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 assuming \([\mathrm{B}]_{0}=0\).

For a zero-order reaction, the rate law is given by: \( rate = k[A]^0 = k\) The relationship between the rate of reaction and the concentration of reactants is given by the integrated rate law. For a zero-order reaction, the integrated rate law is derived as: \( [A] = [A]_0 - kt \) where [A] is the concentration of A at time t, [A]₀ is the initial concentration of A, k is the rate constant, and t is the time in seconds.

Half-life (t₁/₂) is the amount of time it takes for the concentration of a reactant to decrease to one-half of its original concentration. For the half-life of a zero-order reaction, the relation is given as: \( t_{1/2} = \frac{[A]_0}{2k} \) Using the given values, we can calculate the half-life: \( [A]_0 = 1.0 \times 10^{-3} M \) \( k = 5.0 \times 10^{-2} \frac{mol}{L \cdot s} \) \( t_{1/2} = \frac{1.0 \times 10^{-3} M}{2(5.0 \times 10^{-2} \frac{mol}{L \cdot s})} \) Now we calculate the half-life: \( t_{1/2} ≈ 0.01s \)

To find the concentration of B after the given time, we first need to find the concentration of A at that time using the integrated rate law. \( t = 5.0 \times 10^{-3} s \) \( [A] = [A]_0 - kt \) Substituting the given values: \( [A] = 1.0 \times 10^{-3} M - (5.0 \times 10^{-2} \frac{mol}{L \cdot s})(5.0 \times 10^{-3}s) = 1.0 \times 10^{-4} M \) Since the stoichiometry of the reaction is 1:1 it means that for every mole of A consumed, 1 mole of B is produced. Initially, [B]₀ was given as 0. As a result, we can calculate the concentration of B using [B]=[A]₀-[A]. \( [B] = [A]_0 - [A] = 1.0 \times 10^{-3} M - 1.0 \times 10^{-4} M = 9.0 \times 10^{-4} M \) So, the concentration of B after \(5.0 \times 10^{-3} s\) has elapsed is \(9.0 \times 10^{-4} M\).