Question

For the reaction $${\mathrm{Br}}_2(g)+5{\mathrm{F}}_2(g)\rightleftharpoons 2{\mathrm{BrF}}_5(g)$$ the system at equilibrium at a particular temperature is analyzed, and the following concentrations are found: $[\mathrm{Br}{\mathrm{F}}_5(g)]=1.01\times {10}^{-9}M,[{\mathrm{Br}}_2(g)]=2.41\times {10}^{-2}M$ and $[{\mathrm{F}}_2(g)]=8.15\times {10}^{-2}M.$ Calculate the value of $K$ for the reaction at this temperature.

Short answer

The equilibrium constant for the reaction at this temperature is approximately $1.25\times {10}^{-7}$.

Step-by-step solution

Write down the equilibrium constant expression for the given reaction

The equilibrium constant expression for the given reaction is: $$K=\frac{[Br{F}_5{]}^2}{[B{r}_2][F_2{]}^5}$$ where [BrF5], [Br2], and [F2] are the molar concentrations of the compounds BrF5, Br2, and F2, respectively, at equilibrium.

Substitute the given concentrations into the equilibrium constant expression

Now, we'll use the given concentrations to substitute in the expression: $$K=\frac{(1.01\times {10}^{-9}{)}^2}{(2.41\times {10}^{-2})(8.15\times {10}^{-2}{)}^5}$$

Evaluate the expression and find the value of K

To find the value of K, we have to calculate the expression: $$K=\frac{(1.01\times {10}^{-9}{)}^2}{(2.41\times {10}^{-2})(8.15\times {10}^{-2}{)}^5}=\frac{1.0201\times {10}^{-18}}{(2.41\times {10}^{-2})(3.39348\times {10}^{-10})}$$ $$K=\frac{1.0201\times {10}^{-18}}{8.17339\times {10}^{-12}}$$ The final value of K after evaluating the expression is: $$K\approx 1.25\times {10}^{-7}$$ So, the equilibrium constant for the reaction at this temperature is approximately $1.25\times {10}^{-7}$.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as $K$, is a crucial aspect of chemical equilibrium. It gives us insight into the ratio of concentrations of products to reactants at a state where a chemical reaction is in balance. This balance occurs when the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant over time. The equilibrium constant is specific to a particular reaction at a given temperature, as it quantifies how far a reaction will proceed toward products under those conditions.
According to the law of mass action, the equilibrium constant for a reaction is expressed in terms of the concentrations of the products raised to the power of their coefficients in the balanced equation, divided by the concentrations of the reactants raised to their respective coefficients. For the reaction ${\text{Br}}_2+5{\text{F}}_2\rightleftharpoons 2{\text{BrF}}_5$, the equilibrium constant expression is $K=\frac{[{\text{BrF}}_5{]}^2}{[{\text{Br}}_2][{\text{F}}_2{]}^5}$.
It is important to note that a high $K$ value indicates a reaction heavily skewed towards the products, whereas a low $K$ value indicates a significant amount of reactants present at equilibrium.

Chemical Reactions

Chemical reactions involve the transformation of substances, where reactants are converted into products. They are represented by chemical equations, which define the quantity of reactants consumed and products formed based on balanced stoichiometry.
During a reaction, various bonds between atoms in the reactants are broken and new bonds are formed to create the products. The balanced equation for the bromine and fluorine reaction forming bromine pentafluoride is ${\text{Br}}_2+5{\text{F}}_2\rightleftharpoons 2{\text{BrF}}_5$, with each mole of bromine reacting with five moles of fluorine to form two moles of bromine pentafluoride.
This reaction is an example of a dynamic equilibrium where, even though the reactants and products are interconverting, their overall concentrations remain unchanged. Understanding these principles helps predict the direction and extent of reactions.

Molar Concentrations

Molar concentration, also known as molarity, is the measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (mol/L). It is a fundamental concept in chemistry, especially when calculating the equilibrium constant, as it allows chemists to quantitatively describe the composition of the reactants and products in a reaction.
The given problem illustrates how to use molar concentrations to determine the equilibrium constant. For the chemical reaction ${\text{Br}}_2+5{\text{F}}_2\rightleftharpoons 2{\text{BrF}}_5$, molar concentrations of the species are provided: $[{\text{BrF}}_5]=1.01\times {10}^{-9}\text{ M}$, $[{\text{Br}}_2]=2.41\times {10}^{-2}\text{ M}$, and $[{\text{F}}_2]=8.15\times {10}^{-2}\text{ M}$.
These concentrations are plugged into the equilibrium expression $K=\frac{[{\text{BrF}}_5{]}^2}{[{\text{Br}}_2][{\text{F}}_2{]}^5}$ to calculate the equilibrium constant. Thus, molar concentrations are pivotal in evaluating the position of equilibrium and quantifying the extent of reaction progress.