13.29 The conversion of \( \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br} \) into \( \mathrm{C}_{5} \mathrm{H}_{10} \) follows first-order kinetics, with a rate constant of \( 0.385 \mathrm{~h}^{-1} \). If the initial concentration of \( \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br} \) is 0.125 M , find (a) the time at which the concentration will be \( 1.25 \times 10^{-3} \mathrm{M} \); and (b) the concentration after 3.5 h of reaction.

**(a)** For a first-order reaction, the concentration is given by \[ [A] = [A]_0 e^{-kt}. \] We need to find the time \( t \) when the concentration \( [A] \) is \( 1.25 \times 10^{-3} \, \mathrm{M} \). The initial concentration is \( [A]_0 = 0.125 \, \mathrm{M} \) and the rate constant is \( k = 0.385 \, \mathrm{h}^{-1} \). 1. Write the equation with the given values: \[ 1.25 \times 10^{-3} = 0.125 \, e^{-0.385t}. \] 2. Divide both sides by 0.125: \[ \frac{1.25 \times 10^{-3}}{0.125} = e^{-0.385t}. \] \[ 0.01 = e^{-0.385t}. \] 3. Take the natural logarithm of both sides: \[ \ln(0.01) = -0.385t. \] 4. Solve for \( t \): \[ t = \frac{-\ln(0.01)}{0.385}. \] 5. We know \( \ln(0.01) = -4.60517 \), so: \[ t = \frac{-(-4.60517)}{0.385} = \frac{4.60517}{0.385} \approx 11.97 \, \mathrm{h}. \] Thus, the time when the concentration reaches \( 1.25 \times 10^{-3} \, \mathrm{M} \) is approximately \( 12 \, \mathrm{h} \). --- **(b)** To find the concentration after \( 3.5 \, \mathrm{h} \), we use the same first-order decay equation: \[ [A] = 0.125 \, e^{-0.385(3.5)}. \] 1. Calculate the exponent: \[ -0.385 \times 3.5 = -1.3475. \] 2. Compute the exponential term: \[ e^{-1.3475} \approx 0.2596. \] 3. Multiply by the initial concentration: \[ [A] \approx 0.125 \times 0.2596 \approx 0.03245 \, \mathrm{M}. \] Thus, after \( 3.5 \, \mathrm{h} \), the concentration is approximately \( 0.032 \, \mathrm{M} \).