Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is $3.8 \times 10^{-3} \mathrm{M}$ a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

From the given concentrations and time, let's calculate the rate of the reaction. Initial A concentration: \([A]_0 = 1.0 \times 10^{-2} M\) Final A concentration after 8.0 seconds: \([A]_8 = 3.8 \times 10^{-3} M\) Rate of the reaction: \[Rate = -\frac{\Delta [\mathrm{A}]}{\Delta t}\] We have: \(\Delta [\mathrm{A}] = [A]_8 - [A]_0 = 3.8 \times 10^{-3} - 1.0 \times 10^{-2}\) \(\Delta t = 8.0 s\) Now, let's calculate the rate: \(Rate = -\frac{3.8 \times 10^{-3} - 1.0 \times 10^{-2}}{8.0}\)

Using the rate law, we can calculate the value of k: \[Rate = k[\mathrm{A}][\mathrm{B}]^{2} \] We have: \([\mathrm{A}] = 1.0 \times 10^{-2} M\) \([\mathrm{B}] = 3.0 M\) \(Rate\) (calculated in Step 1) Now, solving for k: \(k = \frac{Rate}{[\mathrm{A}][\mathrm{B}]^{2}}\)

To calculate the half-life for this reaction, we must use the following formula: \(t_{1/2} = \frac{1}{k[\mathrm{B}]^{2}}\) We have the value of k (calculated in Step 2) and the concentration of \([\mathrm{B}]\). So, plug in the values and solve for \(t_{1/2}\).

To calculate the concentration of A after 13.0 seconds, we will use the following formula: \([A]_{t} = [A]_{0}\mathrm{e}^{-k[\mathrm{B}]^{2}\mathrm{t}}\) We have: \([A]_{0} = 1.0 \times 10^{-2} M\) \(k\) (calculated in Step 2) \([\mathrm{B}] = 3.0 M\) \(t = 13.0 s\) Now, plug in the values and solve for \([A]_{13}\).

To calculate the concentration of C after 13.0 seconds, we need to use stoichiometry. Recall the balanced equation for the reaction: \[\mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E}\] From the balanced equation: - 1 mole of A reacts with 2 moles of C - So, the number of moles of C reacted is equal to double the number of moles of A reacted. Let \(\Delta [\mathrm{C}]\) be the decrease in concentration for C after 13.0 seconds: \(\Delta [\mathrm{C}] = 2(\Delta [\mathrm{A}])\) To find \([\mathrm{C}]_{13}\), we subtract \(\Delta [\mathrm{C}]\) from the initial concentration of C: \([\mathrm{C}]_{13} = [\mathrm{C}]_{0} - \Delta [\mathrm{C}]\) \([\mathrm{C}]_{13} = 2.0 M - 2(\Delta [\mathrm{A}])\) Here, \(\Delta [\mathrm{A}]\) is the difference in A concentration between time 0 and 13 seconds. We have calculated \([A]_{13}\) in Step 4.