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

Answer : The correct option is, $[I_2]$

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,

$K_{eq}=\frac{[IBr]^2}{[I_2][Br_2]}$

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 $[I_2]$.

$280=\frac{[IBr]^2}{[I_2][I_2]}$

$280=\frac{[IBr]^2}{[I_2]^2}$

$\sqrt{280}=\frac{[IBr]}{[I_2]}$

$[IBr]=16.733\times [I_2]$

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

Therefore, the correct option is, $[I_2]$

Answer 2

B. I'm in the same class