First-Order Vitamin Degradation: Testing a 30-Day Label Claim
A pharmacy student is checking the stability of a liquid vitamin preparation that decomposes by a first-order reaction. From earlier assays, the rate constant at 25 °C is 5.0 × 10^-6 min^-1 and at 35 °C is 2.0 × 10^-5 min^-1. The label says the preparation retains at least 90% of its stated potency for 30 days when stored at 25 °C, but the student suspects refrigeration at 5 °C would extend the shelf life. The preparation is kept in the dark, and no other degradation route is considered. The gas constant R = 8.314 J K^-1 mol^-1. The student has data only at the two temperatures and plans to use the Arrhenius equation. The supervisor asks for an estimate of the activation energy, the time for 10% loss at 5 °C, and a judgement on whether the label claim is supported. The student must also state one assumption that could make the prediction unreliable.
1. Calculate the activation energy from the two rate constants, then estimate the rate constant at 5 °C and the time for 10% loss at that temperature.
Read the worked answer
Use ln(k2/k1) = (Ea/R)(1/T1 - 1/T2).
T1 = 298 K, T2 = 308 K; k1 = 5.0 × 10^-6 min^-1, k2 = 2.0 × 10^-5 min^-1.
ln(k2/k1) = ln 4 = 1.386.
1/298 - 1/308 = 1.089 × 10^-4 K^-1.
Ea = 1.386 × 8.314 / (1.089 × 10^-4) = 1.058 × 10^5 J mol^-1 ≈ 106 kJ mol^-1.
For 5 °C, T = 278 K: ln(k278/k298) = (Ea/R)(1/298 - 1/278).
1/298 - 1/278 = -2.414 × 10^-4 K^-1.
Ea/R = 1.058 × 10^5 / 8.314 = 1.273 × 10^4 K.
Product = -3.073, so k278 = 5.0 × 10^-6 × e^-3.073 = 2.32 × 10^-7 min^-1.
For first order, t90 = ln(100/90)/k = 0.10536 / k.
t90(5 °C) = 0.10536 / (2.32 × 10^-7) = 4.54 × 10^5 min ≈ 315 days.
2. Check the label claim at 25 °C. Does the preparation retain at least 90% potency for 30 days? What maximum rate constant would be needed?
Read the worked answer
At 25 °C, k = 5.0 × 10^-6 min^-1.
t90 = 0.10536 / (5.0 × 10^-6) = 2.11 × 10^4 min.
Convert to days: 2.11 × 10^4 / (24 × 60) = 14.6 days.
Thus 10% loss is predicted in about 15 days at 25 °C, not 30 days. The label claim is not supported by these kinetic data.
For 30 days, t90 = 30 × 24 × 60 = 43,200 min. Required k = 0.10536 / 43,200 = 2.44 × 10^-6 min^-1. The measured k is about twice this value, so decomposition is too fast at 25 °C. Refrigeration at 5 °C would slow degradation markedly, consistent with the estimated 315-day t90 at 5 °C, but the label must be based on the stated storage condition.
3. State one assumption in the Arrhenius extrapolation and explain how it could make the 5 °C prediction unreliable.
Read the worked answer
The calculation assumes that the Arrhenius plot is linear over the range 5-35 °C, so the activation energy obtained from only two temperatures applies at 5 °C. If the degradation mechanism changes with temperature, or if the reaction becomes diffusion-controlled or medium-dependent at low temperature, Ea would not be constant and the extrapolated k at 5 °C could be wrong.
A related assumption is that the reaction remains first order throughout and that no other degradation route, such as oxidation or photolysis, becomes significant. The prediction is therefore a useful estimate, not a substitute for real-time or accelerated stability testing.