Question

Pls help me with this

Answer 1

Answer:

$[I_2]=[Br]=0.31M$

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:

$K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}$

Thus, we solve for x as show below:

$\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M$

Therefore, the concentrations of both bromine and iodine are:

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

Regards!