The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of $5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\( at \)520 \mathrm{nm} .$ (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at $30.0 \mathrm{~min} .\( Calculate the rate constant in units of \)\mathrm{s}^{-1}$. (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

Step by step solution

01

Part (a): Initial concentration

First, apply the Beer-Lambert Law to find the initial concentration: \(A = \epsilon lc\), where: A - the absorbance \( \epsilon \) - the extinction coefficient (\(5.60 \times 10^{3}\, \mathrm{M^{-1}cm^{-1}}\)) l - path length (1.00 cm) c - concentration of the solution (in Molarity) Rearrange the equation to find the concentration: \(c = \frac{A}{\epsilon l}\) Plug the values from the exercise: \(c = \frac{0.605}{(5.60 \times 10^{3}\,\mathrm{M^{-1}cm^{-1}})(1.00\, \mathrm{cm})}\) Calculate the initial concentration: \(c = 1.08 \times 10^{-4}\,\mathrm{M}\)

02

Part (b): Rate constant

Next, apply the first-order reaction kinetics to find the rate constant: \(A_t = A_0e^{-kt}\), where: \(A_t\) - the absorbance at a given time (0.250) \(A_0\) - initial absorbance (0.605) k - rate constant (in s^-1) t - time elapsed (30.0 min = 1800 s) Rearrange the equation to solve for k: \(k = \frac{-\ln{\frac{A_t}{A_0}}}{t}\) Plug values from the exercise and the initial concentration: \(k = \frac{-\ln{\frac{0.250}{0.605}}}{1800\,\mathrm{s}}\) Calculate the rate constant: \(k = 1.14 \times 10^{-3}\,\mathrm{s^{-1}}\)

03

Part (c): Half-life

Now calculate the half-life (t_1/2) using the first-order reaction formula: \(t_{1/2} = \frac{\ln{2}}{k}\) Plug the value of k from the previous part: \(t_{1/2} = \frac{\ln{2}}{1.14 \times 10^{-3}\,\mathrm{s^{-1}}}\) Calculate the half-life: \(t_{1/2} = 607\,\mathrm{s}\)

04

Part (d): Time for the absorbance to fall to 0.100

Finally, calculate the time it takes for the absorbance to fall to 0.100 using the first-order reaction kinetics equation: Rearrange the equation to solve for time: \(t = \frac{-\ln{\frac{A_t}{A_0}}}{k}\) Plug the values from the exercise and the rate constant: \(t = \frac{-\ln{\frac{0.100}{0.605}}}{1.14 \times 10^{-3}\,\mathrm{s^{-1}}}\) Calculate the time: \(t = 4373\,\mathrm{s}\)

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