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In: Chemistry

All of these problems involve preparation and use of a vaccine solution that contains an active...

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 25o C is 8.4 hours. The vaccine should be administered at a concentration of at least 20 μg/L to be fully effective.

1. Calculate the rate constant, k1, at 25o C. Be sure to include the units.

2. Suppose a stock solution of the vaccine with a concentration of 25 μg/L is prepared in a clinic at 8:30 am. The clinic plans to store this solution at 25o 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?

3. The half-life of the active agent is measured at 4o C and found to be 11.5 hours. What is the activation energy for the degradation reaction?

4. A stock solution of the vaccine is prepared at 8:30 am and stored in a refrigerator at 4o 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?

Expert Solution

(1) k1 = 0.693 / t1/2 = 0.693 / 8.4 = 0.0825 h-1 at 25 degree C

(3) k2 = 0.693 / 11.5 = 0.0603 h -1 at 4 degree C

Arrhenius equation,

Log k₂/k₁ = (Ea / 2.303 R)[(1/T₁) - (1/T₂)]

Log(0.0603/0.0825) = (Ea / 2.303x8.314) [(1/298) - (1/277)]

- 0.136 = (Ea) ( - 0.000133)

Ea = 10226 J

Ea = 10.226 kJ

(2) k = 2.303/t log (a/a-x)

0.0825 =2.303 / 3.5 Log ( 25 / a-x)

(a-x) = Remaining amount after completion of 3 and half hours (8.00 to 12.00) = 18.73 micro gram/L

But it is given that for full efective of vaccine it is needed 20 microgram per L at least , So, the vaccine is not effective till the end.

Now, the effective time is,

0.0825 = 2.303 / t Log (25/20)

t = 2.7 h

(4) At 4 degree C

0.0603 = 2.303 / 4 Log (25/a-x)

(a-x) = remaining amount after 4 hours i.e from 8.30 to 12.30 = 19.62 aprroximately = 20 microgram / L

Hence if the vaccine is used at 4 degree C it effectively works up to a maximum of 4 hours.

queen_honey_blossom answered 3 hours ago

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