91Ó°ÊÓ

Based on the information provided in the question, we see that the plot of \(1 /[\mathrm{C}_{4}\mathrm{H}_{6}]\) is linear, while the other plots are curved. This tells us that the reaction is second-order with respect to butadiene, with a rate law of the form: Rate = k\([\mathrm{C}_{4}\mathrm{H}_{6}]^{2}\) So, the rate law for the reaction is: Rate = k\([\mathrm{C}_{4}\mathrm{H}_{6}]^{2}\)

For second-order reactions, the integrated rate law is: \(1 /[\mathrm{C}_{4}\mathrm{H}_{6}]_{t} - 1 /[\mathrm{C}_{4}\mathrm{H}_{6}]_{0} = kt\) We know that at half-life (t = t\(_{1/2}\)), there will be 1/2 of the initial concentration remaining: \([\mathrm{C}_{4}\mathrm{H}_{6}]_{t} = \frac{[\mathrm{C}_{4}\mathrm{H}_{6}]_{0}}{2}\) Substituting this into the integrated rate law and simplifying: \(2 /[\mathrm{C}_{4}\mathrm{H}_{6}]_{0} - 1 /[\mathrm{C}_{4}\mathrm{H}_{6}]_{0} = kt_{1/2}\) Thus, the half-life equation for this second-order reaction is: \(t_{1/2} = \frac{1}{k[\mathrm{C}_{4}\mathrm{H}_{6}]_{0}}\)

Let's define x as the number of half-lives needed for the concentration of butadiene to decrease to 3.1% of its initial concentration. Then, we can calculate x by considering the decrease in concentration after each half-life: \([\mathrm{C}_{4}\mathrm{H}_{6}]_{final} = [\mathrm{C}_{4}\mathrm{H}_{6}]_{0} \left(\frac{1}{2}\right)^{x}\) We want to find the value of x for which: \([\mathrm{C}_{4}\mathrm{H}_{6}]_{final} = 0.031[\mathrm{C}_{4}\mathrm{H}_{6}]_{0}\) Now, plug the final concentration value into the equation: \(0.031[\mathrm{C}_{4}\mathrm{H}_{6}]_{0} = [\mathrm{C}_{4}\mathrm{H}_{6}]_{0} \left(\frac{1}{2}\right)^{x}\) Divide both sides by \([\mathrm{C}_{4}\mathrm{H}_{6}]_{0}\) to get: \(0.031 = \left(\frac{1}{2}\right)^{x}\) To solve for x, we can take the base-2 logarithm of both sides: \(\log_{2}(0.031) = x\) Finally, calculate the value of x: