Question

Determine the equilibrium concentration when 1.00 mole of I2 and 1.00 mole of H2 are put into a 5 liter container. At the reaction temperature, the Keq is 64.0

H2 (g) + I2 (g) 2HI (g)

Answer 1

The equilibrium concentration are :

`{H_(2)}` = 0.04 M

`{I_(2)}` = 0.04 M

`{HI}` = 0.16 M

Step-by-step explanation:

`{H_(2)} + {I_(2)} \rightleftharpoons 2 HI`

it means 1 mole of iodine and hydrogen produce 2 mole of HI

Concentration(C) : Moles per unit volume.It is expressed in Molarity

(M=mol/L )

`concentration = (moles)/(volume)`

Initial moles :

`{H_(2)}` = 1.00

So,
`C = (1)/(V)`

`{I_(2)}` = 1.00

`C = (1)/(V)`

`{HI}` = 0

`C = (0)/(V)`

let during the reaction x moles of both
`{H_(2)}` and
`{I_(2)}` get dissociated , then

At equilibrium ,

`{H_(2)}` = 1.00 - x

`C = (1-x)/(V)`

For iodine

`{I_(2)}` = 1.00 - x

`C = (1-x)/(V)`

1.00 - x mole of hydrogen will produce 2x of HI

`{HI}` = 2x

`C = (2x)/(V)`

`K_(eq) = ([products]^(coefficient))/([reactants]^(coefficient))`

On solving for x , (look at the image)

`{H_(2)}` = 0.04 M

`{I_(2)}` = 0.04 M

`{HI}` = 0.16 M