Consider the reaction $$ 3 \mathrm{~A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}]^{2}[\mathrm{~B}][\mathrm{C}] $$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} M\) a. If after \(3.00 \mathrm{~min},[\mathrm{~A}]=3.26 \times 10^{-5} M\), calculate the value of \(k\). b. Calculate the half-life for this experiment. c. Calculate the concentration of \(\mathrm{B}\) and the concentration of \(\mathrm{A}\) after \(10.0 \mathrm{~min}\).

We are given the rate law and its initial concentrations, as well as the concentration of A after 3 minutes. To find k, we can first find the reaction rate using the given concentrations. The rate law is given as: $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ From the given data, we have: $$ -\frac{(3.26 \times 10^{-5} - 1.00 \times 10^{-4}) \mathrm{M}}{3.00 \mathrm{min}} = k(3.26 \times 10^{-5})^{2}(1.00)(1.00) $$ Divide both sides by \(((3.26 \times 10^{-5})^{2}(1.00)(1.00))\) to solve for k: $$ k = \frac{(-3.42 \times 10^{-5} )}{(3.00 \times 60 (3.26 \times 10^{-5})^2)} $$

Now, compute the value of k using the given information: $$ k = \frac{-3.42 \times 10^{-5}}{(3.00 \times 60 (3.26 \times 10^{-5})^2)} = 4.12 \times 10^{-4} \mathrm{M}^{-3} \mathrm{min}^{-1} $$

To calculate the half-life, we can use the integrated rate law for a second-order reaction: $$ t_{1/2} = \frac{1}{k[\mathrm{B}]_{0}[\mathrm{C}]_{0}} $$ Plugging the values for k, \([\mathrm{B}]_{0}\) and \([\mathrm{C}]_{0}\), we get: $$ t_{1/2} = \frac{1}{(4.12 \times 10^{-4})(1)(1)} = 2426.94 \mathrm{~min} $$

To find the concentrations of A and B after a given time, we will use the integrated rate law for a second-order reaction involving multiple reactants: $$ \frac{1}{[\mathrm{A}]_{t}} - \frac{1}{[\mathrm{A}]_{0}} = kt[\mathrm{B}]_{0}[\mathrm{C}]_{0} $$ Plugging the given values for t, k, and initial concentrations, we can solve for \([\mathrm{A}]_{t}\): $$ \frac{1}{[\mathrm{A}]_{t}} - \frac{1}{(1.00 \times 10^{-4})} = (4.12 \times 10^{-4})(10)(1)(1) $$ To find \([\mathrm{A}]_{t}\), we can rearrange the equation and solve: $$ [\mathrm{A}]_{t} = \frac{1}{\frac{1}{(1.00 \times 10^{-4})} - (4.12 \times 10^{-4})(10)} = 6.45 \times 10^{-5} \mathrm{M} $$ Finally, we can find the concentration of B after 10 minutes using stoichiometry: $$ [\mathrm{B}]_{t} = [\mathrm{B}]_{0} - \frac{1}{3}([\mathrm{A}]_{0} - [\mathrm{A}]_{t}) = 1.00 - \frac{1}{3}(1.00 \times 10^{-4} - 6.45 \times 10^{-5}) = 0.99988 \mathrm{M} $$ The concentration of A after 10 minutes is \(6.45 \times 10^{-5} M\), and the concentration of B is \(0.99988 M\).