Question

The initial concentration of a drug decomposing according to first order kinetics is28 units/mL. The specific decomposition rate, k, obtained from an Arrhenius plot is 2.09 × 10^-5 hr^-1 at room temperature, 25°C. Previous experimentation has shown that when the concentration of the drug falls below 15 units/mL it is not sufficiently potent for use and should be removed from the market. What expiration date should be assigned to this product?

• A. Approximately 31,012.15 hours
• B. Approximately 32,225.56 hours
• C. Approximately 27.216.12 hours
• D. Approximately 29,869.23 hours

Answer 1

The Correct Option is D

To determine the expiration date of the drug based on its decomposition according to first-order kinetics, we apply the first-order kinetics formula:

\(C_t = C_0 e^{-kt}\)

where:

• \(C_t\) is the concentration at time \(t\). 
• \(C_0\) is the initial concentration.
• \(k\) is the rate constant.
• \(t\) is the time.

From the data given:

• Initial concentration, \(C_0 = 28\) units/mL.
• Final concentration, \(C_t = 15\) units/mL (lowest effective concentration).
• Rate constant, \(k = 2.09 \times 10^{-5} \text{ hr}^{-1}\).

Substitute these values into the formula and solve for \(t\):

\(15 = 28 e^{-2.09 \times 10^{-5} t}\)

Rearranging for \(e^{-2.09 \times 10^{-5} t}\):

\(\frac{15}{28} = e^{-2.09 \times 10^{-5} t}\)

Taking the natural logarithm of both sides:

\(\ln{\left(\frac{15}{28}\right)} = -2.09 \times 10^{-5} t\)

Calculate the natural logarithm:

\(\ln{\left(\frac{15}{28}\right)} = -0.6276\)

Substitute this result back into the equation and solve for \(t\):

\(t = \frac{-0.6276}{-2.09 \times 10^{-5}}\)

Compute \(t\):

\(t \approx 29,869.23 \text{ hours}\)

Thus, the expiration date of the drug should be approximately 29,869.23 hours from its initial concentration. Therefore, the correct option is:

• Approximately 29,869.23 hours