Question

The half-life of an anticancer drug is 170 days at 25 oC. Degradation follows the first-order kinetics. When the drug is stored at 5 oC, the degradation rate decreases to one third (1/3) of the rate at 25 oC. What is the shelf life of the drug at 5 oC ?

Answer 1

1694 days

Step-by-step explanation:

In first-order kinetics, the rate is proportional to the amount.

dA/dt = kA

For first-order kinetics, the rate k can be found using the half-life:

t₁,₂ = (ln 2) / k

In other words, the half-life is inversely proportional with the rate.

At the lower temperature, the rate is reduced to a third, so the half-life increases by a factor of 3. Meaning that the new half-life is 170 × 3 = 510 days.

The "shelf life" is the time it takes to reduce the initial amount to 10%. We can solve for this using the half-life equation.

A = A₀ (½)^(t / t₁,₂)

A₀/10 = A₀ (½)^(t / 510)

1/10 = (½)^(t / 510)

ln(1/10) = (t / 510) ln(½)

ln(10) = (t / 510) ln(2)

ln(10) / ln(2) = t / 510

t = 510 ln(10) / ln(2)

t ≈ 1694