Question

To prepare a 17.9 kg dog (assume volume of 17.9 L) for surgery, 0.0215 moles of an anesthetic was administered intravenously. The reaction in which the anesthetic is metabolized displayed a first half life of 64.79 minutes and a second half life of 32.39 minutes. After 24.62 minutes of surgery the drug began to lose its effect, but more surgery was required. How many moles of the anesthetic must be re-administered to restore the original level of the anesthetic?

Answer 1

The reaction is actually zeroth order because the half-life is being halved sequentially:

t½ = [Ao] / 2k

Every time you reach a half-life, the amount of [Ao] will decrease by 2 and based from the equation above, so will the next half-life.

To determine the rate constant of this process we first need to find out the initial concentration of anesthetic:

[Ao] = (0.0215 moles)/(17.9 L) = 1.20x10^-3 M

now use the equation above to determine the value of k:

k = [Ao]/2t½ = (1.20x10^-3 M)/2(64.79 min) = 9.26 x10^-6 M/min

Finally, determine the amount left over after 24.62 min:

[Af] - [Ao] = -kt

[Af] = [Ao] - kt = (1.20x10^-3 M ) - (9.26 x10^-6 M/min )(24.62 min) = 9.72x10^-4 M

No of moles =9.72x10^-4 M * 17.9 L =  0.0174 moles

Thus, moles of the anesthetic must be re-administered to restore the original level of the anesthetic = 0.0215 -   0.0174 moles = 0.0041 moles