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 \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[\mathrm{A}]=3.26 \times 10^{-5} M,\) calculate the value of \(k .\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(B\) and the concentration of A after 10.0 min.

We are given the initial concentration of A, B, and C, as well as the final concentration of A after 3 minutes. We can use this information and the rate law to find the value of k. We can set up the rate equation as follows: $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ Plug in the given values: $$ -\frac{(3.26 \times 10^{-5} \text{ M} - 1.00 \times 10^{-4} \text{ M})}{(3.00 \text{ min})} =k\left(1.00 \times 10^{-4} \text{ M}\right)^{2}(1.00 \mathrm{M})(1.00 \mathrm{M}) $$ Now, we can solve for value of k: $$ k=\frac{1}{12}\left(\frac{7.74 \times 10^{-5} \text{ M}}{(1.00 \times 10^{-4})^{2} \text{ M}^2}\right)= 6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1}. $$ So, the value of k is \(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1}\).

We can use the equation for the half-life of a reaction, defined as: $$ t_{1/2} = \frac{1}{k [\mathrm{A}]_0 [\mathrm{B}]_0 [\mathrm{C}]_0} $$ Plug in the values we have found so far: $$ t_{1/2} = \frac{1}{(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1})(1.00 \times 10^{-4} \mathrm{M})(1.00 \mathrm{M})(1.00 \mathrm{M})} = 15.4 \, \mathrm{min} $$ The half-life for this experiment is 15.4 minutes.

To find the concentration of A and B at 10.0 minutes, we can use the integrated rate law for this reaction, which is: $$ [\mathrm{A}]_{t}=\frac{1}{kt+1 /[\mathrm{A}]_{0}} $$ Plug in the values for A: $$ [\mathrm{A}]_{10}=\frac{1}{(6.48 \times 10^{-3}\, \mathrm{M}^{-2} \mathrm{min}^{-1})(10 \, \mathrm{min})+1 /(1.00 \times 10^{-4}\, \mathrm{M})}= 5.16 \times 10^{-5}\, \mathrm{M} $$ Now, we can find the concentration of B at 10.0 minutes using stoichiometry. Since we started with 1.00 M of B and the stoichiometric ratio of A to B is 3:1, we can find the change in concentration of B: $$ \Delta[\mathrm{B}] = \frac{1}{3}(\Delta[\mathrm{A}]) = -\frac{1}{3}(5.16 \times 10^{-5} \text{ M} - 1.00 \times 10^{-4} \text{ M}) = 1.61\times10^{-5} \mathrm{M} $$ To find the concentration of B at 10.0 minutes, we can subtract this change from the initial concentration of B: $$ [\mathrm{B}]_{10} = [\mathrm{B}]_{0} - 1.61\times10^{-5} \mathrm{M} = 1.00 \mathrm{M} - 1.61\times10^{-5} \mathrm{M}= 0.999984\, \mathrm{M} $$ Therefore, the concentration of A after 10.0 minutes is \(5.16 \times 10^{-5} \mathrm{M}\), and the concentration of B is \(0.999984\, \mathrm{M}\).