Question

The first-order rate constant for the decomposition of a certain hormone in water at \(25^{\circ} \mathrm{C}\) is \(3.42 \times 10^{-4}\) day \(^{-1}\). (a) If a \(0.0200 \mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months, what will its concentration be at the end of that period? (b) How long will it take for the concentration of the solution to drop from \(0.0200 \mathrm{M}\) to \(0.00350 \mathrm{M}\) ? (c) What is the half-life of the hormone?

Short answer

Answer: The concentration of the hormone after 2 months is 0.0181 M. The time it will take for the concentration to drop to 0.00350 M is approximately 124.11 days. The half-life of the hormone is approximately 2026.54 days.

Step-by-step solution

Convert months to days

Since the rate constant is given in days\(^{-1}\), we need to convert 2 months into days: \(2 \times 30 = 60\) days.

Use the first-order equation to find the concentration

Now we can use the equation for the first-order reaction: \(C = C_0e^{-kt}\) Plugging in the values of \(C_0 = 0.0200 \ M\), \(k = 3.42 \times 10^{-4}\) day\(^{-1}\) and \(t = 60\) days: \(C = 0.0200e^{-(3.42 \times 10^{-4})(60)}\) After calculating the concentration, we get: \(C = 0.0181 \mathrm{M}\) #b) Determine the time to reach a lower concentration#

Rearrange the first-order equation for time

To find the time required for the concentration of the hormone to drop from \(0.0200 \mathrm{M}\) to \(0.00350 \mathrm{M}\), we need to find \(t\). Rearrange the above equation for \(t\): \(t = -\frac{1}{k}\ln(\frac{C}{C_0})\)

Plug in values and calculate

Now plug in the values for \(C_0 = 0.0200 \mathrm{M}\), \(C = 0.00350 \mathrm{M}\) and \(k = 3.42 \times 10^{-4} \mathrm{day}^{-1}\): \(t = -\frac{1}{3.42 \times 10^{-4}} \ln(\frac{0.00350}{0.0200})\) After calculating the time, we get: \(t \approx 124.11 \ \mathrm{days}\) #c) Find the half-life of the hormone#

Use the half-life equation for first-order reactions

The half-life of a first-order reaction is given by the equation: \(t_{1/2} = \frac{ln(2)}{k}\)

Calculate the half-life

Plugging in the value of \(k = 3.42 \times 10^{-4} \mathrm{day}^{-1}\), we can calculate the half-life: \(t_{1/2} = \frac{ln(2)}{3.42 \times 10^{-4}}\) After calculating the half-life, we get: \(t_{1/2} \approx 2026.54 \ \mathrm{days}\)