Question

lodine, I2 and bromine, Br2 react in a closed 2.0-L container at 150°C to produce iodine monobromide, IBr, according to the following equation: I2 + Br2 → 2 Br The equilibrium constant, K, at this temperature is 1.2 x 10^2. What are the equilibrium concentrations of all entities in the mixture if the container initially contained 4.00 mol each of iodine and bromine?

Answer 1

The concentration of Bromine is 0.308M

The concentration of Iodine is 0.308M

Step-by-step explanation:

From the question, we are given the following parameters:

`\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}`

Get the concentration of Iodine and Bromine

`\text{Conc}=(Moles)/(Volume)`
`\begin{gathered} _{}\lbrack I_2\rbrack=(4.0)/(2)=2M_{} \\ \lbrack Br_2\rbrack=(4.0)/(2)=2M \end{gathered}`

Given the chemical reaction between Iodine and Bromine;

`\begin{gathered} I_2(g)+Br_2(g)\rightarrow2\text{IBr(g): k = 1.2}*10^2 \\ (2M)\text{ + (2M)}\rightarrow(2xM) \end{gathered}`

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:

`\begin{gathered} k=\frac{conc\text{ of product}}{conc\text{ of reactant}} \\ k=(\lbrack IBr\rbrack^2)/(\lbrack I_2\rbrack\lbrack Br_2\rbrack) \\ 1.2*10^2=((2x)^2)/((2-x)(2-x)) \\ 1.2*10^2=((2x)^2)/((2-x)^2) \end{gathered}`

Simplify the result to get the value of "x"

`\begin{gathered} 1.2*10^2=((2x)/(2-x))^2 \\ 120=((2x)/(2-x))^2 \\ \sqrt[]{120}^{}=(2x)/(2-x) \\ 10.9544=(2x)/(2-x) \end{gathered}`

Cross multiply and simplify

`\begin{gathered} 10.9544(2-x)=2x \\ 21.9089-10.9544x=2x \\ 21.9089=2x+10.9544x \\ 21.9089=12.9544x \\ x=(21.9089)/(12.9544) \\ x=1.692M \end{gathered}`

Get the concentration of the entities in the mixture

`\begin{gathered} \text{ Conc of Br}_2=2-1.692=0.308M \\ \text{ Conc of I}_2=2-1.692=0.308M \end{gathered}`

Hence the equilibrium concentrations of all entities in the mixture are 0.308M of Bromine and 0.308M of Iodine