The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in 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 $[\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .$ 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 depends only on the rate constant (k) and is independent of the concentration of reactants. The rate can be written as: \(rate = -\frac{d[\mathrm {A}]}{dt} = k\) To find the integrated rate law, we need to integrate this expression with respect to time: \(\int_{[\mathrm{A}]_0}^{[\mathrm{A}]} -\frac{d[\mathrm{A}]}{k} = \int_0^t dt\) Now, let's integrate both sides: \(-\frac{[\mathrm{A}] - [\mathrm{A}]_0}{k} = t\) This is the integrated rate law for a zero-order reaction. We can rewrite it to isolate the final concentration of A: \([\mathrm{A}] = -kt + [\mathrm{A}]_0\)

The half-life (t_1/2) is the time it takes for the concentration of a reactant to decrease to half its initial value. We can find the half-life by setting \([\mathrm{A}] = \frac{1}{2}[\mathrm{A}]_0\) in the integrated rate law: \(\frac{1}{2}[\mathrm{A}]_0 = -kt_{1/2} + [\mathrm{A}]_0\) Now, solve for \(t_{1/2}\): \(t_{1/2} = \frac{([\mathrm{A}]_0 - \frac{1}{2}[\mathrm{A}]_0)}{k}\) We are given the rate constant, k, and the initial concentration of A, \([\mathrm{A}]_0\), so we can plug in these values to find the half-life: \(t_{1/2} = \frac{(1.0 \times 10^{-3} \mathrm{M} - \frac{1}{2}(1.0 \times 10^{-3} \mathrm{M}))}{(5.0 \times 10^{-2} \mathrm{mol/L \cdot s})}\) \(t_{1/2} = 1.0 \times 10^{-2} \mathrm{s}\)

The stoichiometry of the reaction is such that for every mole of A consumed, one mole of B is produced: \(\mathrm{A} \longrightarrow \mathrm{B} + \mathrm{C}\) This means that the change in the concentration of B is equal to the change in the concentration of A. To find the concentration of B after 5.0 x 10^{-3} s, we first find the remaining concentration of A at this time using the integrated rate law: \([\mathrm{A}] = -kt + [\mathrm{A}]_0 = - (5.0 \times 10^{-2} \mathrm{mol/L \cdot s})(5.0 \times 10^{-3} \mathrm{s}) + (1.0 \times 10^{-3} \mathrm{M})\) \([\mathrm{A}] = 7.5 \times 10^{-4} \mathrm{M}\) Since the change in the concentration of B is equal to the change in the concentration of A, we can find the concentration of B as: \([\mathrm{B}] = [\mathrm{A}]_0 - [\mathrm{A}] = (1.0 \times 10^{-3} \mathrm{M}) - (7.5 \times 10^{-4} \mathrm{M})\) \([\mathrm{B}] = 2.5 \times 10^{-4} \mathrm{M}\)