Answer:
The concentration of Bromine is 0.308M
The concentration of Iodine is 0.308M
Explanation:
From the question, we are given the following parameters:
[tex]\begin{gathered} \text{Moles of }I_2=4.00\text{moles} \\ \text{Moles of Br}_2=4.00moles \\ \text{Volum}e\text{ of container = 2.0L} \end{gathered}[/tex]
Get the concentration of Iodine and Bromine
[tex]\text{Conc}=\frac{Moles}{Volume}[/tex][tex]\begin{gathered} _{}\lbrack I_2\rbrack=\frac{4.0}{2}=2M_{} \\ \lbrack Br_2\rbrack=\frac{4.0}{2}=2M \end{gathered}[/tex]
Given the chemical reaction between Iodine and Bromine;
[tex]\begin{gathered} I_2(g)+Br_2(g)\rightarrow2\text{IBr(g): k = 1.2}\times10^2 \\ (2M)\text{ + (2M)}\rightarrow(2xM) \end{gathered}[/tex]
From the equation, we can see that there is no concentration for the molecule 2IBr, so we can give it "2x"
According to the equilibrium constant formula:
[tex]\begin{gathered} k=\frac{conc\text{ of product}}{conc\text{ of reactant}} \\ k=\frac{\lbrack IBr\rbrack^2}{\lbrack I_2\rbrack\lbrack Br_2\rbrack} \\ 1.2\times10^2=\frac{(2x)^2}{(2-x)(2-x)} \\ 1.2\times10^2=\frac{(2x)^2}{(2-x)^2} \end{gathered}[/tex]
Simplify the result to get the value of "x"
[tex]\begin{gathered} 1.2\times10^2=(\frac{2x}{2-x})^2 \\ 120=(\frac{2x}{2-x})^2 \\ \sqrt[]{120}^{}=\frac{2x}{2-x} \\ 10.9544=\frac{2x}{2-x} \end{gathered}[/tex]
Cross multiply and simplify
[tex]\begin{gathered} 10.9544(2-x)=2x \\ 21.9089-10.9544x=2x \\ 21.9089=2x+10.9544x \\ 21.9089=12.9544x \\ x=\frac{21.9089}{12.9544} \\ x=1.692M \end{gathered}[/tex]
Get the concentration of the entities in the mixture
[tex]\begin{gathered} \text{ Conc of Br}_2=2-1.692=0.308M \\ \text{ Conc of I}_2=2-1.692=0.308M \end{gathered}[/tex]
Hence the equilibrium concentrations of all entities in the mixture are 0.308M of Bromine and 0.308M of Iodine