Assume that the equilibrium constant for the dissociation of molecular bromine, \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\), at 800 \(\mathrm{K}\) is \(K_{c}=5.4 \times 10^{-3}\). (a) Which species predominates at equilibrium, \(\mathrm{Br}_{2}\) or Br, assuming that the concentration of \(\mathrm{Br}_{2}\) is larger than $5.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$ (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger numeric value of the rate constant, the forward or the reverse reaction?

According to the given equation: \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) We can write the equilibrium constant \(K_c\) expression in terms of the concentrations of the species as: \[K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\] We are given \(K_c=5.4 \times 10^{-3}\), and the concentration of \(\mathrm{Br}_2\) is larger than \(5.4 \times 10^{-3}\ \mathrm{mol} / \mathrm{L}\).

Since we know that the concentration of \(\mathrm{Br}_2\) is larger than \(5.4 \times 10^{-3}\ \mathrm{mol}\ /\ \mathrm{L}\), we have the following inequality: \[ [\mathrm{Br}_2] > 5.4 \times 10^{-3}\ \mathrm{mol}\ /\ \mathrm{L}\] Using the expression for the equilibrium constant, we can write: \[5.4 \times 10^{-3} = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\] We know that \([\mathrm{Br}_2]>5.4 \times 10^{-3}\), so the fraction \(\frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]}\) must be less than 1, which means that: \[ [\mathrm{Br}]^2 < [\mathrm{Br}_2]\] Taking the square root of both sides: \[ [\mathrm{Br}] < \sqrt{[\mathrm{Br}_2]}\] From the above inequality, we can conclude that the concentration of \(\mathrm{Br}_2\) predominates at equilibrium. Therefore, the answer to question (a) is Br₂.

For an elementary reaction, the rate constant for the forward reaction (\(k_f\)) and the reverse reaction (\(k_r\)) are related to the equilibrium constant by: \[K_c = \frac{k_f}{k_r}\] We are given: \[K_c=5.4 \times 10^{-3}\] Since the value of the equilibrium constant \(K_c\) is less than 1, it means that \(k_f < k_r\). Therefore, the reverse reaction has a larger numeric value of the rate constant. Now we have the answers to both parts of the question: (a) The \(\mathrm{Br}_2\) (molecular bromine) predominates at equilibrium. (b) The reverse reaction has a larger numeric value of the rate constant.