Question

Pls help me with this

Answer 1

`[I_2]=[Br]=0.31M`

Step-by-step explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

`I_2+Br_2\rightleftharpoons 2IBr`

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

`K=([IBr]^2)/([I_2][Br_2]) \\\\1.2x10^2=((2x)^2)/((2.0-x)^2)`

Thus, we solve for x as show below:

`√(1.2x10^2) =\sqrt{((2x)^2)/((2.0-x)^2)} \\\\10.95=(2x)/(2.0-x)\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=(21.91)/(12.95) \\\\x=1.69M`

Therefore, the concentrations of both bromine and iodine are:

`[I_2]=[Br]=2.0M-1.69M=0.31M`

Regards!