91Ó°ÊÓ

For first-order reactions, the half-life is given by the equation: \[t_{1/2} = \frac{0.693}{k}\] where \(t_{1/2}\) is the half-life and \(k\) is the rate constant. In this case, the rate constant is given as \(0.271~\text{s}^{-1}\). Plugging the values into the equation, we get: \[t_{1/2} = \frac{0.693}{0.271~\text{s}^{-1}}\] Calculating the half-life, we find: \[t_{1/2} \approx 2.56~\text{s}\] The half-life for this reaction is approximately \(2.56~\text{s}\).

To determine the concentration of \(\text{I}_{2}\) after a certain time, we use the integrated rate law for first-order reactions: \[\ln{\frac{[\text{I}_{2}]_{t}}{[\text{I}_{2}]_{0}}} = -kt\] where \([\text{I}_{2}]_{t}\) is the concentration at time \(t\), \([\text{I}_{2}]_{0}\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time elapsed. In this case, the initial concentration is \(0.050~\text{M}\), the rate constant is \(0.271~\text{s}^{-1}\), and the time elapsed is \(5.12~\text{s}\). Plugging these values into the equation, we get: \[\ln{\frac{[\text{I}_{2}]_{t}}{0.050~\text{M}}} = -0.271~\text{s}^{-1}(5.12~\text{s})\] Now, we just need to solve for \([\text{I}_{2}]_{t}\): \[\ln{\frac{[\text{I}_{2}]_{t}}{0.050~\text{M}}} \approx -1.387\] \[\frac{[\text{I}_{2}]_{t}}{0.050~\text{M}} = e^{-1.387}\] \[ [\text{I}_{2}]_{t} \approx 0.050~\text{M} \cdot e^{-1.387}\] Calculating the concentration, we find: \[[\text{I}_{2}]_{t} \approx 0.016~\text{M}\] After \(5.12~\text{s}\), the concentration of \(\mathrm{I}_{2}(g)\) remaining is approximately \(0.016~\text{M}\).