A particular drug can be sold until 20% of the original drug has undergone first

Question

A particular drug can be sold until 20% of the original drug has undergone first order kinetic reaction. Knowing that k = 1.25 times 10^-2/day, how long will it take before the drug can no longer be sold? 15 days 18 days 25 days 35 days The activation energy of a reaction is 250 kJ/mol. If the rate constant at T_1 = 300 K is k_1 and the rate constant at T_2 = 320 K is k_2, then the reaction is times faster at 320 K than at 300 K. 3 times 10^-29 525 15.0 0.067 A catalyst can act in a chemical reaction to: increase the equilibrium constant

Explanation / Answer

ln ([A]t / [Ao]) = -kt . . .where ([A]t / [Ao]) is the fraction of reactant remaining at time t.

given that ([A]t / [Ao]) = 80%= 0.80

and k = 1.25*10^-2 / days

now calculate the time t as follows:

ln (0.8) = (-1.25*10^-2 / days) t

-0.223= -1.25*10^-2 / days

t = 17.8 day = 18 day

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