Question

An equal number of moles of I2(g) and Br2(g) are placed into a closed container and allowed to establish the following equilibrium:

I2(g) + Br2(g) 2IBr(g)

Keq = 280

Which one of the following relates [IBr] to [I2] at equilibrium?




[I2] = [IBr]




[I2] < [IBr]




[I2] = 2 [IBr]




[I2] = 280 [IBr]

Answer 1

B. I'm in the same class

Answer 2

Answer : The correct option is,
`[I_2]<[IBr]`

Solution : Given,

Moles of
`I_2` = Moles of
`Br_2`

`K_(eq)=280`

The balanced equilibrium reaction is,

`I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)`

The expression for equilibrium constant is,

`K_(eq)=([IBr]^2)/([I_2][Br_2])`

As per question, Moles of
`I_2` = Moles of
`Br_2`

Now put all the given values in above formula, we get the relation between the
`[IBr]` and
`[I_2]`.

`280=([IBr]^2)/([I_2][I_2])`

`280=([IBr]^2)/([I_2]^2)`

`√(280)=([IBr])/([I_2])`

`[IBr]=16.733* [I_2]`

From this we conclude that the concentration of
`IBr` is greater than the concentration of
`I_2`.

Therefore, the correct option is,
`[I_2]<[IBr]`