In medical treatment drugs are ingested, absorbed and eliminated. Let species A
ID: 896136 • Letter: I
Question
In medical treatment drugs are ingested, absorbed and eliminated. Let species A represent the concentration of the drug at its site of introduction (oral ingestion), species B represent the concentration of the drug in the blood and species C represent the concentration of drug eliminated. This process can be described by the following series reaction: In this reaction scheme the delivery of the drug is zeroth order with respect to species A and elimination is first order with respect to species B. (i.e., zeroth order followed by first order kinetics). We are interested in obtaining the concentrations of each species as a function of time; hence, this can be modeled as a batch system. Obtain the net rates of reaction for each species involved in the reaction scheme. Using the results in part (a) formulate the GMB equations for each species. Obtain the integrated expression for C_A as a function of time if the initial concentration of A is equal to C_A0. Sketch the concentration profile of C_A as a function of time. Obtain an expression for the time, t_1, where C_A = 0 and make sure to include this point in your sketch. Obtain the integrated expression for C_B as a function of time up to t = t_1 (i.e., 0 t_1 = t_max species A is no longer present, hence the series reaction can be modeled simply by the single reaction Obtain an integrated expression for C_B as a function of time. To your existing sketch add how C_B changes with time for t > t_max.Explanation / Answer
a) the rate of the first equation is K1 and that of the second reaction is K2.
b)for the first reaction,
since it is a zeroth order reactino,
rate = -d[A]/dt=K1*[A]^0
or -d[A]/dt=K1
for the second reaction,
since it is a first order reaction,
rate = -d[B]/dt=K2*[B]
or -d[B]/dt=K2[B]
c) integrating the first equation, we get
[Ao]-[A]=K1*t1 (integration time from 0 to t1 and the concentation from Ao to A
d)Integrating the second equation,
ln([Bo]/[B])=K2*t2
e)time t when Ca has been depleted = [Ao]/K1
so [B]=[Bo]*e^(-[Ao]*K2/K1)
f)Here Cbo = [Bo]*e^(-[Ao]*K2/K1)
now integrating the second equation,
[B]=[Bo]*e^(-[Ao]*K2/K1) * e^(-K2*(t-([Ao]/K1)))
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