The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in \(\mathrm{A}\) and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} M\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{~s}\) has elapsed.

For a zero-order reaction, the rate law is given by: \[Rate = k[A]^0\] Since \([A]^0 = 1\), the rate law becomes: \[Rate = k\] Now, let's find the integrated rate law by connecting the rate of change of concentration of A to time. We can say: \[-\frac{d[A]}{dt} = k\] Integrating with respect to time, we get: \[\int_{[A]_0}^{[A]} -d[A] = \int_0^t kdt\] The integrated rate law for this zero-order reaction is: \[[A] = [A]_{0} - kt\]

The half-life of a reaction is the time it takes for the reactant concentration to reduce to half its initial value. In this case, we have: Half-life, \(t_{1/2} = \frac{[A]_{0}}{2}\) We can use the integrated rate law from Step 1 to calculate the half-life. Plugging in the given values, we get: \[\frac{[A]_{0}}{2} = [A]_{0} - kt_{1/2}\] Solve for \(t_{1/2}\): \[t_{1/2} = \frac{[A]_{0}}{2k}\] Substitute the values of \([A]_{0}\) and k: \[t_{1/2} = \frac{1.0 \times 10^{-3}}{2 \times 5.0 \times 10^{-2}}\] \[t_{1/2} = 1.0 \times 10^{-4} s\] The half-life of the reaction is \(1.0 \times 10^{-4} s\).

Since the reaction is A → B + C, after a certain time has elapsed, the amount of B produced is equal to the amount of A that has reacted. We can express this as: \[[B] = [A]_{0} - [A]\] We are given that the time elapsed is \(5.0 \times 10^{-3} s\). Using the integrated rate law from Step 1, we can find the concentration of A at this time: \[[A] = [A]_{0} - kt\] \[[A] = 1.0 \times 10^{-3} - 5.0 \times 10^{-2}\times 5.0 \times 10^{-3}\] \[[A] = 1.0 \times 10^{-3} - 2.5 \times 10^{-4}\] \[[A] = 0.75 \times 10^{-3}\] Now, we can find the concentration of B: \[[B] = [A]_{0} - [A]\] \[[B] = 1.0 \times 10^{-3} - 0.75 \times 10^{-3}\] \[[B] = 0.25 \times 10^{-3}\] The concentration of B after \(5.0 \times 10^{-3} s\) has elapsed is \(0.25 \times 10^{-3} M\).