The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant at 520 nm . The reaction occurs in a 1.29-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5700 cm-1M-1 at 520 nm .

a) Calculate the initial concentration of the colored reactant if the absorbance is 0.564 at the beginning of the reaction.

b) The absorbance falls to 0.254 at 30.0 min . Calculate the rate constant in units of s-1.

c) Calculate the half-life of the reaction. (in sec)

d) How long does it take for the absorbance to fall to 0.104? (in sec)

See your post below.

To solve these questions, we can use the Beer-Lambert Law, which relates the absorbance of a sample to the concentration of the absorbing species.

The equation is given as: A = εcl, where A is the absorbance, ε is the molar extinction coefficient, c is the concentration, and l is the path length.

a) We are given the absorbance (A) as 0.564. Using the Beer-Lambert Law, we have A = εcl. Rearranging the equation to solve for the concentration (c), we get: c = A / (εl). Plugging in the values given: A = 0.564, ε = 5700 cm-1M-1, and l = 1.29 cm, we can calculate the concentration (c).

c = 0.564 / (5700 cm-1M-1 * 1.29 cm)
c ≈ 0.087 M

Therefore, the initial concentration of the colored reactant is approximately 0.087 M.

b) We are given the initial absorbance (A₀) as 0.564 and the absorbance at 30.0 min (Aₜ) as 0.254. The rate constant (k) can be found using the equation: ln(A₀/Aₜ) = kt. Rearranging the equation to solve for the rate constant (k), we get: k = ln(A₀/Aₜ) / t. Plugging in the values given: A₀ = 0.564, Aₜ = 0.254, and t = 30.0 min (which must be converted to seconds: 30.0 min * 60 s/min), we can calculate the rate constant (k).

k = ln(0.564/0.254) / (30.0 min * 60 s/min)
k ≈ 0.00331 s-1

Therefore, the rate constant is approximately 0.00331 s-1.

c) The half-life (t₁/₂) of a first-order reaction can be calculated using the equation: t₁/₂ = ln(2) / k. Plugging in the value of the rate constant (k) calculated in part b), we can calculate the half-life (t₁/₂).

t₁/₂ = ln(2) / 0.00331 s-1
t₁/₂ ≈ 209.95 s

Therefore, the half-life of the reaction is approximately 209.95 seconds.

d) To find out how long it takes for the absorbance to fall to 0.104, we can use the equation: ln(A/A₀) = -kt. Rearranging the equation to solve for time (t), we get: t = -ln(A/A₀) / k. Plugging in the values given: A = 0.104, A₀ = 0.564, and k = 0.00331 s-1, we can calculate the time (t).

t = -ln(0.104/0.564) / 0.00331 s-1
t ≈ 603.47 s

Therefore, it takes approximately 603.47 seconds for the absorbance to fall to 0.104.