The active agent in a bottle of sunscreen degrades with first-order kinetics. Th
ID: 998448 • Letter: T
Question
The active agent in a bottle of sunscreen degrades with first-order kinetics. The half-life of the active agent at 25 degrees C is 5.53 x 103 hours. The concentration of the agent should be at least 7.0 g/L to provide sun protection.
(a) Calculate the rate constant, k1, at 25 degrees C. Be sure to include the units.
(b) Suppose you purchase a bottle of sunscreen at the beginning of the summer. The concentration in this bottle is 20 g/L. All summer, and the following year, you store the bottle in your air-conditioned bathroom that has an average temperature of about 25 degrees C. What is the concentration of the active agent next summer, 365 days later? Are you safe using the bottle next summer?
(c) The activation energy for the degradation reaction is 10.3 kJ/mol. Instead of storing the bottle in your bathroom, you leave it in your car, which gets very warm in the Long Island sun and has an average temperature of 38 degrees C throughout the summer. How long will it take for the active agent in the sunscreen to fall below the level required for effectiveness?
Explanation / Answer
ANSWER
(A) Rate constant (k) = 0.069 / t1/2
k = 0.69 / 5.53 x 103 hr = 1.25 X 10-4 hr-1
(B) Concentration after 365 days (C) is given as
365 days = 365 X 24 = 8760 hr
No. of half lives = 8760 / 5530 = 1.58
Amount left after n half lives = (1/2)n X initial concentration
Amount left = (1/2)1.58 X 20g/L = 6.68g/L
Man is not safe if he uses it next summer because the minimum amount for protection is 7g/L.
(C) Use Arrehenius equation to solve the problem
ln k2/k1 = Ea/R [1/T1 - 1/T2]
T = temperature in Kelvin scale
ln k2/k1 = 10300 / 8.314 [1/298K - 1/311K]
ln k2/k1 = 0.124
log k1 /k2 = 0.124 / 2.303 = 0.054
k2 / k1 = antilog (0.054) = 1.32
k2 = k1X 1.32 = 1.25 X 10-4 hr-1 X1.32 = 1.65 X 10-4 hr-1
lnCo/C = kt
ln 20g/L / 7g/L = 1.65 X 10-4 hr-1 X t
1.05 = 1.65 X 10-5 hr-1 X t
t = 1.05 / 1.65 X 10-4 hr-1= 0.636363 X 104 hr = 6363.6hr
After 6363.6hr the active ingredient will be below protection concentration.
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