4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction:
Br₂(l) + Cl₂(g) ⇌ BrCl(g)
At the point of equilibrium there were 82.63 g of Br₂(l) .Compute the value of K_c for this reaction.

Question

Quizplus · Accepted Answer

To calculate the equilibrium constant $K_c$, we first need to determine the equilibrium concentrations of the reactants and products. Given that 82.63 g of $Br_2$ are present at equilibrium, we can calculate the moles of $Br_2$ left:Molar mass of $Br_2 = 79.904 \times 2 = 159.808 \, \text{g/mol}$Moles of $Br_2$ at equilibrium = $\frac{82.63 \, \text{g}}{159.808 \, \text{g/mol}} = 0.517 \, \text{mol}$Since we started with 2.000 mol of $Br_2$, the change in moles of $Br_2$ (and thus $Cl_2$, since they react in a 1:1 ratio) is $2.000 - 0.517 = 1.483 \, \text{mol}$. This is also the amount of $BrCl$ formed, as it is produced from $Br_2$ and $Cl_2$ in a 1:1 ratio.The equilibrium concentrations are then:- $[Br_2] = \frac{0.517 \, \text{mol}}{50.0 \, \text{L}} = 0.01034 \, \text{M}$- $[Cl_2] = \frac{4.000 \, \text{mol} - 1.483 \, \text{mol}}{50.0 \, \text{L}} = 0.05034 \, \text{M}$ (since we started with 4.000 mol of $Cl_2$)- $[BrCl] = \frac{1.483 \, \text{mol}}{50.0 \, \text{L}} = 0.02966 \, \text{M}$The equilibrium constant $K_c$ is given by:$$K_c = \frac{[BrCl]}{[Br_2][Cl_2]} = \frac{0.02966}{0.01034 \times 0.05034} = 0.0699$$Therefore, the correct answer is A) 0.0699.

4000 Mol Chlorine and 2