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 \(1.00-\mathrm{cm}\) 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 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

Calculate Initial Concentration

To find the initial concentration of the colored reactant, we can use the formula A = εcl. We have A = 0.605, ε = 5.60 × 10³ M⁻¹ cm⁻¹, and l = 1 cm. Rearrange the formula to solve for the initial concentration (c₀): c₀ = A / (ε × l) = 0.605 / (5.60 × 10³ M⁻¹ cm⁻¹ × 1 cm) Now, substitute the given values and solve for c₀: c₀ = 0.605 / (5.60 × 10³) = 1.08 × 10⁻⁴ M.

02

Calculate Rate Constant

First, we need to find the concentration after 30 minutes, when the absorbance has fallen to 0.250. We can use the same formula from the previous step: c₁ = A₁ / (ε × l) = 0.250 / (5.60 × 10³ M⁻¹ cm⁻¹ × 1 cm) c₁ = 0.250 / (5.60 × 10³) = 4.46 × 10⁻⁵ M. Now, we can use the first-order reaction equation to find the rate constant k: ln(c₀ / c₁) = kt where t is the time in seconds. In this case, t = 30 min × 60 s/min = 1800 s. Solve for k: k = ln(c₀ / c₁) / t = ln(1.08 × 10⁻⁴ / 4.46 × 10⁻⁵) / 1800 k ≈ 1.135 × 10⁻⁴ s⁻¹.

03

Calculate Half-Life

Now that we have the rate constant k, we can calculate the half-life t_(1/2) using the formula: t_(1/2) = ln(2) / k = ln(2) / 1.135 × 10⁻⁴ t_(1/2) ≈ 6,108 s ≈ 102 min.

04

Time for Absorbance to Fall to 0.100

To find the time it takes for the absorbance to fall to 0.100, first find the final concentration c₂: c₂ = A₂ / (ε × l) = 0.100 / (5.60 × 10³ M⁻¹ cm⁻¹ × 1 cm) c₂ = 0.100 / (5.60 × 10³) = 1.786 × 10⁻⁵ M. Use the first-order reaction equation again and solve for time t: t = ln(c₀ / c₂) / k = ln(1.08 × 10⁻⁴ / 1.786 × 10⁻⁵) / 1.135 × 10⁻⁴ t ≈ 5470 s ≈ 91.17 min.

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