Consider the following hypothetical aqueous reaction: $\mathrm{A}(a q) \rightarrow \mathrm{B}(a q)\(. A flask is charged with \)0.065 \mathrm{~mol}$ of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10 -min interval in units of \(M /\) s. (c) Between \(t=0 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

Since the reaction is A → B, the number of moles of B at each time can be determined by subtracting the number of moles of A remaining from the initial number of moles of A (0.065 mol). At time 0 min: Moles of B = Initial moles of A - Moles of A = 0.065 - 0.065 = 0 mol At time 10 min: Moles of B = 0.065 - 0.051 = 0.014 mol At time 20 min: Moles of B = 0.065 - 0.042 = 0.023 mol At time 30 min: Moles of B = 0.065 - 0.036 = 0.029 mol At time 40 min: Moles of B = 0.065 - 0.031 = 0.034 mol

To find the average rate of disappearance of A, calculate the change in concentration of A (∆[A]) divided by the change in time (∆t) for each time interval. Concentration is obtained by dividing moles by volume (in L). Note: 100.0 mL = 0.100 L At 0-10 min: Average rate = (0.051 - 0.065) mol / (0.100 L * 10 min * (1/60) h) = -0.014 mol / (1/6 h) = -0.084 M/h At 10-20 min: Average rate = (0.042 - 0.051) mol / (0.100 L * 10 min * (1/60) h) = -0.009 mol / (1/6 h) = -0.054 M/h At 20-30 min: Average rate = (0.036 - 0.042) mol / (0.100 L * 10 min * (1/60) h) = -0.006 mol / (1/6 h) = -0.036 M/h At 30-40 min: Average rate = (0.031 - 0.036) mol / (0.100 L * 10 min * (1/60) h) = -0.005 mol / (1/6 h) = -0.030 M/h Convert the average rates from M/h to M/s by multiplying by (1/3600) s/h: At 0-10 min: -0.084 M/h * (1/3600) s/h = -2.33 x 10^-5 M/s At 10-20 min: -0.054 M/h * (1/3600) s/h = -1.5 x 10^-5 M/s At 20-30 min: -0.036 M/h * (1/3600) s/h = -1.0 x 10^-5 M/s At 30-40 min: -0.030 M/h * (1/3600) s/h = -8.33 x 10^-6 M/s

To find the average rate of appearance of B, calculate the change in concentration of B (∆[B]) divided by the change in time (∆t) for the time interval 0-30 min. ∆[B] = (0.029 mol - 0.000 mol) / 0.100 L = 0.29 M ∆t = 30 min * (1/60) h = 0.5 h Average rate of appearance of B = ∆[B] / ∆t = 0.29 M / 0.5 h = 0.58 M/h Convert the average rate from M/h to M/s by multiplying by (1/3600) s/h: 0.58 M/h * (1/3600) s/h = 1.61 x 10^-4 M/s