Question

All of these problems involve preparation and use of a vaccine solution that contains an active agent in an aqueous solution. The active agent in the vaccine degrades in aqueous solutions with first-order kinetics. The half-life of the active agent in water 25 degree C is 8.4 hours. The vaccine should be administered at a concentration of at least 20 mu g/L to be fully effective. Calculate the rate constant, k_1, at 25 degree C. Be sure to include the units. Suppose a stock solution of the vaccine with a concentration of 25 mu g/L is prepared in a clinic at 8:30 am. The clinic plans to store this solution at 25 degree C and use it for all of the patients that are seen for morning appointments between 9:00 am - 12:00 noon. Is this a good practice? Justify your answer. Will the vaccine be effective for all of these patients? If not, at what time will the concentration of the active agent fall below the effective level? The half-life of the active agent is measured at 4 degree C and found to be 11.5 hours. What is the activation energy for the degradation reaction? A stock solution of the vaccine is prepared at 8:30 am and stored in a refrigerator at 4 degree C. It will be used for all of the patients with morning appointments between 9:00 am - 12:00 noon. However, the last patient of the morning was delayed and did not arrive until 12:30 pm. What is the concentration of the active agent in the stock solution at that time? Can the vaccine still be used?

I just need to figure out number 3 and 4. thank you!!

Answer 1

Data.

t_1/2 = 0.693/k

k = rate/[A]

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[A] = 20 ng/L

t_1/2 = 8.4 hrs

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Answer.

3. From half life equation, we know that a first order reaction is independent from initial concentration, this is a measure that approximates the magnitude of a rate constant, the lower the half life, and the greater will be K.

t_1/2 = 0.693/k

@ 25°C = 298 K

30240 s = 0.693/k

k₁ = 2.29x10^-5 s^-1

^--

@ 4°C = 277 K

41400 s = 0.693/k

k₂ = 1.67x10^-5 s^-1

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The dependent of a rate constant regard to temperature is expressed by Arrhenius's equation:

k = Ae^-Ea/RT

Where;

Ea = activation energy (KJ/mol)

A = frequency factor

From this equation we can calculate the relation between two rate constant at different temperatures. Applying ln in both sides:

lnk₁ = lnA - Ea/RT₁

lnk₂ = lnA - Ea/RT₂

by subtracting lnk₂ from lnk₁:

ln k₁/k₂ = Ea/R(T₁-T₂/T₁T₂)

ln(2.29x10^-5 s^-1/1.67x10^-5 s^-1) = Ea/8.314 J/K.mol (298 - 277/298*277)

Ea = 44845.7958 J/mol = 44.85 KJ/mol

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4. If vaccine is prepared at 8:30 and will be used until 12:00 then we have 3 hours and a half (11880 s) for its usage. From rate equation we can obtain:

rate = k[A]

-Δ[A]/Δt = k[A]

From here we can demonstrate:

ln[A]/[Ao] = -kt

ln[A] = -kt + ln [Ao]

ln[A] = -(1.67x10^-5 s^-1)(11880 s) + ln(20 ng/L)

e^ln[A] = e^2.7986

[A] = 16.4228 ng/mol

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The vaccine should not be fully effective because the concentration is lower than the ideal dosage.