The following data were obtained for the dimerization of 1,3 -butadiene, \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{g}),\) at 600 K: \(t=0 \min ,\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0169 \mathrm{M} ; 12.18 \mathrm{min}, 0.0144 \mathrm{M} ; 24.55 \mathrm{min}, 0.0124 \mathrm{M} ; 42.50 \mathrm{min}, 0.0103 \mathrm{M}, 68.05 \min , 0.00845 \mathrm{M}.\) (a) What is the order of this reaction? (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.00423 \mathrm{M} ?\) (d) At what time would \(\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]=0.0050 \mathrm{M} ?\)

In this step, use the concentration and time data provided to ascertain whether the chemical reaction is zeroth, first or second order. Calculate the change in concentration per unit time for any two consecutive data points. If this value remains the same, its order of reaction is zero; if this value variation is linear, then it is first order; if this variation is inversely proportional to the concentration, order of reaction is second. Here, comparing concentration changes over time it could be seen that the rate of reaction decreases linearly over time, therefore, this reaction is first order.

We know that the integrated first order rate law is \(\ln [A] = -kt + \ln [A_0]\), where \([A]\) is the concentration of the reactant, \(k\) is the rate constant, \(t\) is the time, and \([A_0]\) is the initial concentration. By rearranging, we have \(k = (\ln[A_0] - \ln[A])/t)\). From the data, choose two points and substitute the values into this formula to solve for \(k\). Since the reaction is first order and constant irrespective of which points are chosen, picking any two points should yield the same value for \(k\). Let's choose the first and last data points: \(t = 0 \min, [C_4H_6] = 0.0169 M\) and \(t = 68.05 \min, [C_4H_6] = 0.00845 M\). Therefore, calculating \(k = (\ln[0.0169] - \ln[0.00845])/68.05\) gives us \(k = 0.0924 \min^{-1}\).

Use the first-order integrated rate equation with \(A = 0.00423 M\) to calculate the time at which this concentration is achieved. From the formula, \(t = (\ln[A_0] - \ln[A]) / k\). Inserting the values, we get \(t = (\ln[0.0169] - \ln[0.00423])/0.0924\) which results in \(t \approx 102.54\) minutes.

Similarly, substitute \(A = 0.005 M\) into the rate equation and solve for \(t\). On substituting the values, we get \(t = (\ln[0.0169] - \ln[0.005])/0.0924\) which gives us \(t \approx 85.33\) minutes.