For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g)$$ at \(400 \mathrm{K}, K_{c}=7.0 .\) If 0.25 mol of \(\mathrm{Br}_{2}\) and 0.55 \(\mathrm{mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a 3.0 - container at \(400 \mathrm{K},\) what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2},\) and BrCl?

The balanced chemical equation is given in the exercise as: \[\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2\mathrm{BrCl}(g)\]

An ICE table will help us organize the data and determine the change in the concentration of the reactants and products. The table is set up as follows: | | Br₂ | Cl₂ | BrCl | |-----------|-----------|-----------|-----------| | Initial | 0.25 mol | 0.55 mol | 0 mol | | Change | -x mol | -x mol | +2x mol | | Equilibrium| 0.25-x mol | 0.55-x mol | 2x mol | Now we need to find the equilibrium concentrations in terms of x by dividing the moles by the volume of the container (3.0 L). | | Br₂ | Cl₂ | BrCl | |-----------|-----------|-----------|-----------| | Equilibrium| \(\frac{(0.25-x)}{3}\) M | \(\frac{(0.55-x)}{3}\) M | \(\frac{2x}{3}\) M |

The equilibrium constant expression, Kc, can now be written based on the balanced chemical equation and the equilibrium concentrations: \[K_{c} = \frac{[\mathrm{BrCl}]^{2}}{[\mathrm{Br}_{2}] \times [\mathrm{Cl}_{2}]} = \frac{\left(\frac{2x}{3}\right)^{2}}{\frac{(0.25-x)}{3} \times \frac{(0.55-x)}{3}}\] We are given that at 400 K, Kc = 7.0.

Substitute the known value of Kc and solve for x: \[7.0 = \frac{(2x)^{2}}{(0.25-x)(0.55-x)}\] Now, solve the equation to find the value of x. Note that there might be multiple solutions, but we need to choose the one that gives a valid value for the equilibrium concentrations. After solving the equation, we obtain: \(x = 0.046\)

Now that we have calculated the value of x, we can substitute it back into the equilibrium concentrations of the reactants and products. For Br₂: \[[\mathrm{Br}_{2}] = \frac{0.25-x}{3} = \frac{0.25-0.046}{3} = 0.068\: M\] For Cl₂: \[[\mathrm{Cl}_{2}] = \frac{0.55-x}{3} = \frac{0.55-0.046}{3} = 0.168\: M\] For BrCl: \[[\mathrm{BrCl}] = \frac{2x}{3} = \frac{2 \times 0.046}{3} = 0.031\: M\] The equilibrium concentrations of Br₂, Cl₂, and BrCl are 0.068 M, 0.168 M, and 0.031 M, respectively.