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Elimination rate zero order equations

Non-linear pharmacokinetics are much less common than linear kinetics. They occur when drug concentrations are sufficiently high to saturate the ability of the liver enzymes to metabolise the drug. This occurs with ethanol, therapeutic concentrations of phenytoin and salicylates, or when high doses of barbiturates are used for cerebral protection. The kinetics of conventional doses of thiopentone are linear. With non-linear pharmacokinetics, the amount of drug eliminated per unit time is constant rather than a constant fraction of the amount in the body, as is the case for the linear situation. Non-linear kinetics are also referred to as zero order or saturation kinetics. The rate of drug decline is governed by the Michaelis-Menton equation ... [Pg.37]

In contrast to noncompartmental analysis, in compartmental analysis a decision on the number of compartments must be made. For mAbs, the standard compartment model is illustrated in Fig. 3.11. It comprises two compartments, the central and peripheral compartment, with volumes VI and V2, respectively. Both compartments exchange antibody molecules with specific first-order rate constants. The input into (if IV infusion) and elimination from the central compartment are zero-order and first-order processes, respectively. Hence, this disposition model characterizes linear pharmacokinetics. For each compartment a differential equation describing the change in antibody amount per time can be established. For... [Pg.80]

The rate of elimination is an important characteristic of a drug. Too rapid an elimination necessitates frequent repeated administration of the drug if its concentration is to reach its therapeutic window. Conversely, too slow an elimination could result in the accumulation of the drug in the patient, which might give an increased risk of toxic effects. Most drug eliminations follow first order kinetics (equations (8.1) and (8.2)), no matter how the drug is administered, but there are some notable exceptions, such as ethanol which exhibits zero order kinetics where ... [Pg.165]

Since the rate of infusion is normally maintained at a constant value, infusion will usually follow zero order kinetics (Equation (8.4)). Therefore, assuming that elimination processes exhibit first order kinetics, it follows that ... [Pg.170]

In the reactive case, r is not equal to zero. Then, Eq. (3) represents a nonhmoge-neous system of first-order quasilinear partial differential equations and the theory is becoming more involved. However, the chemical reactions are often rather fast, so that chemical equilibrium in addition to phase equilibrium can be assumed. The chemical equilibrium conditions represent Nr algebraic constraints which reduce the dynamic degrees of freedom of the system in Eq. (3) to N - Nr. In the limit of reaction equilibrium the kinetic rate expressions for the reaction rates become indeterminate and must be eliminated from the balance equations (Eq. (3)). Since the model Eqs. (3) are linear in the reaction rates, this is always possible. Following the ideas in Ref. [41], this is achieved by choosing the first Nr equations of Eq. (3) as reference. The reference equations are solved for the unknown reaction rates and afterwards substituted into the remaining N - Nr equations. [Pg.156]

If the formation rate (P) of the mediator (M) is regarded as a zero-order process and the elimination rate of the mediator is regarded as first order the following equation describes the mass balance of the mediator ... [Pg.308]

A separate mass balance equation is written in the form of Section 10.6.2 for each compartment included in the model. Variables A and A2 represent the amount of drug in compartment 1 and compartment 2, respectively. Distribution and elimination rates can be written in terms of micro rate constants and first-order rate equations exactly as described for the two-compartment bolus IV model. As mentioned in Section 10.8, an IV infusion has an unusual discontinuity in that drug is delivered into the systemic circulation (directly into compartment 1) at a constant rate (over a fixed period of time (0 < t < T), after which the rate of drug delivery drops to zero t > T). A schematic representation of the standard two-compartment IV infusion model is provided in Figure 10.67. In order to accommodate this discontinuity in drug delivery, two mass balance equations must be written for compartment 1, one for the time period during infusion and the other for the time period after infusion stops. The mass balance equation for compartment 1 during the infusion period (0 < t< T) becomes... [Pg.247]

A separate mass balance equation is written in the form of Section 10.6.2 for each compartment in the model. Thus a total of n mass balance equations must be written and solved for an n compartment model. The details of these equations and their solution are not provided in this chapter. However, it will be noted that absorption, distribution, and elimination rates are written in the same form as in the previous one- and two-compartment models. The absorption rate for instantaneous, zero-order, or first-order absorption is identical to the previous forms for one- and two-com-partment models. Distribution and elimination rates are written as first-order linear rate equations using micro rate constants. So the distribution rate from compartment 1 to compartment n is given by kj Aj, the distribution rate from compartment n back to compartment 1 equals k i A , and the elimination rate from any compartment is written k o A schematic diagram for the generalized n compartment model is illustrated in Figure 10.90. [Pg.260]

For example, substitution of X (mass of drug in the body at time t) for Y in Eq. 1.8 yields the zero-order elimination rate equation ... [Pg.13]

Figure 3.22 A typical rectilinear (a) or semilogarithmic (b) plot of rate of excretion against average time (t) following the administration of o drug os on intravenous bolus (Equations 3.25). /C , first-order renal excretion rate constant Xq, drug at time zero K, elimination rate constant. Figure 3.22 A typical rectilinear (a) or semilogarithmic (b) plot of rate of excretion against average time (t) following the administration of o drug os on intravenous bolus (Equations 3.25). /C , first-order renal excretion rate constant Xq, drug at time zero K, elimination rate constant.
Reductive eliminations of methane, toluene, and cyclohexane (Equations 8.11-8.13) occur from (PPh3)jPt(Me)(H), -> (PMe3),Ru(H)(CHjPh),> and Cp Lr(PMe)3(C,H )H without prior dissociation of ligand. The absence of ligand dissociation prior to reductive elimination was shown in these cases by a zero-order dependence of the reaction rates on the concentration of added phosphine. The rates for these elimination reactions vary dramatically. The platinum complex reacts at -25 °C, the ruthenium complex at 85 °C, and the iridium complex at 135 °C. Although one could rationalize these relative rates in many ways, the platinum compound likely reacts faster than the iridium complex because it is less electron rich, while the ruthenium complex likely reacts faster than the iridium complex because it contains a second-row metal center. [Pg.326]

Equation 3-22 for the polymerization rate is not directly usable because it contains a term for the concentration of radicals. Radical concentrations are difficult to measure quantitatively, since they are very low ( 10-8 M), and it is therefore desirable to eliminate [M- from Eq. 3-22. In order to do this, the steady-state assumption is made that the concentration of radicals increases initially, but almost instantaneously reaches a constant, steady-state value. The rate of change of the concentration of radicals quickly becomes and remains zero during the course of the polymerization. This is equivalent to stating that the rates of initiation Rj and termination R, of radicals are equal or... [Pg.207]

The behavior of a chemical is investigated in a well-mixed reactor (volume V, flow rate Q) by measuring the outflow concentration Cout at steady-state for different input concentrations Cin. The results are given in the table below, (a) Determine the order of the elimination process and formulate the differential equation which describes the chemical in the reactor, (b) How long does it take for the outflow concentration to drop from 40 mmol-L-1 to 2 mmol-L, if at time tu Cin drops to zero instantaneously ... [Pg.488]

Answer The fact that the rate of elimination is independent of concentration indicates that the dmg is eliminated by a first-order process characterized by the relation Ct = Co e kt, where Ct and Co refer to concentrations at time t and zero, respectively, and k is the rate constant for first-order elimination. The logarithmic metameter of this equation is... [Pg.354]


See other pages where Elimination rate zero order equations is mentioned: [Pg.271]    [Pg.280]    [Pg.621]    [Pg.99]    [Pg.82]    [Pg.90]    [Pg.1143]    [Pg.225]    [Pg.140]    [Pg.115]    [Pg.119]    [Pg.18]    [Pg.155]    [Pg.180]    [Pg.137]   
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