)

(

2 L

0.5 mol

)

2

K

c

=11.1

Answer 2

Answer:

The equilibrium constant of decomposition reaction of HI is equal to 0.25.

Explanation:

The chemical reaction of heating of hydrogen iodide:

                              $2HI$    ⇄      $H_{2}$    +      $I_{2}$

Initial:                    1mole           -----            ------

At equilibrium:     0.5mol         0.25           0.25

Equilibrium constant is the ratio of concentration of products to the concentration of reactants.

At equilibrium, the equilibrium constant:

    $K_{eq}= \frac{[H_{2}][I_{2}]}{[HI]^{2}}$

⇒   $K_{eq}=\frac{(0.25)(0.25)}{(0.5)^{2}}$

⇒   $K_{eq}=\frac{1}{4}$

∴    $K_{eq}=0.25$

Therefore, the equilibrium constant is equal to 0.25.